answer:<think>Alright, so I've got these two sub-problems to solve, both related to insurance claims modeling. Let me take them one at a time.Starting with Sub-problem 1: The number of claims ( N(t) ) at time ( t ) follows a Poisson process with rate ( lambda ). The average number of claims per year is 1000. I need to find the probability that there will be more than 1100 claims in a year.Okay, so Poisson processes model the number of events happening in a fixed interval of time. The key property is that the number of events in non-overlapping intervals are independent, and the number of events in any interval follows a Poisson distribution.Given that the average number of claims per year is 1000, that means ( lambda ) is 1000 claims per year. So, ( N(t) ) is a Poisson process with rate ( lambda = 1000 ).We need ( P(N(1) > 1100) ), where 1 represents one year. Since ( N(1) ) follows a Poisson distribution with parameter ( lambda = 1000 ), we can write:( P(N(1) > 1100) = 1 - P(N(1) leq 1100) )Calculating this directly might be tricky because the Poisson distribution with such a large ( lambda ) can be cumbersome. I remember that for large ( lambda ), the Poisson distribution can be approximated by a normal distribution with mean ( mu = lambda ) and variance ( sigma^2 = lambda ). So, ( mu = 1000 ) and ( sigma = sqrt{1000} approx 31.6228 ).Using the normal approximation, we can standardize the variable:( Z = frac{X - mu}{sigma} = frac{1100 - 1000}{31.6228} approx frac{100}{31.6228} approx 3.1623 )So, ( P(N(1) > 1100) approx P(Z > 3.1623) ). Looking at the standard normal distribution table, a Z-score of 3.16 corresponds to a cumulative probability of approximately 0.9992. Therefore, the probability that Z is greater than 3.16 is ( 1 - 0.9992 = 0.0008 ).But wait, I should consider the continuity correction since we're approximating a discrete distribution with a continuous one. So, instead of ( P(N(1) > 1100) ), we should use ( P(N(1) geq 1100.5) ). Let me recalculate that.( Z = frac{1100.5 - 1000}{31.6228} approx frac{100.5}{31.6228} approx 3.177 )Looking up 3.177 in the Z-table, the cumulative probability is about 0.9992 (since 3.17 is 0.9992 and 3.18 is 0.9993). So, the probability that Z is greater than 3.177 is approximately ( 1 - 0.9992 = 0.0008 ). Hmm, not much difference. So, the probability is roughly 0.08%.Alternatively, maybe using the exact Poisson formula? But with ( lambda = 1000 ), calculating ( P(N(1) > 1100) ) exactly would involve summing from 1101 to infinity, which is computationally intensive. So, the normal approximation is a practical approach here.Alternatively, another approximation for Poisson is the normal distribution, so I think the answer is approximately 0.08%.Moving on to Sub-problem 2: The claim amounts ( X_i ) are independent and identically distributed with an exponential distribution with mean ( mu ). The total claim amount ( S(T) = sum_{i=1}^{N(T)} X_i ) over a year must not exceed 1,200,000 for the company to remain profitable. We need to determine the probability that the company will remain profitable, given that ( N(t) ) follows the Poisson process with rate ( lambda ) from Sub-problem 1.So, ( N(T) ) is Poisson with ( lambda = 1000 ), and each ( X_i ) is exponential with mean ( mu ). The total claim amount ( S(T) ) is the sum of ( N(T) ) exponential random variables. I remember that the sum of independent exponential variables with the same rate parameter follows a gamma distribution. Specifically, if each ( X_i ) is exponential with rate ( beta ), then ( S(T) ) given ( N(T) = n ) is gamma distributed with shape ( n ) and rate ( beta ).But in our case, the mean of each ( X_i ) is ( mu ), so the rate ( beta = 1/mu ). Therefore, given ( N(T) = n ), ( S(T) ) has a gamma distribution with parameters ( n ) and ( beta = 1/mu ).But since ( N(T) ) itself is a random variable, ( S(T) ) is a compound distribution, specifically a Poisson-gamma distribution, which is also known as a Tweedie distribution. However, calculating probabilities for such a distribution can be complex.Alternatively, perhaps we can use the law of total probability. The probability that ( S(T) leq 1,200,000 ) is equal to the expected value of the probability that ( S(T) leq 1,200,000 ) given ( N(T) ).Mathematically, that is:( P(S(T) leq 1,200,000) = E[P(S(T) leq 1,200,000 | N(T))] )Given that ( N(T) = n ), ( S(T) ) is the sum of ( n ) exponential variables, which is gamma distributed as I mentioned. The CDF of a gamma distribution is not straightforward, but perhaps we can use the normal approximation again, especially since ( n ) is large (on average 1000).Wait, but ( N(T) ) is Poisson with mean 1000, so it's already a large number. So, for each ( n ), ( S(T) ) given ( N(T) = n ) is approximately normal with mean ( nmu ) and variance ( nmu^2 ) (since variance of exponential is ( mu^2 )).Therefore, for each ( n ), ( S(T) ) is approximately ( N(nmu, nmu^2) ). So, the probability ( P(S(T) leq 1,200,000 | N(T) = n) ) can be approximated by:( Pleft( Z leq frac{1,200,000 - nmu}{sqrt{n}mu} right) )Where ( Z ) is the standard normal variable.Therefore, the overall probability is:( Eleft[ Phileft( frac{1,200,000 - N(T)mu}{sqrt{N(T)}mu} right) right] )Where ( Phi ) is the standard normal CDF, and the expectation is taken over ( N(T) ) which is Poisson with ( lambda = 1000 ).This seems complicated because we have to compute the expectation of a function of a Poisson random variable. Maybe we can approximate it by considering the mean and variance of ( N(T) ).Alternatively, perhaps we can model ( S(T) ) as a compound Poisson process and use the Central Limit Theorem for compound processes.Wait, another approach: Since both ( N(T) ) and ( X_i ) are involved, and ( N(T) ) is Poisson, perhaps we can use the fact that the total claim amount ( S(T) ) has a compound Poisson distribution.The mean of ( S(T) ) is ( E[S(T)] = E[N(T)] E[X_i] = 1000 mu ).The variance of ( S(T) ) is ( Var(S(T)) = E[N(T)] Var(X_i) + Var(N(T)) (E[X_i])^2 ). Since ( X_i ) are independent of ( N(T) ), this is applicable.Given that ( Var(X_i) = mu^2 ) (since exponential distribution variance is ( mu^2 )), and ( Var(N(T)) = lambda = 1000 ).Therefore,( Var(S(T)) = 1000 mu^2 + 1000 mu^2 = 2000 mu^2 )So, ( S(T) ) has mean ( 1000 mu ) and variance ( 2000 mu^2 ). Therefore, the standard deviation is ( sqrt{2000} mu approx 44.721 mu ).If we assume that ( S(T) ) is approximately normally distributed (which might be a stretch, but given the large number of claims, it could be reasonable), then we can write:( P(S(T) leq 1,200,000) approx Phileft( frac{1,200,000 - 1000 mu}{sqrt{2000} mu} right) )But wait, we don't know ( mu ). The problem states that the total claim amount must not exceed 1,200,000. It doesn't specify ( mu ), so perhaps we need to express the probability in terms of ( mu ), or maybe ( mu ) is given implicitly?Wait, let me check the problem statement again."Assume the claim amounts ( X_i ) are independent and identically distributed with an exponential distribution with mean ( mu ). If the total claim amount ( S(T) = sum_{i=1}^{N(T)} X_i ) over a year must not exceed 1,200,000 for the company to remain profitable, determine the probability that the company will remain profitable given that ( N(t) ) follows the Poisson process with the rate ( lambda ) from Sub-problem 1."Hmm, it doesn't give ( mu ). So, perhaps we need to express the probability in terms of ( mu ), or maybe ( mu ) is given in Sub-problem 1? Wait, in Sub-problem 1, the average number of claims is 1000, but nothing about the claim sizes. So, ( mu ) is just the mean of each claim, which is given as an exponential distribution with mean ( mu ). So, unless ( mu ) is provided, we can't compute a numerical probability.Wait, maybe I missed something. Let me check again.No, the problem only gives the average number of claims per year as 1000, and the total claim amount must not exceed 1,200,000. So, perhaps we can express the probability in terms of ( mu ), but the problem says "determine the probability", implying it should be a numerical answer. So, maybe ( mu ) is given in the problem? Wait, no, it's just stated as an exponential distribution with mean ( mu ). So, perhaps we need to express the probability as a function of ( mu ), but the problem doesn't specify. Hmm.Alternatively, maybe the mean ( mu ) is related to the total claim amount? Wait, if the total claim amount is 1,200,000, and the average number of claims is 1000, then the average claim size would be ( mu = 1,200,000 / 1000 = 1200 ). But wait, that would be the case if the total claim amount is exactly 1,200,000, but in reality, it's a random variable. So, perhaps the expected total claim amount is ( 1000 mu ), and we need the probability that ( S(T) leq 1,200,000 ). So, if ( 1000 mu ) is the mean, then ( mu = 1200 ) would make the mean total claim amount equal to 1,200,000. But that would mean the probability that ( S(T) leq 1,200,000 ) is 0.5, which doesn't make sense because the distribution is skewed.Wait, maybe I'm overcomplicating. Let me think again.We have ( S(T) ) is the sum of ( N(T) ) exponential variables. Each ( X_i ) has mean ( mu ), so the expected total claim amount is ( E[S(T)] = E[N(T)] E[X_i] = 1000 mu ). The total claim amount must not exceed 1,200,000. So, if ( 1000 mu ) is the expected total, then to find the probability that ( S(T) leq 1,200,000 ), we can standardize it.Assuming ( S(T) ) is approximately normal (due to the Central Limit Theorem, since ( N(T) ) is large), then:( P(S(T) leq 1,200,000) approx Phileft( frac{1,200,000 - 1000 mu}{sqrt{2000} mu} right) )But without knowing ( mu ), we can't compute this. So, perhaps the problem expects us to express the probability in terms of ( mu ), or maybe ( mu ) is given implicitly?Wait, maybe I misread the problem. Let me check again."Assume the claim amounts ( X_i ) are independent and identically distributed with an exponential distribution with mean ( mu ). If the total claim amount ( S(T) = sum_{i=1}^{N(T)} X_i ) over a year must not exceed 1,200,000 for the company to remain profitable, determine the probability that the company will remain profitable given that ( N(t) ) follows the Poisson process with the rate ( lambda ) from Sub-problem 1."No, it doesn't give ( mu ). So, perhaps the problem expects us to express the probability in terms of ( mu ), but the question says "determine the probability", which suggests a numerical answer. Maybe I need to find ( mu ) such that the probability is a certain value? But the problem doesn't specify that.Alternatively, perhaps the total claim amount is fixed at 1,200,000, so the mean claim size ( mu ) is 1,200,000 / 1000 = 1200. So, if ( mu = 1200 ), then the expected total claim amount is 1,200,000, and we can compute the probability that ( S(T) leq 1,200,000 ). But in that case, since the distribution is skewed, the probability would be less than 0.5.Wait, but if ( mu = 1200 ), then the variance of ( S(T) ) is ( 2000 times (1200)^2 = 2000 times 1,440,000 = 2,880,000,000 ), so the standard deviation is ( sqrt{2,880,000,000} approx 53,665.63 ).Then, the Z-score would be:( Z = frac{1,200,000 - 1000 times 1200}{53,665.63} = frac{1,200,000 - 1,200,000}{53,665.63} = 0 )So, ( P(S(T) leq 1,200,000) approx Phi(0) = 0.5 ). But that seems counterintuitive because the exponential distribution is skewed, so the median is less than the mean. Wait, but when we sum many exponentials, the Central Limit Theorem makes the sum approximately normal, so the median would be close to the mean. So, maybe the probability is approximately 0.5.But that seems too simplistic. Alternatively, maybe the problem expects us to use the exact distribution.Wait, another approach: The total claim amount ( S(T) ) given ( N(T) = n ) is gamma distributed with shape ( n ) and rate ( 1/mu ). The CDF of a gamma distribution is given by the regularized gamma function. However, calculating this for each ( n ) and then taking the expectation over ( N(T) ) is quite involved.Alternatively, perhaps we can use the moment generating function or characteristic function, but that might be too complex.Wait, maybe we can use the fact that the sum of exponentials is gamma, and then use the normal approximation for the gamma distribution. For large ( n ), the gamma distribution can be approximated by a normal distribution with mean ( nmu ) and variance ( nmu^2 ). So, for each ( n ), ( S(T) ) is approximately ( N(nmu, nmu^2) ). Then, the probability ( P(S(T) leq 1,200,000 | N(T) = n) ) is approximately ( Phileft( frac{1,200,000 - nmu}{sqrt{n}mu} right) ).Then, the overall probability is:( Eleft[ Phileft( frac{1,200,000 - N(T)mu}{sqrt{N(T)}mu} right) right] )But ( N(T) ) is Poisson with ( lambda = 1000 ). So, we need to compute:( sum_{n=0}^{infty} frac{e^{-1000} 1000^n}{n!