answer:<think>Okay, so I have this statistics problem here about female and male athletes' scores in a mixed-gender league. The goal is to challenge stereotypes by analyzing their performance. Let me try to figure out the two parts step by step.Starting with the first question: What's the probability that a randomly selected female athlete's score is higher than a randomly selected male athlete's score? Hmm, okay. Both distributions are normal, so I remember that when dealing with the difference between two normal variables, the result is also normal. Let me denote the female athlete's score as X and the male athlete's score as Y. So, X ~ N(μ_F, σ_F²) which is N(75, 10²), and Y ~ N(μ_M, σ_M²) which is N(70, 12²). I need to find P(X > Y). I think this can be transformed into finding the probability that X - Y > 0. Since X and Y are independent, the distribution of X - Y will also be normal. The mean of X - Y should be μ_F - μ_M, which is 75 - 70 = 5. The variance of X - Y is σ_F² + σ_M² because variances add when subtracting independent variables. So that's 10² + 12² = 100 + 144 = 244. Therefore, the standard deviation is sqrt(244). Let me compute that: sqrt(244) is approximately 15.62, but I'll keep it exact for now.So, X - Y ~ N(5, 244). To find P(X - Y > 0), I can standardize this. Let Z = (X - Y - 5)/sqrt(244). Then Z follows a standard normal distribution, N(0,1). So, P(X - Y > 0) = P(Z > (0 - 5)/sqrt(244)) = P(Z > -5/sqrt(244)). But since the standard normal distribution is symmetric, P(Z > -a) = P(Z < a). So, this is equal to Φ(5/sqrt(244)). Let me compute 5/sqrt(244). sqrt(244) is sqrt(4*61) = 2*sqrt(61), so 5/(2*sqrt(61)). That's approximately 5/(2*7.81) ≈ 5/15.62 ≈ 0.32. So, Φ(0.32). But the question asks for the answer in terms of Φ, so I don't need to compute the numerical value. So, the probability is Φ(5/sqrt(244)). Alternatively, since 5/sqrt(244) can be written as 5/sqrt(10² + 12²), which is 5/sqrt(244). So, I think that's the answer.Wait, let me double-check. The difference X - Y has mean 5 and standard deviation sqrt(244). So, the Z-score is (0 - 5)/sqrt(244) = -5/sqrt(244). So, P(Z > -5/sqrt(244)) is indeed Φ(5/sqrt(244)). Yeah, that makes sense.Moving on to the second question: She forms a mixed-gender team of 5 female and 5 male athletes selected randomly. What is the expected average score and the standard deviation of the average score?Okay, so the team has 10 athletes: 5 female and 5 male. The average score would be the total score divided by 10. So, I need to find E[average] and SD[average].First, the expected value. The expected total score is the sum of the expected scores of each athlete. Since each female athlete has an expected score of 75 and each male athlete has an expected score of 70, the total expected score is 5*75 + 5*70. Let me compute that: 5*75 is 375, and 5*70 is 350. So, total expected score is 375 + 350 = 725. Therefore, the expected average is 725 / 10 = 72.5.Now, the standard deviation of the average. Since the athletes are selected randomly, their scores are independent. The variance of the total score is the sum of the variances of each individual score. For each female athlete, the variance is 10² = 100, and for each male athlete, it's 12² = 144. So, the total variance is 5*100 + 5*144. Let me calculate that: 5*100 is 500, and 5*144 is 720. So, total variance is 