} Phileft( frac{1,200,000 - nmu}{sqrt{n}mu} right) )This is a complex sum to compute, especially since ( n ) can be from 0 to infinity. However, since ( N(T) ) is Poisson with mean 1000, the probability mass is concentrated around ( n = 1000 ). So, maybe we can approximate this sum by considering ( n ) around 1000.Let me define ( n = 1000 + k ), where ( k ) is small relative to 1000. Then, we can approximate the sum by expanding around ( n = 1000 ).But this might get too involved. Alternatively, perhaps we can use the delta method or a Taylor expansion to approximate the expectation.Let me denote ( Z_n = frac{1,200,000 - nmu}{sqrt{n}mu} ). Then, ( Phi(Z_n) ) is the probability that ( S(T) leq 1,200,000 ) given ( N(T) = n ).We can write ( Z_n = frac{1,200,000}{sqrt{n}mu} - frac{nmu}{sqrt{n}mu} = frac{1,200,000}{sqrt{n}mu} - sqrt{n} ).Let me define ( a = frac{1,200,000}{mu} ), so ( Z_n = frac{a}{sqrt{n}} - sqrt{n} ).Then, ( Phi(Z_n) = Phileft( frac{a}{sqrt{n}} - sqrt{n} right) ).Now, we need to compute ( E[Phi(Z_n)] ) where ( n ) is Poisson(1000).This seems complicated, but perhaps we can use a Taylor expansion of ( Phi(Z_n) ) around ( n = 1000 ).Let me denote ( n = 1000 + delta ), where ( delta ) is small. Then, we can expand ( Z_n ) in terms of ( delta ).But this might not be the best approach. Alternatively, perhaps we can use the fact that for large ( n ), ( Z_n ) can be approximated by a function of ( n ), and then use the properties of the Poisson distribution.Wait, another idea: Since ( N(T) ) is Poisson with mean 1000, and ( S(T) ) given ( N(T) = n ) is approximately normal with mean ( nmu ) and variance ( nmu^2 ), then the overall distribution of ( S(T) ) is a mixture of normals. The expectation of the CDF of a normal variable can be approximated using the Edgeworth expansion or other methods, but this is getting too advanced.Alternatively, perhaps we can use the fact that ( S(T) ) is approximately normal with mean ( 1000mu ) and variance ( 2000mu^2 ), as I calculated earlier. Then, the probability ( P(S(T) leq 1,200,000) ) is approximately ( Phileft( frac{1,200,000 - 1000mu}{sqrt{2000}mu} right) ).But without knowing ( mu ), we can't compute this. So, perhaps the problem expects us to express the probability in terms of ( mu ), but the question says "determine the probability", which suggests a numerical answer. Maybe I need to find ( mu ) such that the probability is a certain value? But the problem doesn't specify that.Wait, perhaps I made a mistake earlier. Let me think again.The problem says: "the total claim amount ( S(T) = sum_{i=1}^{N(T)} X_i ) over a year must not exceed 1,200,000 for the company to remain profitable". So, the company wants ( P(S(T) leq 1,200,000) ). We need to find this probability.But without knowing ( mu ), we can't compute it numerically. So, perhaps the problem expects us to express the probability in terms of ( mu ), but the question says "determine the probability", which is confusing. Alternatively, maybe ( mu ) is given in the problem, but I missed it.Wait, in Sub-problem 1, the average number of claims is 1000, but nothing about the claim sizes. So, ( mu ) is just the mean of each claim, which is given as an exponential distribution with mean ( mu ). So, unless ( mu ) is provided, we can't compute a numerical probability.Wait, maybe the problem expects us to assume that the mean claim size ( mu ) is such that the expected total claim amount is 1,200,000. So, ( E[S(T)] = 1000mu = 1,200,000 ), which gives ( mu = 1200 ). Then, we can compute the probability that ( S(T) leq 1,200,000 ).If ( mu = 1200 ), then ( S(T) ) has mean ( 1000 times 1200 = 1,200,000 ) and variance ( 2000 times (1200)^2 = 2,880,000,000 ), so standard deviation ( sqrt{2,880,000,000} approx 53,665.63 ).Then, the Z-score is:( Z = frac{1,200,000 - 1,200,000}{53,665.63} = 0 )So, ( P(S(T) leq 1,200,000) approx Phi(0) = 0.5 ).But wait, that seems too simplistic because the distribution is skewed. However, for a large number of claims, the Central Limit Theorem suggests that the distribution of ( S(T) ) is approximately normal, so the median would be close to the mean, making the probability approximately 0.5.But let me verify this with a more precise approach. If ( S(T) ) is approximately normal with mean 1,200,000 and standard deviation 53,665.63, then ( P(S(T) leq 1,200,000) = 0.5 ).Therefore, the probability that the company will remain profitable is approximately 50%.But wait, that seems too high because the exponential distribution is skewed to the right, meaning that the total claim amount could exceed the mean more often than not. However, when we sum many exponentials, the Central Limit Theorem makes the distribution symmetric around the mean, so the median is close to the mean, making the probability approximately 0.5.Alternatively, maybe the exact probability is slightly less than 0.5 because the gamma distribution is skewed, but for large ( n ), the normal approximation is good.So, perhaps the answer is approximately 50%.But let me think again. If ( mu = 1200 ), then each claim is exponential with mean 1200, so the distribution of each ( X_i ) is skewed. However, when we sum 1000 of them, the Central Limit Theorem makes the sum approximately normal, so the total claim amount ( S(T) ) is approximately normal with mean 1,200,000 and standard deviation 53,665.63. Therefore, the probability that ( S(T) leq 1,200,000 ) is indeed approximately 0.5.But wait, the problem says "the total claim amount must not exceed 1,200,000". So, if the expected total is 1,200,000, the probability that it doesn't exceed that is about 50%. That seems correct.Alternatively, if the company wants a higher probability, say 95%, they would need to set the total claim limit higher than the mean. But in this case, the limit is set at the mean, so the probability is about 50%.Therefore, the probability that the company will remain profitable is approximately 50%.But wait, let me check if this makes sense. If the total claim amount is exactly the mean, then the probability that it's less than or equal to the mean is 0.5 for a symmetric distribution. Since the sum of exponentials is approximately normal for large ( n ), this makes sense.So, to summarize:Sub-problem 1: Using normal approximation, the probability is approximately 0.08%.Sub-problem 2: Assuming ( mu = 1200 ), the probability is approximately 50%.But wait, in Sub-problem 2, I assumed ( mu = 1200 ) because the expected total claim amount is 1,200,000. But the problem doesn't specify that the expected total claim amount is 1,200,000. It just says the total must not exceed 1,200,000. So, perhaps ( mu ) is not 1200, but we need to find the probability in terms of ( mu ).Wait, the problem says: "the total claim amount ( S(T) = sum_{i=1}^{N(T)} X_i ) over a year must not exceed 1,200,000 for the company to remain profitable". So, the company wants ( P(S(T) leq 1,200,000) ). We need to find this probability, given that ( N(t) ) is Poisson with ( lambda = 1000 ) and each ( X_i ) is exponential with mean ( mu ).So, we can't assume ( mu = 1200 ). Therefore, the probability is a function of ( mu ). But the problem says "determine the probability", which suggests a numerical answer. So, perhaps I missed something.Wait, maybe the problem expects us to use the fact that the total claim amount is 1,200,000, and given that ( N(T) ) is Poisson with mean 1000, we can find ( mu ) such that ( E[S(T)] = 1000 mu = 1,200,000 ), which gives ( mu = 1200 ). Then, as before, the probability is approximately 0.5.But the problem doesn't state that the expected total claim amount is 1,200,000. It just says the total must not exceed that amount. So, perhaps the company is setting a limit of 1,200,000, and we need to find the probability that the total claims are below that limit, given the distribution of claims.Therefore, without knowing ( mu ), we can't compute a numerical probability. So, perhaps the problem expects us to express the probability in terms of ( mu ), but the question says "determine the probability", which is confusing.Alternatively, maybe the problem expects us to use the fact that the total claim amount is 1,200,000, and given that ( N(T) ) is Poisson with mean 1000, we can find ( mu ) such that the probability is a certain value. But the problem doesn't specify that.Wait, perhaps the problem is expecting us to use the fact that the total claim amount is 1,200,000, and given that the number of claims is Poisson with mean 1000, we can find the probability that the total claim amount is less than or equal to 1,200,000. But without knowing ( mu ), we can't compute this.Wait, maybe the problem is expecting us to assume that the mean claim size ( mu ) is such that the expected total claim amount is 1,200,000, which would make ( mu = 1200 ). Then, as before, the probability is approximately 0.5.But the problem doesn't state that the expected total claim amount is 1,200,000. It just says the total must not exceed that amount. So, perhaps the problem is expecting us to express the probability in terms of ( mu ), but the question says "determine the probability", which is confusing.Alternatively, maybe I'm overcomplicating. Let me try to proceed with the assumption that ( mu = 1200 ), as that would make the expected total claim amount equal to the limit, and then the probability is approximately 0.5.Therefore, the answers are:Sub-problem 1: Approximately 0.08%Sub-problem 2: Approximately 50%But let me double-check Sub-problem 1.Given ( N(1) ) ~ Poisson(1000), we want ( P(N(1) > 1100) ). Using normal approximation with continuity correction:( Z = frac{1100.5 - 1000}{sqrt{1000}} approx frac{100.5}{31.6228} approx 3.177 )Looking up 3.177 in the Z-table, the cumulative probability is approximately 0.9992, so ( P(Z > 3.177) = 1 - 0.9992 = 0.0008 ), which is 0.08%.Yes, that seems correct.For Sub-problem 2, assuming ( mu = 1200 ), the probability is approximately 0.5.But wait, another thought: If ( mu ) is not 1200, then the probability would be different. For example, if ( mu ) is less than 1200, the expected total claim amount would be less than 1,200,000, making the probability higher than 0.5. Conversely, if ( mu ) is greater than 1200, the probability would be less than 0.5.But since the problem doesn't specify ( mu ), I think the only way to answer is to express the probability in terms of ( mu ), but the question says "determine the probability", which suggests a numerical answer. Therefore, perhaps the problem expects us to assume ( mu = 1200 ), making the probability approximately 0.5.Alternatively, maybe the problem expects us to use the exact distribution, but that's too complex without knowing ( mu ).In conclusion, I think the answers are:Sub-problem 1: Approximately 0.08%Sub-problem 2: Approximately 50%</think>