500 + 720 = 1220. Therefore, the variance of the average score is 1220 divided by (10)^2, which is 1220 / 100 = 12.2. Hence, the standard deviation is sqrt(12.2). Let me compute that: sqrt(12.2) is approximately 3.49, but again, I can leave it as sqrt(12.2) or simplify it. Wait, 12.2 is 61/5, so sqrt(61/5) is sqrt(61)/sqrt(5). That's approximately 7.81/2.24 ≈ 3.49, which matches my earlier approximation. So, the standard deviation of the average is sqrt(12.2).Let me recap: The expected average is 72.5, and the standard deviation is sqrt(12.2). That seems right because when you average independent variables, the variance scales by 1/n, so the standard deviation scales by 1/sqrt(n). Wait, just to make sure, the total variance is 5*(10²) + 5*(12²) = 500 + 720 = 1220. Then, the average is total divided by 10, so the variance is 1220 / (10)^2 = 12.2. So, yes, that's correct.I think that's all. So, summarizing:1. The probability is Φ(5/sqrt(244)).2. The expected average is 72.5, and the standard deviation is sqrt(12.2).Final Answer1. The probability is boxed{Phileft(frac{5}{sqrt{244}}right)}.2. The expected average score is boxed{72.5} and the standard deviation of the average score is boxed{sqrt{12.2}}.</think>

answer:<think>Okay, so I have this problem about aviation safety, and I need to figure out two probabilities. Let me take it step by step.First, the problem is divided into two parts. The first part is about calculating the probability that the next 20 flights, each lasting 5 hours, will experience at least one turbulence event of the same magnitude. The second part is about determining the probability that a similar turbulence event in the future would exceed the 3.0g threshold, which could compromise the aircraft's structural integrity.Starting with the first part. The dataset says that turbulence events of the same magnitude occur approximately once every 1,000 flight hours. So, if I have 10,000 flight hours, there are about 10 such events. That gives me a rate of 1 event per 1,000 hours.Now, the investigator is looking at the next 20 flights, each lasting 5 hours. So, the total flight hours to consider are 20 times 5, which is 100 hours. So, in 100 flight hours, how likely is it that at least one turbulence event occurs?Hmm, this sounds like a Poisson probability problem. The Poisson distribution is used to model the number of events occurring in a fixed interval of time or space, given the average rate of occurrence. The formula for the Poisson probability is:P(k; λ) = (λ^k * e^(-λ)) / k!Where:- P(k; λ) is the probability of k events occurring,- λ is the average number of events in the interval,- e is the base of the natural logarithm.In this case, the average number of events (λ) in 100 hours would be 100 / 1000 = 0.1. So, λ = 0.1.We need the probability of at least one event, which is 1 minus the probability of zero events. So, P(at least 1) = 1 - P(0; 0.1).Calculating P(0; 0.1):P(0; 0.1) = (0.1^0 * e^(-0.1)) / 0! = (1 * e^(-0.1)) / 1 = e^(-0.1)I remember that e^(-0.1) is approximately 0.9048. So, P(0; 0.1) ≈ 0.9048.Therefore, P(at least 1) = 1 - 0.9048 = 0.0952, or about 9.52%.Wait, let me double-check. The rate is 1 per 1000 hours, so in 100 hours, it's 0.1. So, yes, the Poisson parameter λ is 0.1. The probability of zero events is e^(-0.1), which is roughly 0.9048. So, 1 - 0.9048 is indeed approximately 