answer:<think>Alright, so I have this problem about a celebrity chef named Alex who wants to manage their social media posts to balance their public persona with their partner's privacy. There are two sub-problems here, and I need to solve both. Let me start with the first one.Sub-problem 1:Alex wants to post content such that the number of posts involving their partner is at most 1/10th of the total posts in a month. They plan to include their partner in exactly 5 posts that month. I need to find the minimum total number of posts ( n ) they need to make to satisfy this condition.Okay, so let's break this down. The number of posts with their partner is 5. This number should be at most 1/10th of the total posts ( n ). So, mathematically, this can be written as:[ 5 leq frac{1}{10}n ]I need to solve for ( n ). Let me rewrite that inequality:[ 5 leq frac{n}{10} ]To solve for ( n ), I can multiply both sides by 10:[ 5 times 10 leq n ][ 50 leq n ]So, ( n ) must be at least 50. But since the number of posts has to be a whole number, and 50 is already a whole number, the minimum ( n ) is 50. Let me double-check that.If Alex makes 50 posts in total, then 1/10th of that is 5. So, 5 posts involving their partner is exactly 1/10th of the total. That satisfies the condition. If they made fewer than 50 posts, say 49, then 1/10th of 49 is 4.9, which is less than 5. So, 5 posts would exceed the 1/10th limit. Therefore, 50 is indeed the minimum number of posts needed.Sub-problem 2:Now, over the next year, Alex plans to increase their total number of posts by 5% each month, compounded monthly. They start with ( n ) posts from Sub-problem 1 in the first month. I need to derive a general expression for the total number of posts ( P(t) ) in the ( t )-th month and find the number of posts in the 12th month.Alright, so starting with ( n = 50 ) posts in the first month, each subsequent month they increase by 5%. This sounds like a geometric sequence where each term is multiplied by a common ratio each time.In general, for a geometric sequence, the ( t )-th term is given by:[ P(t) = P(1) times r^{t-1} ]Where ( P(1) ) is the first term, ( r ) is the common ratio, and ( t ) is the term number.In this case, the common ratio ( r ) is 1 plus the growth rate. Since it's a 5% increase each month, the growth rate is 0.05, so:[ r = 1 + 0.05 = 1.05 ]Therefore, the expression for ( P(t) ) is:[ P(t) = 50 times (1.05)^{t-1} ]Let me verify this formula. For the first month (( t = 1 )):[ P(1) = 50 times (1.05)^{0} = 50 times 1 = 50 ]That's correct. For the second month (( t = 2 )):[ P(2) = 50 times (1.05)^{1} = 50 times 1.05 = 52.5 ]Hmm, 52.5 posts? Since the number of posts should be a whole number, maybe we need to consider rounding. But the problem doesn't specify whether to round or not, so perhaps we can keep it as a decimal for the general expression.But let me check the problem statement again. It says "derive a general expression for the total number of posts ( P(t) )". It doesn't specify rounding, so I think it's acceptable to leave it as a decimal. So, the expression is correct.Now, the question also asks for the total number of posts in the 12th month. So, we need to compute ( P(12) ).Using the formula:[ P(12) = 50 times (1.05)^{12 - 1} ][ P(12) = 50 times (1.05)^{11} ]I need to calculate ( (1.05)^{11} ). Let me compute this step by step.First, I can compute ( (1.05)^{11} ). Alternatively, I can use logarithms or a calculator, but since I don't have a calculator here, maybe I can approximate it.Alternatively, I remember that ( (1.05)^{12} ) is approximately 1.795856, which is a known value for 5% over 12 periods. But since we need ( (1.05)^{11} ), that would be ( (1.05)^{12} / 1.05 ).So,[ (1.05)^{11} = frac{(1.05)^{12}}{1.05} approx frac{1.795856}{1.05} approx 1.71034 ]Wait, let me verify that. If ( (1.05)^{12} approx 1.795856 ), then dividing by 1.05 gives approximately 1.71034.Alternatively, I can compute ( (1.05)^{11} ) step by step:Compute ( (1.05)^1 = 1.05 )( (1.05)^2 = 1.1025 )( (1.05)^3 = 1.157625 )( (1.05)^4 = 1.21550625 )( (1.05)^5 = 1.2762815625 )( (1.05)^6 = 1.3400956406 )( (1.05)^7 = 1.4071004226 )( (1.05)^8 = 1.4774554437 )( (1.05)^9 = 1.5513282159 )( (1.05)^{10} = 1.6288946267 )( (1.05)^{11} = 1.7103393580 )So, approximately 1.7103393580.Therefore,[ P(12) = 50 times 1.7103393580 approx 50 times 1.7103393580 ]Calculating that:50 multiplied by 1.7103393580.First, 50 times 1 is 50.50 times 0.7103393580 is:0.7103393580 * 50 = 35.5169679So, total is 50 + 35.5169679 = 85.5169679.So, approximately 85.5169679.Since the number of posts can't be a fraction, but the problem doesn't specify rounding, so perhaps we can leave it as a decimal or round it to the nearest whole number.If we round to the nearest whole number, 85.5169679 is approximately 86.But let me check if my calculation for ( (1.05)^{11} ) was accurate.Wait, when I computed step by step, I got approximately 1.7103393580. Let me verify that.Yes, each step was multiplying by 1.05:Starting from 1.05^1 = 1.051.05^2 = 1.05*1.05 = 1.10251.05^3 = 1.1025*1.05 = 1.1576251.05^4 = 1.157625*1.05 = 1.215506251.05^5 = 1.21550625*1.05 = 1.27628156251.05^6 = 1.2762815625*1.05 = 1.34009564061.05^7 = 1.3400956406*1.05 ≈ 1.40710042261.05^8 = 1.4071004226*1.05 ≈ 1.47745544371.05^9 = 1.4774554437*1.05 ≈ 1.55132821591.05^{10} = 1.5513282159*1.05 ≈ 1.62889462671.05^{11} = 1.6288946267*1.05 ≈ 1.7103393580Yes, that seems correct. So, 1.7103393580 is accurate.Therefore, 50 * 1.7103393580 ≈ 85.5169679.So, approximately 85.5169679 posts. Since you can't have a fraction of a post, but the problem doesn't specify rounding, so maybe we can present it as approximately 85.52 or just keep it as 85.5169679.But let me see if the problem expects an exact value or an approximate decimal. Since it's compounded monthly, the exact value would be 50*(1.05)^11, which is an exact expression, but if they want a numerical value, then approximately 85.52.But let me check if 50*(1.05)^11 is exactly equal to 85.5169679 or if I need to compute it more accurately.Wait, let me compute 50*1.7103393580:50 * 1.7103393580= 50 * 1 + 50 * 0.7103393580= 50 + 35.5169679= 85.5169679So, yes, that's accurate.Alternatively, if I use logarithms or a calculator, I can get a more precise value, but since I don't have a calculator here, I think 85.5169679 is precise enough.But let me think again. Maybe I can express it as an exact fraction or something, but 1.05 is 21/20, so 1.05^11 is (21/20)^11, which is a huge fraction, but probably not necessary here.So, I think the answer is approximately 85.52 posts. But since the number of posts should be a whole number, maybe Alex would have to make 86 posts in the 12th month.But the problem doesn't specify whether to round up or down, so perhaps it's acceptable to leave it as a decimal. Alternatively, maybe we can express it as an exact value.Wait, let me check the problem statement again. It says "derive a general expression for the total number of posts ( P(t) ) Alex will make in the ( t )-th month." So, the expression is 50*(1.05)^{t-1}, which is exact. Then, for the 12th month, it's 50*(1.05)^11, which is approximately 85.5169679.But maybe the problem expects an exact value in terms of exponents, but I think they want a numerical value. So, rounding to two decimal places, it's approximately 85.52.Alternatively, if we consider that the number of posts must be an integer, then 85.5169679 would round to 86.But the problem doesn't specify, so I think it's safer to present both the exact expression and the approximate value.Wait, but in the first sub-problem, the answer was an integer, 50. So, maybe in the second sub-problem, they expect an integer as well.So, perhaps I should round 85.5169679 to the nearest whole number, which is 86.Alternatively, maybe Alex can make a fraction of a post, but that doesn't make sense in reality. So, 86 is the correct number.But let me think again. If Alex starts with 50 posts, and each month increases by 5%, then in the second month, it's 52.5, which is 52 or 53? But the problem didn't specify rounding in the first place, so maybe they allow fractional posts for the sake of the model.But in reality, you can't have half a post, but in mathematical terms, it's acceptable. So, perhaps the answer is 85.5169679, which is approximately 85.52.But let me check if I can compute it more accurately.Alternatively, I can use the formula for compound interest, which is similar to this problem.The formula is:[ A = P(1 + r)^t ]But in this case, it's monthly, so for each month, it's compounded. So, for the ( t )-th month, it's:[ P(t) = P(1) times (1 + r)^{t-1} ]Which is what I used earlier.So, for ( t = 12 ), it's:[ P(12) = 50 times (1.05)^{11} ]As calculated earlier, approximately 85.5169679.So, I think that's the answer.Summary:Sub-problem 1: The minimum total number of posts ( n ) is 50.Sub-problem 2: The general expression is ( P(t) = 50 times (1.05)^{t-1} ), and in the 12th month, the number of posts is approximately 85.52, which would round to 86 if we consider whole posts.But since the problem didn't specify rounding, I think it's better to present the exact value as 85.5169679 or approximately 85.52.Wait, but in the first sub-problem, the answer was an integer, so maybe in the second sub-problem, they expect an integer as well. So, 86.Alternatively, maybe they accept the decimal value. I think I'll go with the exact expression and the approximate decimal value.Final AnswerSub-problem 1: The minimum total number of posts ( n ) is boxed{50}.Sub-problem 2: The total number of posts in the 12th month is approximately boxed{86}.Wait, but earlier I calculated it as approximately 85.5169679, which is closer to 85.52. So, if we round to the nearest whole number, it's 86. But if we consider that the number of posts must be a whole number, then 86 is the correct answer.Alternatively, if we keep it as a decimal, it's approximately 85.52, but since the problem didn't specify, I think 86 is acceptable.But let me check if 50*(1.05)^11 is exactly 85.5169679.Yes, because 1.05^11 is approximately 1.710339358, and 50*1.710339358 is 85.5169679.So, if we round to the nearest whole number, it's 86.Therefore, I think the answer is 86.Final AnswerSub-problem 1: boxed{50}Sub-problem 2: boxed{86}</think>

answer:<think>Okay, so I have this problem about a literary agent who helps an author plan their career. The agent has identified a demand function for books on royal figures, which is given by D(p) = 1500 - 20p. Here, p is the price of the book in dollars, and D(p) is the number of books demanded at that price. The cost function for producing these books is C(q) = 500 + 10q, where q is the quantity of books produced.There are two parts to this problem. The first part is to determine the price p that maximizes the agent's profit. Profit is given by total revenue minus total cost. Total revenue R(p) is p multiplied by D(p), and total cost is given by C(q). The second part is similar, but the demand function changes to D2(p) = 1200 - 25p, and I need to find the new optimal price p that maximizes profit with this new demand function.Let me start with the first part.Problem 1: Determine the price p that maximizes profit.First, I need to recall that profit is total revenue minus total cost. So, profit π(p) = R(p) - C(q). But since q is the quantity produced, and in a competitive market, the quantity produced is equal to the quantity demanded, so q = D(p). Therefore, I can express profit as a function of p.So, let me write that down:Profit π(p) = R(p) - C(q)But R(p) = p * D(p) and C(q) = 500 + 10q, and q = D(p). So substituting, we get:π(p) = p * D(p) - (500 + 10 * D(p))Given D(p) = 1500 - 20p, so substituting that in:π(p) = p*(1500 - 20p) - (500 + 10*(1500 - 20p))Let me compute each part step by step.First, compute R(p) = p*(1500 - 20p):R(p) = 1500p - 20p²Then, compute C(q) = 500 + 10*(1500 - 20p):C(q) = 500 + 15000 - 200pC(q) = 15500 - 200pTherefore, profit π(p) = R(p) - C(q) = (1500p - 20p²) - (15500 - 200p)Let me simplify this:π(p) = 1500p - 20p² - 15500 + 200pCombine like terms:1500p + 200p = 1700pSo, π(p) = -20p² + 1700p - 15500Now, this is a quadratic function in terms of p. Since the coefficient of p² is negative (-20), the parabola opens downward, meaning the vertex is the maximum point. So, the maximum profit occurs at the vertex of this parabola.The vertex of a parabola given by ax² + bx + c is at p = -b/(2a). So, let's compute that.Here, a = -20, b = 1700.So, p = -1700 / (2*(-20)) = -1700 / (-40) = 42.5So, p = 42.50 is the price that maximizes profit.Wait, let me double-check my calculations.Compute π(p) again:π(p) = p*(1500 - 20p) - (500 + 10*(1500 - 20p))First term: 1500p - 20p²Second term: 500 + 15000 - 200p = 15500 - 200pSubtracting: 1500p - 20p² - 15500 + 200pCombine 1500p + 200p = 1700pSo, π(p) = -20p² + 1700p - 15500Yes, that seems correct.Then, vertex at p = -b/(2a) = -1700/(2*(-20)) = 1700/40 = 42.5Yes, so p = 42.50.Alternatively, I can take the derivative of π(p) with respect to p, set it equal to zero, and solve for p.Let me do that as a check.π(p) = -20p² + 1700p - 15500dπ/dp = -40p + 1700Set derivative equal to zero:-40p + 1700 = 0-40p = -1700p = (-1700)/(-40) = 42.5Same result. So, that seems correct.Therefore, the optimal price is 42.50.Problem 2: New demand function D2(p) = 1200 - 25p. Find the new optimal price p.Alright, so now the demand function changes to D2(p) = 1200 - 25p. The cost function remains the same: C(q) = 500 + 10q.Again, we need to find the price p that maximizes profit. So, similar approach: express profit as a function of p, then find its maximum.So, profit π(p) = R(p) - C(q)Again, R(p) = p * D2(p) = p*(1200 - 25p)C(q) = 500 + 10*q, and q = D2(p) = 1200 - 25pSo, π(p) = p*(1200 - 25p) - (500 + 10*(1200 - 25p))Let me compute each part.First, compute R(p):R(p) = 1200p - 25p²Then, compute C(q):C(q) = 500 + 10*(1200 - 25p) = 500 + 12000 - 250p = 12500 - 250pTherefore, profit π(p) = (1200p - 25p²) - (12500 - 250p)Simplify:π(p) = 1200p - 25p² - 12500 + 250pCombine like terms:1200p + 250p = 1450pSo, π(p) = -25p² + 1450p - 12500Again, this is a quadratic function in p, opening downward (since coefficient of p² is -25). So, the maximum occurs at the vertex.Compute p = -b/(2a). Here, a = -25, b = 1450.So, p = -1450 / (2*(-25)) = -1450 / (-50) = 29So, p = 29 is the new optimal price.Wait, let me verify with calculus as well.Compute derivative of π(p):π(p) = -25p² + 1450p - 12500dπ/dp = -50p + 1450Set derivative to zero:-50p + 1450 = 0-50p = -1450p = (-1450)/(-50) = 29Same result. So, p = 29.Therefore, the new optimal price is 29.Wait, let me just double-check the profit function again.Starting from π(p) = R(p) - C(q)R(p) = p*(1200 -25p) = 1200p -25p²C(q) = 500 +10*(1200 -25p) = 500 +12000 -250p = 12500 -250pSo, π(p) = (1200p -25p²) - (12500 -250p) = 1200p -25p² -12500 +250p = (1200p +250p) -25p² -12500 = 1450p -25p² -12500Yes, that's correct.So, the calculations seem consistent.Therefore, the optimal price after the competitor enters is 29.Summary:1. Original demand function: Optimal price is 42.50.2. New demand function: Optimal price is 29.I think that's it. I don't see any mistakes in my calculations, and both methods (vertex formula and calculus) give the same results.Final Answer1. The optimal price is boxed{42.50} dollars.2. The new optimal price is boxed{29} dollars.</think>

answer:<think>Alright, so I've got these two graph theory problems to solve, both related to directed graphs. Let me take them one at a time and think through each step carefully.Starting with the first problem:1. Prove that a strongly connected directed graph G with n vertices and m edges has a Hamiltonian cycle if and only if the in-degree and out-degree of every vertex are equal.Hmm, okay. So, I know that a Hamiltonian cycle is a cycle that visits every vertex exactly once and returns to the starting vertex. A strongly connected graph means that there's a directed path from any vertex to any other vertex.The statement is an "if and only if," so I need to prove both directions:- If G has a Hamiltonian cycle, then the in-degree and out-degree of every vertex are equal.- If the in-degree and out-degree of every vertex are equal, then G has a Hamiltonian cycle.Let me tackle the first direction.Forward Direction (⇒):Assume that G has a Hamiltonian cycle. Since it's a cycle, each vertex must have exactly one incoming edge and one outgoing edge within the cycle. So, for each vertex v, the in-degree d^-(v) = 1 and the out-degree d^+(v) = 1. But wait, the graph might have more edges than just the Hamiltonian cycle. So, does that mean the total in-degree and out-degree are equal?Wait, no. The total in-degree and out-degree for each vertex in the entire graph must be equal because for the entire graph, the sum of all in-degrees equals the sum of all out-degrees, which is equal to m. But if the graph has a Hamiltonian cycle, each vertex has at least one in and one out, but maybe more.But the problem states that the in-degree and out-degree of every vertex are equal. So, if G has a Hamiltonian cycle, does that force the in-degree and out-degree to be equal for each vertex?Wait, maybe not necessarily. Because the graph could have additional edges beyond the Hamiltonian cycle, so a vertex could have more in-degrees or more out-degrees. Hmm, so perhaps my initial thought is wrong.Wait, maybe I need to think about the graph being strongly connected and having equal in-degree and out-degree for each vertex. That's a condition for something else... Oh, right! That's the condition for a directed graph to have an Eulerian circuit. But here, we're talking about a Hamiltonian cycle.Wait, so maybe the forward direction isn't straightforward. Let me think again.If G has a Hamiltonian cycle, then it's a cycle that includes every vertex. In that cycle, each vertex has in-degree 1 and out-degree 1. However, the graph might have additional edges. So, the total in-degree and out-degree for each vertex could be higher, but they don't necessarily have to be equal.Wait, so perhaps the forward direction isn't necessarily true? But the problem says "if and only if," so both directions must hold.Wait, maybe I need to reconsider. Perhaps the problem is saying that G has a Hamiltonian cycle if and only if the in-degree equals the out-degree for every vertex. So, maybe it's not about the entire graph's in and out degrees, but specifically that each vertex has equal in and out degrees.Wait, but in a strongly connected graph, if each vertex has equal in-degree and out-degree, does that imply a Hamiltonian cycle? Or is it the other way around?Wait, I think I might be confusing Eulerian circuits with Hamiltonian cycles. Let me recall:- A directed graph has an Eulerian circuit if and only if it is strongly connected and every vertex has equal in-degree and out-degree.- A Hamiltonian cycle is a cycle that visits every vertex exactly once, which is a different concept.So, the problem is stating that in a strongly connected graph, having equal in-degree and out-degree for every vertex is equivalent to having a Hamiltonian cycle.Wait, that doesn't sound right because having equal in and out degrees is about Eulerian circuits, not Hamiltonian cycles.Wait, maybe the problem is misstated? Or perhaps I'm misunderstanding.Wait, let me check the problem again:"Prove that there exists a Hamiltonian cycle in G if and only if the in-degree and out-degree of every vertex are equal."Hmm, so it's saying that the existence of a Hamiltonian cycle is equivalent to all vertices having equal in and out degrees.But I thought that was the condition for an Eulerian circuit, not a Hamiltonian cycle.Wait, perhaps in a strongly connected graph, if all vertices have equal in and out degrees, then it's sufficient for a Hamiltonian cycle? Or maybe it's necessary?Wait, let me think of an example.Take a directed cycle with 3 vertices: each vertex has in-degree 1 and out-degree 1. It's strongly connected and has a Hamiltonian cycle. So, that satisfies the condition.But what if I have a strongly connected graph where each vertex has in-degree and out-degree equal to 2. Does that necessarily have a Hamiltonian cycle? Hmm, not necessarily. For example, consider two separate directed cycles, but wait, no, the graph is strongly connected, so it can't be two separate cycles.Wait, no, in a strongly connected graph, you can't have two separate cycles because you can go from one cycle to the other. So, maybe in that case, it's more complex.Wait, perhaps I need to think about the Ghouila-Houri theorem, which states that a strongly connected directed graph with n vertices has a Hamiltonian cycle if every vertex has out-degree at least n/2.But that's a different condition.Wait, so maybe the problem is incorrect? Or perhaps I'm missing something.Wait, let me try to think about the forward direction again.If G has a Hamiltonian cycle, then for each vertex, the number of times it's entered equals the number of times it's exited in the cycle, which is once each. But the graph might have more edges, so the in-degree and out-degree could be higher, but not necessarily equal.Wait, so perhaps the forward direction isn't necessarily true. So, maybe the problem is misstated.Wait, but the problem says "if and only if," so both directions must hold. So, perhaps the problem is correct, but I'm missing something.Wait, let me think about the reverse direction.Reverse Direction (⇐):Assume that for every vertex v in G, d^-(v) = d^+(v). We need to show that G has a Hamiltonian cycle.Wait, but as I thought earlier, this is the condition for an Eulerian circuit, not necessarily a Hamiltonian cycle.Wait, perhaps in a strongly connected graph where each vertex has equal in and out degrees, it's possible to have a Hamiltonian cycle, but I don't think it's guaranteed.Wait, for example, take a graph with 4 vertices, each with in-degree and out-degree 2. It's strongly connected, but does it have a Hamiltonian cycle?Wait, let me construct such a graph. Suppose we have vertices A, B, C, D.Each has in-degree and out-degree 2. So, for example, A could have edges to B and C, and incoming edges from B and C. Similarly for others.But does this graph necessarily have a Hamiltonian cycle? I'm not sure.Wait, perhaps it's not necessarily true. So, maybe the problem is incorrect, or perhaps I'm misunderstanding.Wait, maybe the problem is referring to a regular tournament graph, where each vertex has the same in-degree and out-degree, which is (n-1)/2 for odd n, but that's only possible if n is odd.Wait, but the problem doesn't specify that n is odd, so maybe that's not it.Wait, perhaps the problem is correct, and I need to think differently.Wait, maybe in a strongly connected graph where each vertex has equal in-degree and out-degree, the graph is Eulerian, but not necessarily Hamiltonian. So, perhaps the problem is incorrect, or perhaps I'm missing a key theorem.Wait, maybe I should look up the conditions for a directed graph to have a Hamiltonian cycle.Wait, I recall that a theorem by Ghouila-Houri states that a strongly connected directed graph with n vertices has a Hamiltonian cycle if every vertex has out-degree at least n/2.But that's a different condition.Alternatively, there's the theorem by Chvásal, which is a more general condition, but it's more complex.Wait, perhaps the problem is using a different approach. Maybe it's considering that if in-degree equals out-degree for all vertices, then the graph is Eulerian, and perhaps in some cases, Eulerian implies Hamiltonian.But I don't think that's generally true.Wait, maybe in a strongly connected Eulerian graph, there exists a Hamiltonian cycle. Is that true?Wait, no, I don't think so. For example, consider a graph with two vertices, each with two edges going both ways. So, each vertex has in-degree 2 and out-degree 2. It's strongly connected and Eulerian, but it doesn't have a Hamiltonian cycle because it's just two vertices with multiple edges.Wait, but in that case, it's not a simple graph, but the problem doesn't specify whether the graph is simple or not.Wait, but even in a simple graph, if you have a graph where each vertex has in-degree and out-degree equal, it's Eulerian, but not necessarily Hamiltonian.Wait, for example, take a graph with three vertices, each with in-degree and out-degree 2. So, each vertex has two incoming and two outgoing edges. But in a simple graph, each vertex can only have two edges, so it's a complete directed graph, which is a tournament. In that case, it's strongly connected and has a Hamiltonian cycle.Wait, but if n is 4, and each vertex has in-degree and out-degree 2, it's possible to have a graph that's strongly connected but doesn't have a Hamiltonian cycle.Wait, let me try to construct such a graph.Take four vertices: A, B, C, D.Each has in-degree and out-degree 2.Let me connect A to B and C, B to C and D, C to D and A, D to A and B.Wait, does this graph have a Hamiltonian cycle?Let me see: A→B→C→D→A. That's a cycle, but does it include all vertices? Yes, it's a Hamiltonian cycle.Wait, maybe in this case, it does have a Hamiltonian cycle.Wait, maybe in a strongly connected Eulerian graph, you always have a Hamiltonian cycle. But I'm not sure.Wait, let me think of another example. Suppose we have a graph with four vertices, each with in-degree and out-degree 2, but arranged in a way that there's no Hamiltonian cycle.Wait, maybe if the graph is composed of two separate cycles, but since it's strongly connected, that's not possible.Wait, no, if it's strongly connected, you can't have two separate cycles because you can go from one to the other.Wait, maybe I'm overcomplicating this.Wait, perhaps the problem is correct, and the condition that in-degree equals out-degree for all vertices in a strongly connected graph implies the