0.0952. So, about a 9.5% chance of experiencing at least one turbulence event in the next 20 flights.Okay, that seems reasonable. I think that's the answer for the first part.Now, moving on to the second part. During the turbulence event, the aircraft experienced a sudden vertical acceleration of 2.5g. The structural components are designed to withstand up to 3.0g. The vertical accelerations during turbulence are normally distributed with a mean of 2.0g and a standard deviation of 0.5g. I need to find the probability that a similar turbulence event would exceed 3.0g.So, this is a normal distribution problem. We have a mean (μ) of 2.0g and a standard deviation (σ) of 0.5g. We need to find P(X > 3.0), where X is the vertical acceleration.To find this probability, I can standardize the value and use the Z-table. The formula for Z-score is:Z = (X - μ) / σPlugging in the numbers:Z = (3.0 - 2.0) / 0.5 = 1.0 / 0.5 = 2.0So, Z = 2.0. Now, I need to find the probability that Z is greater than 2.0. From the standard normal distribution table, the area to the left of Z=2.0 is approximately 0.9772. Therefore, the area to the right (which is what we need) is 1 - 0.9772 = 0.0228, or 2.28%.Let me verify that. Yes, Z=2.0 corresponds to about 97.72% cumulative probability, so the tail probability is 2.28%. That seems correct.Alternatively, using a calculator or more precise Z-table, the exact value might be slightly different, but 2.28% is a commonly accepted approximation.So, the probability that a similar turbulence event would exceed 3.0g is approximately 2.28%.Putting it all together, the first probability is about 9.5%, and the second is about 2.28%.I think that's it. I don't see any mistakes in my calculations, but let me just recap.For the first part:- Total flight hours: 20 * 5 = 100 hours.- Turbulence rate: 1 per 1000 hours, so λ = 0.1.- Poisson probability of at least one event: 1 - e^(-0.1) ≈ 9.5%.For the second part:- Normal distribution with μ=2.0, σ=0.5.- Z-score for 3.0g is 2.0.- Probability above Z=2.0 is about 2.28%.Yes, that all adds up. I think I'm confident with these results.Final Answer1. The probability is boxed{0.0952}.2. The probability is boxed{0.0228}.</think>

answer:<think>Okay, so I have this problem about a record store owner and a graphic designer creating limited edition vinyl album covers. The cost for each design is given by the equation ( C = 50n + 25k^2 ), where ( n ) is the number of colors and ( k ) is the complexity. There are two sub-problems here. Let me tackle them one by one.Sub-problem 1:The store owner wants to produce 10 unique album covers with a total budget of 12,000. The complexity ( k ) is directly proportional to the number of colors ( n ), so ( k = cn ), where ( c ) is the proportionality constant we need to find. Also, we need to determine the number of colors ( n ) used in each design to maximize the number of unique designs within the budget.Alright, so first, let's break this down. The total cost for 10 album covers is 12,000. So, the cost per album cover is ( frac{12,000}{10} = 1,200 ) dollars.Given that ( k = cn ), we can substitute this into the cost equation. So, the cost per design becomes:( C = 50n + 25k^2 = 50n + 25(cn)^2 = 50n + 25c^2n^2 ).Since each design costs 1,200, we have:( 50n + 25c^2n^2 = 1,200 ).But wait, is this per design or total? The total budget is 12,000 for 10 designs, so each design is 1,200. So yes, the equation above is correct.So, we have:( 25c^2n^2 + 50n - 1,200 = 0 ).This is a quadratic equation in terms of ( n ). But we have two variables here: ( c ) and ( n ). So, we need another equation or condition to solve for both.Wait, the problem says that ( k ) is directly proportional to ( n ), so ( k = cn ). That's the only relationship given. So, maybe we need to express ( c ) in terms of ( n ) or vice versa.But since we have only one equation and two variables, perhaps we need to find ( c ) such that the number of colors ( n ) is maximized? Or maybe the number of unique designs is maximized, but since the number of designs is fixed at 10, perhaps we need to maximize something else.Wait, the problem says: "determine the number of colors ( n ) used in each design to maximize the number of unique designs within the budget." Hmm, but the number of designs is fixed at 10. So maybe it's about maximizing the number of colors per design without exceeding the budget.Alternatively, perhaps the problem is to find ( c ) such that with the given budget, the number of colors ( n ) is as large as possible. Or maybe it's about ensuring that the total cost doesn't exceed the budget, so we need to find ( c ) such that ( n ) is feasible.Wait, maybe I need to approach it differently. Let's see.Total cost for 10 designs is 10*C = 12,000, so C = 1,200 per design.So, ( 50n + 25k^2 = 1,200 ).Given ( k = cn ), substitute:( 50n + 25c^2n^2 = 1,200 ).Let's write this as:( 25c^2n^2 + 50n - 1,200 = 0 ).This is a quadratic in ( n ):( 25c^2n^2 + 50n - 1,200 = 0 ).To solve for ( n ), we can use the quadratic formula:( n = frac{-b pm sqrt{b^2 - 4ac}}{2a} ).Where ( a = 25c^2 ), ( b = 50 ), and ( c = -1,200 ).Plugging in:( n = frac{-50 pm sqrt{50^2 - 4*25c^2*(-1,200)}}{2*25c^2} ).Simplify:( n = frac{-50 pm sqrt{2,500 + 120,000c^2}}{50c^2} ).Since ( n ) must be positive, we take the positive root:( n = frac{-50 + sqrt{2,500 + 120,000c^2}}{50c^2} ).Hmm, this seems a bit complicated. Maybe there's another approach.Alternatively, since we have ( k = cn ), we can express ( c = k/n ). So, substituting back into the cost equation:( 50n + 25k^2 = 1,200 ).But ( k = cn ), so:( 50n + 25c^2n^2 = 1,200 ).Let's factor out 25n:( 25n(2 + c^2n) = 1,200 ).So,( n(2 + c^2n) = 48 ).This is a quadratic in ( n ):( c^2n^2 + 2n - 48 = 0 ).Again, using quadratic formula:( n = frac{-2 pm sqrt{4 + 192c^2}}{2c^2} ).Simplify:( n = frac{-2 + sqrt{4 + 192c^2}}{2c^2} ).Since ( n ) must be positive, we take the positive root.But this still leaves us with ( c ) as a variable. So, perhaps we need to find ( c ) such that ( n ) is maximized? Or perhaps the problem is to find ( c ) such that the equation has a real solution for ( n ). But that might not be the case.Wait, maybe I misinterpreted the problem. It says, "find the proportionality constant ( c ) such that ( k = cn ), and determine the number of colors ( n ) used in each design to maximize the number of unique designs within the budget."Wait, the number of unique designs is fixed at 10. So, perhaps the goal is to maximize ( n ) given the budget constraint. So, we need to maximize ( n ) such that the total cost is within 12,000.So, perhaps we can express ( c ) in terms of ( n ) and then find the maximum ( n ) possible.From ( k = cn ), and the cost equation:( 50n + 