existence of a Hamiltonian cycle.But I'm not entirely sure. Maybe I need to think about it differently.Wait, perhaps the problem is using the fact that in a strongly connected graph where each vertex has equal in-degree and out-degree, the graph is Eulerian, and perhaps in such a graph, an Eulerian circuit can be used to find a Hamiltonian cycle.But I don't see a direct connection.Wait, maybe I'm missing a theorem here. Let me try to recall.Wait, I think I might have confused the conditions. Let me think again.If a directed graph is strongly connected and every vertex has equal in-degree and out-degree, then it's Eulerian, meaning it has an Eulerian circuit.But having an Eulerian circuit doesn't necessarily imply a Hamiltonian cycle.Wait, so perhaps the problem is incorrect, or perhaps I'm misunderstanding the problem.Wait, maybe the problem is actually about Eulerian circuits, not Hamiltonian cycles. But no, the problem clearly states Hamiltonian cycle.Wait, perhaps the problem is correct, and I need to think about it differently.Wait, maybe in a strongly connected graph where each vertex has equal in-degree and out-degree, the graph has a Hamiltonian cycle.But I'm not sure. Let me try to find a counterexample.Suppose n=4, each vertex has in-degree and out-degree 2.Construct the graph as follows:A→B, A→CB→C, B→DC→D, C→AD→A, D→BNow, does this graph have a Hamiltonian cycle?Let's see: A→B→C→D→A. That's a cycle, but does it include all vertices? Yes, it's a Hamiltonian cycle.Wait, so in this case, it does have a Hamiltonian cycle.Wait, maybe all such graphs have Hamiltonian cycles. Let me try another example.Take n=5, each vertex has in-degree and out-degree 2.Construct the graph as follows:A→B, A→CB→C, B→DC→D, C→ED→E, D→AE→A, E→BNow, does this graph have a Hamiltonian cycle?Let me try to find one: A→B→C→D→E→A. That's a cycle, but it's a Hamiltonian cycle.Wait, so maybe in such cases, it's always possible to find a Hamiltonian cycle.Wait, perhaps the condition that the graph is strongly connected and each vertex has equal in-degree and out-degree implies that it's Hamiltonian.But I'm not sure. Let me check.Wait, I think there's a theorem called the Ghouila-Houri theorem which states that a strongly connected directed graph with n vertices has a Hamiltonian cycle if every vertex has out-degree at least n/2.But in our case, if each vertex has out-degree equal to in-degree, which could be less than n/2.Wait, for example, if n=4, each vertex has out-degree 2, which is equal to n/2, so it satisfies the Ghouila-Houri condition.But if n=5, each vertex has out-degree 2, which is less than n/2 (which is 2.5), so it doesn't satisfy the Ghouila-Houri condition, but in my example, it still had a Hamiltonian cycle.Wait, so maybe the condition of equal in-degree and out-degree in a strongly connected graph is sufficient for a Hamiltonian cycle, even if it doesn't meet the Ghouila-Houri condition.Wait, but I'm not sure. Maybe I need to think about it differently.Wait, perhaps the problem is correct, and I need to proceed with the proof.So, for the forward direction, if G has a Hamiltonian cycle, then in the cycle, each vertex has in-degree 1 and out-degree 1. But the graph might have more edges, so the total in-degree and out-degree could be higher, but they must be equal because the sum of in-degrees equals the sum of out-degrees, which is m.Wait, but the problem states that the in-degree and out-degree of every vertex are equal, not just that the sum is equal.Wait, so if G has a Hamiltonian cycle, does that imply that each vertex has equal in-degree and out-degree?No, because the graph could have additional edges that make some vertices have higher in-degree or out-degree.Wait, so perhaps the forward direction isn't necessarily true.Wait, maybe the problem is misstated, or perhaps I'm misunderstanding.Wait, perhaps the problem is saying that G has a Hamiltonian cycle if and only if the in-degree equals the out-degree for every vertex, but that might not be correct.Wait, maybe I should look up the problem or similar theorems.Wait, I think I'm stuck here. Maybe I should proceed to the second problem and come back to this one later.Problem 2:Suppose that for each vertex v_i ∈ V, the in-degree d^-(v_i) and out-degree d^+(v_i) are both equal to k. Given that k is a constant, determine the minimum value of k such that the graph G contains a Eulerian circuit. Hint: Use the concepts of Eulerian circuits and degree sequences in directed graphs.Okay, so for a directed graph to have a Eulerian circuit, it must be strongly connected, and every vertex must have equal in-degree and out-degree.But in this problem, we're given that each vertex has in-degree and out-degree equal to k, so the condition for a Eulerian circuit is satisfied in terms of degrees.But the problem is asking for the minimum value of k such that G contains a Eulerian circuit.Wait, but if the graph is strongly connected and each vertex has equal in-degree and out-degree, then it has a Eulerian circuit regardless of k, as long as k ≥ 1.Wait, but maybe the problem is not assuming strong connectivity, but just that each vertex has in-degree and out-degree equal to k.Wait, the problem says "the graph G contains a Eulerian circuit," so it must be strongly connected and have equal in and out degrees.But since the problem states that each vertex has in-degree and out-degree equal to k, which satisfies the degree condition for a Eulerian circuit. So, the only other condition is strong connectivity.But the problem is asking for the minimum k such that G contains a Eulerian circuit. So, perhaps it's considering the case where the graph might not be strongly connected unless k is sufficiently large.Wait, but the problem doesn't specify anything about the graph being strongly connected, so maybe we need to ensure that the graph is strongly connected as well.Wait, but if k is too small, the graph might not be strongly connected.Wait, for example, if k=1, then each vertex has one incoming and one outgoing edge. But the graph could be a collection of cycles, which are strongly connected components, but not necessarily the entire graph being strongly connected.Wait, so to ensure that the graph is strongly connected, we might need a higher k.Wait, but the problem is asking for the minimum k such that G contains a Eulerian circuit. So, perhaps the minimum k is 1, but only if the graph is strongly connected.Wait, but without the graph being strongly connected, even if each vertex has equal in and out degrees, it might not have a Eulerian circuit.Wait, so maybe the problem is assuming that the graph is strongly connected, and then the minimum k is 1.But I'm not sure. Let me think again.Wait, the problem says "determine the minimum value of k such that the graph G contains a Eulerian circuit."So, perhaps it's considering that for a graph with n vertices, each with in-degree and out-degree k, what's the minimum k that guarantees the existence of a Eulerian circuit.But in that case, the graph must be strongly connected, and each vertex has equal in and out degrees.But the minimum k would be 1, but only if the graph is strongly connected.Wait, but if k=1, the graph could be a single cycle, which is strongly connected and has a Eulerian circuit.But if k=1 and the graph is not strongly connected, then it might consist of multiple cycles, each of which is strongly connected, but the entire graph isn't.Wait, so perhaps the problem is considering that the graph is strongly connected, and then the minimum k is 1.But the problem doesn't specify that the graph is strongly connected, so maybe the answer is that k must be at least 1, but the graph must also be strongly connected.Wait, but the problem is asking for the minimum k such that G contains a Eulerian circuit, so perhaps it's considering that the graph is strongly connected, and then k can be 1.Wait, but I'm not sure. Maybe I need to think about it differently.Wait, perhaps the problem is considering that for a directed graph with n vertices, each with in-degree and out-degree k, the minimum k such that the graph is strongly connected and thus has a Eulerian circuit.In that case, the minimum k would be 1, because a single cycle (k=1) is strongly connected and has a Eulerian circuit.But if k=0, the graph has no edges, so it doesn't have a Eulerian circuit.Wait, but k must be at least 1.Wait, but the problem says "k is a constant," so perhaps the minimum k is 1.But I'm not sure. Maybe I should think about it in terms of the number of edges.Wait, the number of edges m is n*k, since each vertex has out-degree k.For a Eulerian circuit, the graph must be strongly connected and each vertex has equal in and out degrees, which is satisfied here.So, the minimum k is 1, provided that the graph is strongly connected.But if k=1, the graph could be a single cycle, which is strongly connected, so it has a Eulerian circuit.If k=1 and the graph is not strongly connected, it might have multiple cycles, but each cycle is strongly connected, but the entire graph isn't, so it doesn't have a Eulerian circuit.Wait, but the problem is asking for the minimum k such that G contains a Eulerian circuit. So, perhaps the answer is that k must be at least 1, but the graph must also be strongly connected.But the problem doesn't specify anything about the graph being strongly connected, so maybe the answer is that k must be at least 1, but the graph must also be strongly connected.Wait, but the problem is asking for the minimum k, so perhaps it's 1, assuming the graph is strongly connected.Wait, but without the graph being strongly connected, even with k=1, it might not have a Eulerian circuit.Wait, maybe the problem is considering that the graph is strongly connected, and then the minimum k is 1.I think I'm overcomplicating this. Let me try to answer.For a directed graph to have a Eulerian circuit, it must be strongly connected and every vertex must have equal in-degree and out-degree.In this problem, each vertex has in-degree and out-degree equal to k, so the degree condition is satisfied.The only other condition is strong connectivity.So, the minimum k is 1, provided that the graph is strongly connected.But if the graph isn't strongly connected, even with k=1, it might not have a Eulerian circuit.But the problem is asking for the minimum k such that G contains a Eulerian circuit. So, perhaps the answer is that k must be at least 1, but the graph must also be strongly connected.But since the problem doesn't specify strong connectivity, maybe the answer is that k must be at least 1, but the graph must also be strongly connected.Wait, but the problem is asking for the minimum k, so perhaps it's 1, assuming the graph is strongly connected.Alternatively, maybe the problem is considering that for a graph with n vertices, the minimum k is 1, because a single cycle (k=1) suffices.But I'm not entirely sure. Maybe the answer is k=1.Wait, but let me think of a specific example.If n=3, and k=1, then the graph could be a directed cycle, which is strongly connected and has a Eulerian circuit.If n=4, and k=1, the graph could be a single cycle, which is strongly connected and has a Eulerian circuit.So, in that case, k=1 is sufficient.But if the graph isn't strongly connected, even with k=1, it might not have a Eulerian circuit.But the problem is asking for the minimum k such that G contains a Eulerian circuit, so perhaps it's 1, assuming the graph is strongly connected.Wait, but the problem doesn't specify that the graph is strongly connected, so maybe the answer is that k must be at least 1, but the graph must also be strongly connected.But since the problem is asking for the minimum k, and not considering strong connectivity, maybe the answer is k=1.Wait, but I'm not sure. Maybe I should proceed with k=1.But wait, in the first problem, the graph is strongly connected, so maybe in the second problem, it's also strongly connected.Wait, the second problem doesn't specify that G is strongly connected, so perhaps the answer is that k must be at least 1, but the graph must also be strongly connected.But the problem is asking for the minimum k, so perhaps the answer is k=1, assuming the graph is strongly connected.Wait, but without the graph being strongly connected, even with k=1, it might not have a Eulerian circuit.So, perhaps the answer is that k must be at least 1, and the graph must be strongly connected.But the problem is only asking for the minimum k, so maybe it's 1.I think I'll go with k=1 as the minimum value.</think>

answer:<think>Okay, so I have two problems here about an actor's earnings from films. Let me start with the first one.Problem 1:The revenue function is given by ( R(t) = 10^6 e^{-0.1t} ) dollars per week. The actor's earnings are ( E(t) = alpha int_{0}^{t} R(tau) , dtau ). I need to find the total earnings by the end of the 10th week, so ( t = 10 ).First, I should compute the integral of ( R(tau) ) from 0 to 10. Let's write that out:( E(10) = alpha int_{0}^{10} 10^6 e^{-0.1tau} , dtau )I can factor out the constants, so:( E(10) = alpha times 10^6 times int_{0}^{10} e^{-0.1tau} , dtau )The integral of ( e^{kt} ) is ( frac{1}{k} e^{kt} ), so here, ( k = -0.1 ). Therefore, the integral becomes:( int e^{-0.1tau} dtau = frac{1}{-0.1} e^{-0.1tau} + C = -10 e^{-0.1tau} + C )Now, evaluating from 0 to 10:( [-10 e^{-0.1 times 10}] - [-10 e^{-0.1 times 0}] = (-10 e^{-1}) - (-10 e^{0}) = -10 e^{-1} + 10 times 1 = 10(1 - e^{-1}) )So, plugging back into ( E(10) ):( E(10) = alpha times 10^6 times 10(1 - e^{-1}) = alpha times 10^7 (1 - e^{-1}) )But wait, the problem doesn't specify the value of ( alpha ). Hmm, maybe I missed something? Let me check the problem statement again.Oh, it says the actor receives a percentage of the gross revenue, but in the function ( E(t) = alpha int R(tau) dtau ), ( alpha ) is the constant percentage. So, maybe ( alpha ) is given? Wait, no, in the first problem, it's only given ( R(t) ), and we need to compute ( E(t) ). But without knowing ( alpha ), I can't get a numerical value. Hmm, maybe ( alpha ) is given as part of the problem? Wait, no, looking back, the first problem only gives ( R(t) ) and asks for ( E(t) ) by week 10. So perhaps ( alpha ) is a constant that we can leave in terms of, or maybe it's given in the problem? Wait, no, the problem says "the actor's earnings from each film is a function of ( R(t) ), given by ( E(t) = alpha int R(tau) dtau )". So, ( alpha ) is a constant percentage, but it's not specified here. Hmm, that seems odd. Maybe I need to assume ( alpha ) is given? Wait, no, in the first problem, I think ( alpha ) is a constant, but it's not provided. Wait, maybe I misread the problem.Wait, no, the first problem says "the actor receives a percentage of the gross revenue", so ( alpha ) is that percentage. But since it's not given, perhaps it's just left as a variable? But the problem says "find the total earnings ( E(t) )", so maybe they just want the expression in terms of ( alpha ). Hmm, but the problem doesn't specify, so maybe I need to compute it in terms of ( alpha ). Alternatively, maybe ( alpha ) is given in the problem, but I think I might have missed it.Wait, no, looking back, the first problem only gives ( R(t) ) and says "find the total earnings ( E(t) ) by the end of the 10th week". So, perhaps ( alpha ) is given in the problem, but I think it's not. Wait, maybe I misread the problem. Let me check again.Wait, the first problem says "the actor's earnings from each film is a function of ( R(t) ), given by ( E(t) = alpha int R(tau) dtau ), where ( alpha ) is a constant percentage representing the actor's share." So, ( alpha ) is a constant, but it's not given in the first problem. Hmm, that's confusing. Maybe I need to leave the answer in terms of ( alpha )?Wait, but the problem says "find the total earnings ( E(t) )", so perhaps it's expecting a numerical value, but without ( alpha ), I can't compute it numerically. Maybe I made a mistake earlier.Wait, perhaps ( alpha ) is given in the problem, but I think I might have missed it. Wait, no, the first problem only gives ( R(t) ) and asks for ( E(t) ) by week 10. So, maybe ( alpha ) is 10%, or some standard percentage? But that's not stated. Hmm, maybe I need to proceed with ( alpha ) as a variable.Wait, but the problem says "the actor's earnings from each film is a function of ( R(t) ), given by ( E(t) = alpha int R(tau) dtau )", so ( alpha ) is a constant, but it's not provided. Therefore, perhaps the answer is expressed in terms of ( alpha ). So, in that case, I can write ( E(10) = alpha times 10^7 (1 - e^{-1}) ). But maybe I can compute the numerical value of the integral and leave ( alpha ) as a factor.Wait, let's compute ( 10^7 (1 - e^{-1}) ). Since ( e^{-1} ) is approximately 0.3679, so ( 1 - 0.3679 = 0.6321 ). Therefore, ( 10^7 times 0.6321 = 6,321,000 ). So, ( E(10) = alpha times 6,321,000 ).But the problem doesn't specify ( alpha ), so maybe that's the answer? Or perhaps I'm supposed to find ( alpha ) from some other information? Wait, no, the first problem only gives ( R(t) ) and asks for ( E(t) ) by week 10. So, perhaps the answer is ( 6,321,000 alpha ). But maybe I'm missing something.Wait, perhaps ( alpha ) is given in the problem, but I think it's not. Wait, no, the problem says "the actor's earnings from each film is a function of ( R(t) ), given by ( E(t) = alpha int R(tau) dtau )", so ( alpha ) is a constant percentage, but it's not specified. Therefore, I think the answer is ( E(10) = 6,321,000 alpha ). But maybe I should leave it in terms of ( alpha ) as ( 10^7 (1 - e^{-1}) alpha ).Wait, but perhaps I made a mistake in the integral calculation. Let me double-check:( int_{0}^{10} e^{-0.1tau} dtau = left[ -10 e^{-0.1tau} right]_0^{10} = (-10 e^{-1}) - (-10 e^{0}) = -10 e^{-1} + 10 = 10(1 - e^{-1}) ). That seems correct.So, ( E(10) = alpha times 10^6 times 10 (1 - e^{-1}) = alpha times 10^7 (1 - e^{-1}) ). So, yes, that's correct. Therefore, the total earnings are ( 10^7 (1 - e^{-1}) alpha ). Since ( 1 - e^{-1} ) is approximately 0.6321, so ( 10^7 times 0.6321 = 6,321,000 ). Therefore, ( E(10) = 6,321,000 alpha ).But the problem doesn't specify ( alpha ), so perhaps that's the answer. Alternatively, maybe ( alpha ) is given in the problem, but I think it's not. Wait, no, the problem says "the actor receives a percentage of the gross revenue", so ( alpha ) is that percentage, but it's not given. Therefore, perhaps the answer is expressed in terms of ( alpha ). So, I think that's the case.But wait, maybe I misread the problem. Let me check again.Wait, the problem says "the actor's earnings from each film is a function of ( R(t) ), given by ( E(t) = alpha int R(tau) dtau ), where ( alpha ) is a constant percentage representing the actor's share." So, ( alpha ) is a constant, but it's not given in the first problem. Therefore, I think the answer is ( E(10) = 10^7 (1 - e^{-1}) alpha ), which is approximately ( 6,321,000 alpha ).But maybe the problem expects a numerical value, so perhaps I need to assume ( alpha ) is 10%, or some other value. Wait, but the problem doesn't specify, so I think I should leave it in terms of ( alpha ). Therefore, the total earnings are ( 10^7 (1 - e^{-1}) alpha ) dollars.Alternatively, maybe I made a mistake in the integral. Let me check again.Wait, the integral of ( e^{-0.1tau} ) from 0 to 10 is indeed ( 10(1 - e^{-1}) ). So, ( E(10) = alpha times 10^6 times 10(1 - e^{-1}) = alpha times 10^7 (1 - e^{-1}) ). So, that's correct.Therefore, the answer is ( E(10) = 10^7 (1 - e^{-1}) alpha ), which is approximately ( 6,321,000 alpha ).Wait, but maybe I should write it as ( 10^7 (1 - e^{-1}) alpha ) without approximating, since ( e^{-1} ) is an exact value.So, to sum up, the total earnings by the end of the 10th week are ( 10^7 (1 - e^{-1}) alpha ) dollars.But wait, the problem says "find the total earnings ( E(t) )", so maybe they just want the expression, but I think it's more likely that they expect a numerical value. But without ( alpha ), I can't compute it numerically. Therefore, perhaps I need to leave it in terms of ( alpha ).Alternatively, maybe I misread the problem, and ( alpha ) is given in the problem. Let me check again.Wait, the problem says "the actor's earnings from each film is a function of ( R(t) ), given by ( E(t) = alpha int R(tau) dtau ), where ( alpha ) is a constant percentage representing the actor's share." So, ( alpha ) is a constant, but it's not given in the first problem. Therefore, I think the answer is ( E(10) = 10^7 (1 - e^{-1}) alpha ).But maybe the problem expects me to compute it in terms of ( alpha ), so perhaps that's the answer.Wait, but in the second problem, ( alpha ) is given as 0.01, so maybe in the first problem, ( alpha ) is also given, but I think it's not. Wait, no, in the first problem, ( alpha ) is not given, so I think the answer is ( 10^7 (1 - e^{-1}) alpha ).Alternatively, maybe I made a mistake in the integral. Let me check again.Wait, the integral of ( e^{-0.1tau} ) from 0 to 10 is:( int_{0}^{10} e^{-0.1tau} dtau = left[ -10 e^{-0.1tau} right]_0^{10} = (-10 e^{-1}) - (-10 e^{0}) = -10 e^{-1} + 10 = 10(1 - e^{-1}) ). That's correct.So, ( E(10) = alpha times 10^6 times 10(1 - e^{-1}) = alpha times 10^7 (1 - e^{-1}) ).Therefore, the total earnings are ( 10^7 (1 - e^{-1}) alpha ) dollars.But since ( alpha ) is a percentage, perhaps it's given as a decimal, like 0.1 for 10%, but since it's not given, I think that's the answer.Wait, but the problem says "the actor receives a percentage of the gross revenue", so ( alpha ) is that percentage, but it's not specified. Therefore, I think the answer is ( E(10) = 10^7 (1 - e^{-1}) alpha ).Alternatively, maybe I should write it as ( E(10) = 10^7 (1 - e^{-1}) times alpha ), but that's the same thing.So, I think that's the answer for the first problem.Problem 2:Now, the second problem is a bit more involved. The revenue function is given by ( R(t) = a t^k e^{-bt} ), where ( a ), ( b ), and ( k ) are constants. The actor's earnings by the end of the 10th week are 2,000,000, and ( alpha = 0.01 ). Also, ( R(t) ) achieves its maximum at ( t = 5 ).So, I need to find the values of ( a ), ( b ), and ( k ).First, let's note that the maximum of ( R(t) ) occurs at ( t = 5 ). To find the maximum, we can take the derivative of ( R(t) ) with respect to ( t ) and set it equal to zero at ( t = 5 ).So, let's compute ( R'(t) ):( R(t) = a t^k e^{-bt} )Using the product rule, the derivative is:( R'(t) = a [k t^{k-1} e^{-bt} + t^k (-b) e^{-bt}] = a e^{-bt} t^{k-1} (k - b t) )Setting ( R'(5) = 0 ):( a e^{-5b} 5^{k-1} (k - 5b) = 0 )Since ( a ), ( e^{-5b} ), and ( 5^{k-1} ) are all positive (assuming ( a > 0 ), ( b > 0 ), ( k > 0 )), the only way for this product to be zero is if ( k - 5b = 0 ). Therefore:( k = 5b )So, that's one equation: ( k = 5b ).Next, we know that the actor's earnings by week 10 are 2,000,000, and ( alpha = 0.01 ). So, using the earnings function:( E(10) = alpha int_{0}^{10} R(t) dt = 0.01 int_{0}^{10} a t^k e^{-bt} dt = 2,000,000 )So, let's write that equation:( 0.01 times a int_{0}^{10} t^k e^{-bt} dt = 2,000,000 )Divide both sides by 0.01:( a int_{0}^{10} t^k e^{-bt} dt = 200,000,000 )So, ( a times int_{0}^{10} t^k e^{-bt} dt = 2 times 10^8 )Now, we have two unknowns: ( a ) and ( b ) (since ( k = 5b )). So, we need another equation, but I think we only have one equation so far. Wait, no, actually, we have ( k = 5b ), so we can express everything in terms of ( b ).So, let's substitute ( k = 5b ) into the integral:( a int_{0}^{10} t^{5b} e^{-bt} dt = 2 times 10^8 )Hmm, this integral looks like the incomplete gamma function. Recall that:( int_{0}^{x} t^{n} e^{-bt} dt = frac{Gamma(n+1, bx)}{b^{n+1}}} )But I'm not sure if that's helpful here. Alternatively, perhaps we can use integration by parts or look for a substitution.Alternatively, perhaps we can make a substitution to simplify the integral. Let me try substituting ( u = bt ), so ( t = u/b ), ( dt = du/b ). Then, when ( t = 0 ), ( u = 0 ), and when ( t = 10 ), ( u = 10b ).So, substituting:( int_{0}^{10} t^{5b} e^{-bt} dt = int_{0}^{10b} left( frac{u}{b} right)^{5b} e^{-u} times frac{du}{b} )Simplify:( = frac{1}{b^{5b + 1}} int_{0}^{10b} u^{5b} e^{-u} du )Hmm, that's still complicated, but perhaps we can express it in terms of the gamma function. The integral ( int_{0}^{x} u^{n} e^{-u} du ) is the lower incomplete gamma function ( gamma(n+1, x) ). So, we can write:( int_{0}^{10b} u^{5b} e^{-u} du = gamma(5b + 1, 10b) )Therefore, the integral becomes:( frac{1}{b^{5b + 1}} gamma(5b + 1, 10b) )So, plugging back into our equation:( a times frac{1}{b^{5b + 1}} gamma(5b + 1, 10b) = 2 times 10^8 )This seems quite complex, and I don't think we can solve for ( b ) analytically here. Maybe we need to make an assumption or find a way to simplify.Alternatively, perhaps we can assume that ( 10b ) is large enough that the integral ( int_{0}^{10b} u^{5b} e^{-u} du ) is approximately equal to the complete gamma function ( Gamma(5b + 1) ), since for large ( x ), ( gamma(n, x) ) approaches ( Gamma(n) ). But ( 10b ) may not be that large, depending on ( b ).Alternatively, perhaps we can use the fact that ( R(t) ) has a maximum at ( t = 5 ), and perhaps the revenue function is symmetric around that point or something, but I don't think that's necessarily the case.Alternatively, maybe we can assume that ( b ) is small, so that ( 10b ) is not too large, but I don't know.Alternatively, perhaps we can make an educated guess for ( b ). Let's think about the revenue function ( R(t) = a t^k e^{-bt} ). The maximum occurs at ( t = 5 ), so perhaps ( b ) is such that the exponential decay isn't too rapid. Let's try to assume a value for ( b ) and see if we can find a consistent solution.Alternatively, perhaps we can use the fact that ( R(t) ) has a maximum at ( t = 5 ), so the derivative is zero there, which we already used to get ( k = 5b ).Alternatively, perhaps we can use the fact that the integral from 0 to 10 of ( t^k e^{-bt} dt ) can be expressed in terms of the gamma function, but I think that's what I did earlier.Wait, maybe I can use the substitution ( u = bt ), so ( t = u/b ), ( dt = du/b ), as before. Then, the integral becomes:( int_{0}^{10} t^{5b} e^{-bt} dt = frac{1}{b^{5b + 1}} int_{0}^{10b} u^{5b} e^{-u} du )Now, if ( 10b ) is large, say, much larger than ( 5b ), then the integral ( int_{0}^{10b} u^{5b} e^{-u} du ) is approximately ( Gamma(5b + 1) ), since the gamma function is the integral from 0 to infinity. So, if ( 10b ) is large, then ( gamma(5b + 1, 10b) approx Gamma(5b + 1) ).Therefore, the integral becomes approximately:( frac{Gamma(5b + 1)}{b^{5b + 1}} )So, then, our equation becomes:( a times frac{Gamma(5b + 1)}{b^{5b + 1}} = 2 times 10^8 )But this is still complicated because ( Gamma(5b + 1) ) is a function that's not easy to invert. However, perhaps we can make an assumption about ( b ) to simplify.Alternatively, perhaps we can assume that ( 5b ) is an integer, which would make ( Gamma(5b + 1) = (5b)! ). Let's try that.Suppose ( 5b ) is an integer. Let's let ( 5b = n ), where ( n ) is a positive integer. Then, ( b = n/5 ), and ( k = 5b = n ).So, substituting into the equation:( a times frac{n!}{(n/5)^{n + 1}} = 2 times 10^8 )Simplify:( a = 2 times 10^8 times frac{(n/5)^{n + 1}}{n!} )Now, we can try different integer values of ( n ) to see which gives a reasonable ( a ).Let's try ( n = 5 ):Then, ( b = 1 ), ( k = 5 ).Compute ( a ):( a = 2 times 10^8 times frac{(5/5)^{5 + 1}}{5!} = 2 times 10^8 times frac{1^6}{120} = 2 times 10^8 / 120 ≈ 1,666,666.67 )So, ( a ≈ 1,666,666.67 ), ( b = 1 ), ( k = 5 ).Let me check if this works.Compute ( R(t) = a t^k e^{-bt} = 1,666,666.67 t^5 e^{-t} ).Compute the integral ( int_{0}^{10} t^5 e^{-t} dt ). The exact value of this integral is ( gamma(6, 10) ), which is approximately equal to ( Gamma(6) = 120 ) since 10 is large. But let's compute it more accurately.Using integration by parts or a calculator, ( int_{0}^{10} t^5 e^{-t} dt ≈ 120 times (1 - e^{-10} times (1 + 10 + 10^2/2! + ... + 10^5/5!)) ). But since ( e^{-10} ) is very small, the integral is approximately 120.Therefore, ( E(10) = 0.01 times a times 120 ≈ 0.01 times 1,666,666.67 times 120 ≈ 0.01 times 200,000,000 = 2,000,000 ), which matches the given earnings.Therefore, ( a ≈ 1,666,666.67 ), ( b = 1 ), ( k = 5 ).Wait, but let me check if ( R(t) ) actually has a maximum at ( t = 5 ) with these values.Compute ( R'(t) = a e^{-t} t^4 (5 - t) ). Setting ( R'(t) = 0 ), we get ( t = 5 ), which is correct. So, yes, the maximum is at ( t = 5 ).Therefore, the values are ( a = 1,666,666.67 ), ( b = 1 ), ( k = 5 ).But let me check if ( n = 5 ) is the only possible solution. Let's try ( n = 10 ), so ( b = 2 ), ( k = 10 ).Then, ( a = 2 times 10^8 times frac{(10/5)^{10 + 1}}{10!} = 2 times 10^8 times frac{2^{11}}{3,628,800} )Compute ( 2^{11} = 2048 ), so:( a = 2 times 10^8 times 2048 / 3,628,800 ≈ 2 times 10^8 times 0.564 ≈ 112,800,000 )Then, compute ( int_{0}^{10} t^{10} e^{-2t} dt ). This integral is more complex, but let's approximate it.Using the substitution ( u = 2t ), ( t = u/2 ), ( dt = du/2 ), so:( int_{0}^{20} (u/2)^{10} e^{-u} times du/2 = frac{1}{2^{11}} int_{0}^{20} u^{10} e^{-u} du )The integral ( int_{0}^{20} u^{10} e^{-u} du ) is approximately ( Gamma(11) = 10! = 3,628,800 ), since 20 is large.Therefore, the integral is approximately ( 3,628,800 / 2048 ≈ 1,771.56 )Then, ( E(10) = 0.01 times a times 1,771.56 ≈ 0.01 times 112,800,000 times 1,771.56 ). Wait, that's way too large, because 112,800,000 * 1,771.56 is enormous, and multiplying by 0.01 would still be way more than 2,000,000. So, this suggests that ( n = 10 ) is not a valid solution.Therefore, ( n = 5 ) seems to be the correct choice.Alternatively, let's try ( n = 4 ), so ( b = 4/5 = 0.8 ), ( k = 4 ).Then, ( a = 2 times 10^8 times frac{(4/5)^{4 + 1}}{4!} = 2 times 10^8 times frac{(0.8)^5}{24} ≈ 2 times 10^8 times 0.32768 / 24 ≈ 2 times 10^8 times 0.013653 ≈ 27,306,000 )Then, compute ( int_{0}^{10} t^4 e^{-0.8t} dt ). Let's approximate this integral.Using substitution ( u = 0.8t ), ( t = u/0.8 ), ( dt = du/0.8 ), so:( int_{0}^{8} (u/0.8)^4 e^{-u} times du/0.8 = frac{1}{0.8^5} times frac{1}{0.8} int_{0}^{8} u^4 e^{-u} du = frac{1}{0.8^6} gamma(5, 8) )Compute ( 0.8^6 ≈ 0.262144 ), so ( 1/0.262144 ≈ 3.8147 )Now, ( gamma(5, 8) ) is the lower incomplete gamma function. The value of ( gamma(5,8) ) can be approximated using series expansion or tables. Alternatively, since 8 is not extremely large, but for ( n = 5 ), ( gamma(5,8) ) is close to ( Gamma(5) = 24 ), but let's compute it more accurately.Using the recurrence relation for the incomplete gamma function:( gamma(n, x) = (n-1)! e^{-x} sum_{k=0}^{n-1} frac{x^k}{k!} )For ( n = 5 ), ( x = 8 ):( gamma(5,8) = 4! e^{-8} sum_{k=0}^{4} frac{8^k}{k!} = 24 e^{-8} (1 + 8 + 32 + 80/3 + 16) )Wait, let me compute the sum:( sum_{k=0}^{4} frac{8^k}{k!} = 1 + 8 + 32 + 80/3 + 16 )Wait, 8^0/0! = 18^1/1! = 88^2/2! = 64/2 = 328^3/3! = 512/6 ≈ 85.33338^4/4! = 4096/24 ≈ 170.6667Wait, no, 8^3 is 512, divided by 6 is approximately 85.33338^4 is 4096, divided by 24 is approximately 170.6667So, sum is 1 + 8 + 32 + 85.3333 + 170.6667 ≈ 1 + 8 = 9, +32 = 41, +85.3333 ≈ 126.3333, +170.6667 ≈ 297So, ( gamma(5,8) ≈ 24 e^{-8} times 297 )Compute ( e^{-8} ≈ 0.00033546 )So, ( 24 times 0.00033546 ≈ 0.008051 )Multiply by 297: ( 0.008051 times 297 ≈ 2.391 )Therefore, ( gamma(5,8) ≈ 2.391 )So, the integral becomes:( 3.8147 times 2.391 ≈ 9.12 )Therefore, ( E(10) = 0.01 times a times 9.12 ≈ 0.01 times 27,306,000 times 9.12 ≈ 0.01 times 249,300,000 ≈ 2,493,000 ), which is higher than the required 2,000,000. So, this suggests that ( n = 4 ) is not the correct choice.Similarly, trying ( n = 6 ), ( b = 6/5 = 1.2 ), ( k = 6 ):( a = 2 times 10^8 times frac{(6/5)^{6 + 1}}{6!} = 2 times 10^8 times frac{(1.2)^7}{720} ≈ 2 times 10^8 times 3.58318 / 720 ≈ 2 times 10^8 times 0.004976 ≈ 995,200 )Then, compute ( int_{0}^{10} t^6 e^{-1.2t} dt ). Again, using substitution ( u = 1.2t ), ( t = u/1.2 ), ( dt = du/1.2 ):( int_{0}^{12} (u/1.2)^6 e^{-u} times du/1.2 = frac{1}{1.2^7} times frac{1}{1.2} int_{0}^{12} u^6 e^{-u} du = frac{1}{1.2^8} gamma(7, 12) )Compute ( 1.2^8 ≈ 4.2998 ), so ( 1/4.2998 ≈ 0.2327 )Now, ( gamma(7,12) ) is approximately ( Gamma(7) = 6! = 720 ), since 12 is large. So, the integral is approximately ( 0.2327 times 720 ≈ 167.3 )Then, ( E(10) = 0.01 times a times 167.3 ≈ 0.01 times 995,200 times 167.3 ≈ 0.01 times 166,000,000 ≈ 1,660,000 ), which is less than 2,000,000. So, this suggests that ( n = 6 ) is too high.Therefore, between ( n = 5 ) and ( n = 6 ), the correct value is likely ( n = 5 ), as it gives exactly 2,000,000 when approximated.Therefore, the values are ( a = 1,666,666.67 ), ( b = 1 ), ( k = 5 ).But let me check with ( n = 5 ) more accurately.Compute ( a = 1,666,666.67 ), ( b = 1 ), ( k = 5 ).Compute ( int_{0}^{10} t^5 e^{-t} dt ). As I mentioned earlier, this is approximately 120, since ( Gamma(6) = 120 ), and the integral from 0 to 10 is very close to that.Therefore, ( E(10) = 0.01 times 1,666,666.67 times 120 ≈ 0.01 times 200,000,000 = 2,000,000 ), which matches exactly.Therefore, the correct values are ( a = 1,666,666.67 ), ( b = 1 ), ( k = 5 ).But let me express ( a ) as a fraction. Since ( a = 2 times 10^8 / 120 = (2 times 10^8) / 120 = (2 times 10^8) / (12 times 10) ) = (2 times 10^7) / 12 = (10^7) / 6 ≈ 1,666,666.67 ). So, ( a = 10^7 / 6 ).Therefore, the exact values are ( a = frac{10^7}{6} ), ( b = 1 ), ( k = 5 ).So, to sum up, the values are ( a = frac{10^7}{6} ), ( b = 1 ), ( k = 5 ).Final Answer1. The total earnings by the end of the 10th week are boxed{6321000 alpha} dollars.2. The values of the constants are ( a = boxed{dfrac{10^7}{6}} ), ( b = boxed{1} ), and ( k = boxed{5} ).</think>

answer:<think>Okay, so I have this differential equation to solve for Sub-problem 1. Let me write it down again to make sure I have it right:[ frac{dP}{dt} = kPleft(1 - frac{P}{L}right) - cP ]Given the constants: ( k = 0.5 ), ( L = 100 ), and ( c = 0.1 ). The initial condition is ( P(0) = 10 ).Hmm, this looks like a logistic growth equation but with an additional term subtracting ( cP ). So, it's a modified logistic equation. I remember that the standard logistic equation is:[ frac{dP}{dt} = rPleft(1 - frac{P}{K}right) ]Where ( r ) is the growth rate and ( K ) is the carrying capacity. In this case, we have an extra term ( -cP ), which probably represents some kind of decay or inhibition due to regulations.Let me rewrite the given differential equation with the given constants plugged in:[ frac{dP}{dt} = 0.5Pleft(1 - frac{P}{100}right) - 0.1P ]Simplify the equation:First, distribute the 0.5P:[ frac{dP}{dt} = 0.5P - frac{0.5P^2}{100} - 0.1P ]Simplify the terms:0.5P - 0.1P is 0.4P, so:[ frac{dP}{dt} = 0.4P - frac{0.5P^2}{100} ]Simplify the second term:0.5 divided by 100 is 0.005, so:[ frac{dP}{dt} = 0.4P - 0.005P^2 ]So, the differential equation becomes:[ frac{dP}{dt} = 0.4P - 0.005P^2 ]This is a first-order ordinary differential equation, and it's separable. So, I can write it as:[ frac{dP}{0.4P - 0.005P^2} = dt ]Let me factor out P from the denominator:[ frac{dP}{P(0.4 - 0.005P)} = dt ]So, now it's set up for partial fractions. Let me set:[ frac{1}{P(0.4 - 0.005P)} = frac{A}{P} + frac{B}{0.4 - 0.005P} ]Multiply both sides by ( P(0.4 - 0.005P) ):[ 1 = A(0.4 - 0.005P) + BP ]Now, let's solve for A and B. Let me expand the right side:[ 1 = 0.4A - 0.005AP + BP ]Group the terms with P:[ 1 = 0.4A + P(-0.005A + B) ]Since this must hold for all P, the coefficients of like terms must be equal on both sides. So, the constant term on the left is 1, and on the right, it's 0.4A. The coefficient of P on the left is 0, and on the right, it's (-0.005A + B).So, we have two equations:1. ( 0.4A = 1 )2. ( -0.005A + B = 0 )Solving the first equation:( A = 1 / 0.4 = 2.5 )Then, substitute A into the second equation:( -0.005 * 2.5 + B = 0 )Calculate:-0.005 * 2.5 = -0.0125So,-0.0125 + B = 0 => B = 0.0125Therefore, the partial fractions decomposition is:[ frac{1}{P(0.4 - 0.005P)} = frac{2.5}{P} + frac{0.0125}{0.4 - 0.005P} ]So, going back to the integral:[ int left( frac{2.5}{P} + frac{0.0125}{0.4 - 0.005P} right) dP = int dt ]Let me compute each integral separately.First integral:[ int frac{2.5}{P} dP = 2.5 ln|P| + C_1 ]Second integral:Let me rewrite the denominator:0.4 - 0.005P = 0.005(80 - P)Wait, 0.4 / 0.005 = 80, so yes, 0.4 = 0.005*80.So,[ frac{0.0125}{0.4 - 0.005P} = frac{0.0125}{0.005(80 - P)} = frac{0.0125 / 0.005}{80 - P} = frac{2.5}{80 - P} ]So, the second integral becomes:[ int frac{2.5}{80 - P} dP ]Let me make a substitution: let u = 80 - P, then du = -dP, so -du = dP.Thus,[ int frac{2.5}{u} (-du) = -2.5 ln|u| + C_2 = -2.5 ln|80 - P| + C_2 ]Putting it all together, the left side integral is:2.5 ln|P| - 2.5 ln|80 - P| + CWhich can be written as:2.5 ln|P / (80 - P)| + CThe right side integral is:[ int dt = t + C' ]So, combining both sides:2.5 ln|P / (80 - P)| = t + CWhere C is the constant of integration.Now, let's solve for P. First, divide both sides by 2.5:ln|P / (80 - P)| = (t + C) / 2.5Exponentiate both sides:|P / (80 - P)| = e^{(t + C)/2.5} = e^{t/2.5} * e^{C/2.5}Let me denote e^{C/2.5} as another constant, say, K. So,P / (80 - P) = K e^{t/2.5}Since P and 80 - P are positive (assuming P < 80, which makes sense because L is 100, but let's see), so we can drop the absolute value.So,P / (80 - P) = K e^{t/2.5}Now, solve for P:Multiply both sides by (80 - P):P = K e^{t/2.5} (80 - P)Expand the right side:P = 80 K e^{t/2.5} - K e^{t/2.5} PBring all terms with P to the left:P + K e^{t/2.5} P = 80 K e^{t/2.5}Factor out P:P (1 + K e^{t/2.5}) = 80 K e^{t/2.5}Therefore,P = [80 K e^{t/2.5}] / [1 + K e^{t/2.5}]We can factor out e^{t/2.5} in the denominator:P = [80 K e^{t/2.5}] / [e^{t/2.5}(K + e^{-t/2.5})] = 80 K / (K + e^{-t/2.5})But let me write it as:P(t) = 80 K / (K + e^{-t/2.5})Alternatively, I can write it as:P(t) = 80 / (1 + (1/K) e^{-t/2.5})But let's keep it as P(t) = 80 K / (K + e^{-t/2.5})Now, we need to find K using the initial condition P(0) = 10.At t = 0:P(0) = 10 = 80 K / (K + e^{0}) = 80 K / (K + 1)So,10 = 80 K / (K + 1)Multiply both sides by (K + 1):10(K + 1) = 80 K10K + 10 = 80KSubtract 10K:10 = 70KSo,K = 10 / 70 = 1/7 ≈ 0.142857Therefore, the solution is:P(t) = 80 * (1/7) / [ (1/7) + e^{-t/2.5} ]Simplify numerator and denominator:Numerator: 80/7Denominator: (1/7) + e^{-t/2.5} = (1 + 7 e^{-t/2.5}) / 7So,P(t) = (80/7) / [ (1 + 7 e^{-t/2.5}) / 7 ] = (80/7) * (7 / (1 + 7 e^{-t/2.5})) = 80 / (1 + 7 e^{-t/2.5})So, the solution simplifies to:P(t) = 80 / (1 + 7 e^{-t/2.5})Let me double-check the steps to make sure I didn't make a mistake.Starting from the partial fractions, I think that was correct. Then integrating, I had to adjust for the substitution, which I did. Then solving for P, I think that was okay. Then applying the initial condition, which gave K = 1/7. Plugging back in, the algebra seems correct.So, the solution is P(t) = 80 / (1 + 7 e^{-t/2.5})Wait, but the original equation had L = 100, but in the solution, the carrying capacity seems to be 80. That's interesting. Let me think about that.Looking back at the differential equation:After simplifying, we had:dP/dt = 0.4P - 0.005P^2Which can be written as:dP/dt = P(0.4 - 0.005P)So, the equilibrium points are when dP/dt = 0, which is at P = 0 and P = 0.4 / 0.005 = 80. So, the carrying capacity in this modified logistic equation is 80, not 100. That makes sense because the term -cP reduces the effective carrying capacity.So, even though L was given as 100, the presence of the -cP term changes the carrying capacity to a lower value, which is 80 in this case.Therefore, the solution P(t) approaches 80 as t approaches infinity, which is consistent with our result.So, Sub-problem 1 is solved, and the solution is:P(t) = 80 / (1 + 7 e^{-t/2.5})Now, moving on to Sub-problem 2: Determine the time t when P(t) reaches 90% of L, which is 90% of 100, so 90 units per hour.Wait, but in our solution, the carrying capacity is 80, so 90% of L (which is 100) is 90, but our P(t) approaches 80. So, is it possible for P(t) to reach 90? Since the carrying capacity is 80, P(t) can't exceed 80. So, maybe there's a misunderstanding here.Wait, let me check the original problem statement again.The problem says: "determine the time t when the productivity P(t) reaches 90% of the carrying capacity L."But in our case, L is 100, but the effective carrying capacity due to the equation is 80. So, 90% of L is 90, but our P(t) can't reach 90 because it asymptotically approaches 80. So, maybe the question is referring to 90% of the effective carrying capacity, which is 80. So, 90% of 80 is 72.Alternatively, perhaps I made a mistake in interpreting L. Let me go back.Wait, in the original differential equation, L is the carrying capacity, but in the equation, after simplifying, the carrying capacity became 80. So, maybe the question is using L as 100, so 90% of 100 is 90. But since our solution can't reach 90, maybe I need to reconsider.Wait, perhaps I made a mistake in the partial fractions or the integration. Let me double-check.Original DE:dP/dt = 0.4P - 0.005P^2Which is dP/dt = P(0.4 - 0.005P)So, the equilibrium points are P=0 and P=80. So, the carrying capacity is 80, not 100. So, 90% of L (100) is 90, but our P(t) can't reach 90. So, perhaps the question is expecting 90% of the effective carrying capacity, which is 80, so 72.Alternatively, maybe I misapplied the constants. Let me check the original equation again.Original DE:dP/dt = kP(1 - P/L) - cPWith k=0.5, L=100, c=0.1.So, substituting:dP/dt = 0.5P(1 - P/100) - 0.1PWhich is 0.5P - 0.005P^2 - 0.1P = 0.4P - 0.005P^2Yes, that's correct. So, the carrying capacity is 80, not 100. So, 90% of L (100) is 90, but since P(t) approaches 80, it can't reach 90. So, perhaps the question is misworded, or maybe I need to interpret it differently.Wait, maybe the question is referring to 90% of the effective carrying capacity, which is 80. So, 90% of 80 is 72. So, perhaps the question wants the time when P(t) reaches 72.Alternatively, maybe I need to consider that the carrying capacity is still 100, but the equation is modified. Wait, no, in the equation, the carrying capacity is determined by the equilibrium point, which is when dP/dt = 0, so P=0 or P=80.So, perhaps the question is expecting 90% of 80, which is 72. Let me proceed with that.So, set P(t) = 72 and solve for t.Given P(t) = 80 / (1 + 7 e^{-t/2.5})Set 72 = 80 / (1 + 7 e^{-t/2.5})Multiply both sides by (1 + 7 e^{-t/2.5}):72(1 + 7 e^{-t/2.5}) = 80Divide both sides by 72:1 + 7 e^{-t/2.5} = 80 / 72 = 10 / 9 ≈ 1.1111Subtract 1:7 e^{-t/2.5} = 10/9 - 1 = 1/9Divide both sides by 7:e^{-t/2.5} = (1/9) / 7 = 1 / 63Take natural logarithm of both sides:- t / 2.5 = ln(1/63) = -ln(63)Multiply both sides by -1:t / 2.5 = ln(63)So,t = 2.5 ln(63)Calculate ln(63):ln(63) ≈ 4.1431So,t ≈ 2.5 * 4.1431 ≈ 10.3578 monthsSo, approximately 10.36 months.But let me verify the steps again.Starting from P(t) = 80 / (1 + 7 e^{-t/2.5})Set P(t) = 72:72 = 80 / (1 + 7 e^{-t/2.5})Multiply both sides by denominator:72(1 + 7 e^{-t/2.5}) = 8072 + 504 e^{-t/2.5} = 80Subtract 72:504 e^{-t/2.5} = 8Divide by 504:e^{-t/2.5} = 8 / 504 = 1 / 63Yes, that's correct.So, t = 2.5 ln(63) ≈ 2.5 * 4.1431 ≈ 10.3578 months.So, approximately 10.36 months.Alternatively, if the question insists on 90% of L=100, which is 90, but since our P(t) can't reach 90, maybe we need to consider that. But in that case, the equation would have no solution because P(t) approaches 80. So, perhaps the question is referring to 90% of the effective carrying capacity, which is 80, so 72.Alternatively, maybe I made a mistake in the solution, and the carrying capacity is still 100. Let me check the original DE again.Original DE:dP/dt = kP(1 - P/L) - cPWith k=0.5, L=100, c=0.1.So, dP/dt = 0.5P(1 - P/100) - 0.1P= 0.5P - 0.005P^2 - 0.1P= 0.4P - 0.005P^2So, the equilibrium points are P=0 and P=80, as before.So, the carrying capacity is indeed 80, not 100. So, 90% of L=100 is 90, which is beyond the carrying capacity. So, the productivity can't reach 90. Therefore, perhaps the question is referring to 90% of the effective carrying capacity, which is 80, so 72.Alternatively, maybe the question is misworded, and they meant 90% of the effective carrying capacity, which is 80. So, 72.Alternatively, perhaps I made a mistake in solving the DE. Let me check again.Wait, another approach: maybe I can write the DE in terms of the original L=100 and see if the solution can reach 90.But from the DE, the equilibrium is at P=80, so P(t) can't exceed 80. So, 90 is beyond that. Therefore, the time when P(t) reaches 90 doesn't exist. So, perhaps the question is expecting 90% of the effective carrying capacity, which is 72.Alternatively, maybe I made a mistake in the partial fractions or the integration. Let me check.Wait, when I did the partial fractions, I had:1 / [P(0.4 - 0.005P)] = A/P + B/(0.4 - 0.005P)Then, 1 = A(0.4 - 0.005P) + BPThen, 1 = 0.4A + P(-0.005A + B)So, 0.4A = 1 => A=2.5Then, -0.005A + B = 0 => B=0.005A=0.0125So, that's correct.Then, integrating:Integral of [2.5/P + 0.0125/(0.4 - 0.005P)] dP = integral dtWhich became:2.5 ln P - 2.5 ln(80 - P) = t + CWait, hold on, when I did the substitution for the second integral, I think I might have made a mistake.Wait, the second integral was:Integral of [0.0125 / (0.4 - 0.005P)] dPI rewrote 0.4 - 0.005P as 0.005(80 - P), so:0.0125 / (0.005(80 - P)) = 2.5 / (80 - P)So, the integral becomes:Integral of [2.5 / (80 - P)] dPLet u = 80 - P, du = -dPSo, integral becomes:-2.5 Integral of [1/u] du = -2.5 ln|u| + C = -2.5 ln|80 - P| + CSo, that part is correct.So, the left side integral is:2.5 ln P - 2.5 ln(80 - P) + C = t + C'Which simplifies to:2.5 ln(P / (80 - P)) = t + CSo, exponentiate both sides:P / (80 - P) = e^{(t + C)/2.5} = K e^{t/2.5}, where K = e^{C/2.5}Then, solving for P:P = K e^{t/2.5} (80 - P)P = 80 K e^{t/2.5} - K e^{t/2.5} PBring terms with P to the left:P + K e^{t/2.5} P = 80 K e^{t/2.5}Factor P:P (1 + K e^{t/2.5}) = 80 K e^{t/2.5}So,P = [80 K e^{t/2.5}] / [1 + K e^{t/2.5}]Which can be written as:P = 80 / [1 + (1/K) e^{-t/2.5}]Then, applying initial condition P(0) = 10:10 = 80 / [1 + (1/K)]So,10(1 + 1/K) = 8010 + 10/K = 8010/K = 70So,K = 10 / 70 = 1/7Therefore,P(t) = 80 / [1 + 7 e^{-t/2.5}]Yes, that's correct.So, the solution is correct, and the carrying capacity is 80. Therefore, 90% of L=100 is 90, which is beyond the carrying capacity, so P(t) can't reach 90. Therefore, perhaps the question is referring to 90% of the effective carrying capacity, which is 80, so 72.Alternatively, maybe the question is expecting to use L=100 in the solution, but that would require a different approach.Wait, perhaps I can consider the original DE without simplifying, but that would complicate things. Alternatively, maybe I can adjust the equation to have L=100 as the carrying capacity despite the -cP term.Wait, let me think about that. The standard logistic equation with a harvesting term is:dP/dt = rP(1 - P/K) - hWhere h is the harvesting rate. In our case, it's dP/dt = kP(1 - P/L) - cP, which is similar but with harvesting proportional to P, not constant.In such cases, the equilibrium points are found by setting dP/dt = 0:0 = kP(1 - P/L) - cPFactor P:0 = P [k(1 - P/L) - c]So, P=0 or k(1 - P/L) - c = 0Solving for P:k(1 - P/L) = c1 - P/L = c/kP/L = 1 - c/kSo,P = L(1 - c/k)Given k=0.5, c=0.1, L=100:P = 100(1 - 0.1/0.5) = 100(1 - 0.2) = 100*0.8 = 80So, that's consistent with our earlier result. So, the carrying capacity is indeed 80, not 100. Therefore, 90% of L=100 is 90, which is beyond the carrying capacity, so P(t) can't reach 90. Therefore, the time when P(t) reaches 90 doesn't exist.But the question says: "determine the time t when the productivity P(t) reaches 90% of the carrying capacity L."Wait, but L is 100, so 90% of L is 90. But since P(t) can't reach 90, maybe the question is expecting to use the effective carrying capacity, which is 80, so 90% of 80 is 72.Alternatively, perhaps the question is misworded, and they meant 90% of the effective carrying capacity. So, I think I should proceed with 72.So, solving for t when P(t) = 72:72 = 80 / (1 + 7 e^{-t/2.5})Multiply both sides by denominator:72(1 + 7 e^{-t/2.5}) = 8072 + 504 e^{-t/2.5} = 80Subtract 72:504 e^{-t/2.5} = 8Divide by 504:e^{-t/2.5} = 8 / 504 = 1 / 63Take natural log:- t / 2.5 = ln(1/63) = -ln(63)Multiply both sides by -1:t / 2.5 = ln(63)So,t = 2.5 ln(63)Calculate ln(63):ln(63) ≈ 4.1431So,t ≈ 2.5 * 4.1431 ≈ 10.3578 monthsSo, approximately 10.36 months.Therefore, the time when P(t) reaches 90% of the carrying capacity (assuming 90% of the effective carrying capacity, which is 80, so 72) is approximately 10.36 months.Alternatively, if the question insists on 90% of L=100, which is 90, then P(t) can't reach 90, so the time is undefined. But since the question asks for 90% of L, which is 100, and given that the solution approaches 80, it's impossible. Therefore, perhaps the question is referring to 90% of the effective carrying capacity, which is 80, so 72.So, I think the answer is approximately 10.36 months.Let me just recap:Sub-problem 1: Solved the DE and found P(t) = 80 / (1 + 7 e^{-t/2.5})Sub-problem 2: Found t when P(t) = 72, which is 90% of the effective carrying capacity (80). The time is approximately 10.36 months.I think that's the correct approach.</think>