25k^2 = 1,200 ).Substitute ( k = cn ):( 50n + 25c^2n^2 = 1,200 ).Let's solve for ( c^2 ):( 25c^2n^2 = 1,200 - 50n ).So,( c^2 = frac{1,200 - 50n}{25n^2} = frac{48 - 2n}{n^2} ).Therefore,( c = sqrt{frac{48 - 2n}{n^2}} = sqrt{frac{48 - 2n}{n^2}} = sqrt{frac{48}{n^2} - frac{2}{n}} ).But ( c ) must be a real number, so the expression under the square root must be non-negative:( frac{48 - 2n}{n^2} geq 0 ).Since ( n^2 ) is always positive, the numerator must be non-negative:( 48 - 2n geq 0 ).So,( 2n leq 48 ).Thus,( n leq 24 ).So, the maximum possible ( n ) is 24. But we need to check if this is feasible.If ( n = 24 ), then:( c^2 = frac{48 - 2*24}{24^2} = frac{48 - 48}{576} = 0 ).So, ( c = 0 ). But if ( c = 0 ), then ( k = 0 ), which would mean the design has no complexity, which might not make sense. So, perhaps ( n ) can't be 24 because that would require ( k = 0 ), which might not be practical.So, let's try to find the maximum ( n ) such that ( c ) is positive.Wait, but as ( n ) increases, ( c ) decreases. So, the maximum ( n ) before ( c ) becomes zero is 24, but as I said, that would make ( c = 0 ), which might not be acceptable.Alternatively, perhaps we need to find ( c ) such that ( n ) is as large as possible without making ( c ) negative or zero.Wait, maybe I'm overcomplicating. Let's think differently.We have:( 50n + 25k^2 = 1,200 ).And ( k = cn ).So, substituting:( 50n + 25c^2n^2 = 1,200 ).Let's divide both sides by 25:( 2n + c^2n^2 = 48 ).So,( c^2n^2 + 2n - 48 = 0 ).This is a quadratic in ( n ):( c^2n^2 + 2n - 48 = 0 ).To have real solutions for ( n ), the discriminant must be non-negative:( (2)^2 - 4*c^2*(-48) geq 0 ).So,( 4 + 192c^2 geq 0 ).Which is always true since ( c^2 ) is non-negative. So, for any ( c ), there is a solution for ( n ).But we need to find ( c ) such that ( n ) is maximized. Wait, but ( n ) is dependent on ( c ). So, perhaps we need to express ( n ) in terms of ( c ) and then find the maximum ( n ).From the quadratic equation:( n = frac{-2 pm sqrt{4 + 192c^2}}{2c^2} ).Taking the positive root:( n = frac{-2 + sqrt{4 + 192c^2}}{2c^2} ).We can simplify this:( n = frac{-2 + 2sqrt{1 + 48c^2}}{2c^2} = frac{-1 + sqrt{1 + 48c^2}}{c^2} ).To maximize ( n ), we can take the derivative of ( n ) with respect to ( c ) and set it to zero.Let me denote ( n(c) = frac{-1 + sqrt{1 + 48c^2}}{c^2} ).Let me compute the derivative ( n'(c) ).First, let me write ( n(c) = frac{sqrt{1 + 48c^2} - 1}{c^2} ).Let me set ( f(c) = sqrt{1 + 48c^2} - 1 ) and ( g(c) = c^2 ), so ( n(c) = f(c)/g(c) ).Using the quotient rule:( n'(c) = frac{f'(c)g(c) - f(c)g'(c)}{[g(c)]^2} ).Compute ( f'(c) ):( f'(c) = frac{1}{2}(1 + 48c^2)^{-1/2} * 96c = frac{48c}{sqrt{1 + 48c^2}} ).Compute ( g'(c) = 2c ).So,( n'(c) = frac{left(frac{48c}{sqrt{1 + 48c^2}}right)c^2 - (sqrt{1 + 48c^2} - 1)(2c)}{c^4} ).Simplify numerator:First term: ( frac{48c^3}{sqrt{1 + 48c^2}} ).Second term: ( -2c(sqrt{1 + 48c^2} - 1) ).So, numerator:( frac{48c^3}{sqrt{1 + 48c^2}} - 2csqrt{1 + 48c^2} + 2c ).Let me factor out ( 2c ):( 2c left( frac{24c^2}{sqrt{1 + 48c^2}} - sqrt{1 + 48c^2} + 1 right) ).Set numerator equal to zero for critical points:( 2c left( frac{24c^2}{sqrt{1 + 48c^2}} - sqrt{1 + 48c^2} + 1 right) = 0 ).Since ( c neq 0 ) (otherwise ( k = 0 )), we have:( frac{24c^2}{sqrt{1 + 48c^2}} - sqrt{1 + 48c^2} + 1 = 0 ).Let me denote ( s = sqrt{1 + 48c^2} ). Then, ( s^2 = 1 + 48c^2 ), so ( c^2 = frac{s^2 - 1}{48} ).Substitute into the equation:( frac{24 * frac{s^2 - 1}{48}}{s} - s + 1 = 0 ).Simplify:( frac{(s^2 - 1)/2}{s} - s + 1 = 0 ).Which is:( frac{s^2 - 1}{2s} - s + 1 = 0 ).Multiply through by 2s to eliminate denominators:( (s^2 - 1) - 2s^2 + 2s = 0 ).Simplify:( s^2 - 1 - 2s^2 + 2s = 0 ).Combine like terms:( -s^2 + 2s - 1 = 0 ).Multiply both sides by -1:( s^2 - 2s + 1 = 0 ).This factors as:( (s - 1)^2 = 0 ).So, ( s = 1 ).But ( s = sqrt{1 + 48c^2} = 1 ).So,( sqrt{1 + 48c^2} = 1 ).Square both sides:( 1 + 48c^2 = 1 ).Thus,( 48c^2 = 0 ).So, ( c = 0 ).But ( c = 0 ) would mean ( k = 0 ), which we already saw leads to ( n = 24 ), but that's not practical because complexity can't be zero. So, perhaps the maximum ( n ) occurs at ( c ) approaching zero, but in reality, ( c ) must be positive.Alternatively, maybe the maximum ( n ) is achieved when ( c ) is as small as possible, but that might not make sense in the context of the problem.Wait, perhaps I made a mistake in the derivative approach. Maybe instead of maximizing ( n ), we need to find ( c ) such that the total cost is exactly 12,000 for 10 designs, and then find ( n ) accordingly.Wait, the problem says: "find the proportionality constant ( c ) such that ( k = cn ), and determine the number of colors ( n ) used in each design to maximize the number of unique designs within the budget."Wait, the number of unique designs is fixed at 10, so perhaps the goal is to maximize ( n ) given the budget constraint. So, we need to find the maximum ( n ) such that the total cost for 10 designs is 12,000.So, let's go back to the equation:( 50n + 25k^2 = 1,200 ).With ( k = cn ), so:( 50n + 25c^2n^2 = 1,200 ).We can write this as:( 25c^2n^2 + 50n - 1,200 = 0 ).Let me divide both sides by 25:( c^2n^2 + 2n - 48 = 0 ).This is a quadratic in ( n ):( c^2n^2 + 2n - 48 = 0 ).To solve for ( n ):( n = frac{-2 pm sqrt{4 + 192c^2}}{2c^2} ).Taking the positive root:( n = frac{-2 + sqrt{4 + 192c^2}}{2c^2} ).We can factor out 4 under the square root:( n = frac{-2 + 2sqrt{1 + 48c^2}}{2c^2} = frac{-1 + sqrt{1 + 48c^2}}{c^2} ).Now, to maximize ( n ), we can consider ( c ) as a variable and find the value of ( c ) that maximizes ( n ).Let me denote ( f(c) = frac{-1 + sqrt{1 + 48c^2}}{c^2} ).To find the maximum, take the derivative ( f'(c) ) and set it to zero.Compute ( f'(c) ):Let me write ( f(c) = frac{sqrt{1 + 48c^2} - 1}{c^2} ).Let me set ( u = sqrt{1 + 48c^2} ), so ( u = (1 + 48c^2)^{1/2} ).Then, ( f(c) = frac{u - 1}{c^2} ).Compute ( du/dc = frac{1}{2}(1 + 48c^2)^{-1/2} * 96c = frac{48c}{sqrt{1 + 48c^2}} ).Now, using the quotient rule for ( f(c) ):( f'(c) = frac{(du/dc)*c^2 - (u - 1)*2c}{c^4} ).Substitute ( du/dc ):( f'(c) = frac{left(frac{48c}{sqrt{1 + 48c^2}}right)c^2 - (u - 1)2c}{c^4} ).Simplify numerator:First term: ( frac{48c^3}{sqrt{1 + 48c^2}} ).Second term: ( -2c(u - 1) ).So, numerator:( frac{48c^3}{sqrt{1 + 48c^2}} - 2c(u - 1) ).Factor out 2c:( 2cleft(frac{24c^2}{sqrt{1 + 48c^2}} - (u - 1)right) ).Set numerator equal to zero:( 2cleft(frac{24c^2}{sqrt{1 + 48c^2}} - (u - 1)right) = 0 ).Since ( c neq 0 ), we have:( frac{24c^2}{sqrt{1 + 48c^2}} - (u - 1) = 0 ).But ( u = sqrt{1 + 48c^2} ), so:( frac{24c^2}{sqrt{1 + 48c^2}} - (sqrt{1 + 48c^2} - 1) = 0 ).Let me denote ( s = sqrt{1 + 48c^2} ), so ( s^2 = 1 + 48c^2 ), which implies ( c^2 = frac{s^2 - 1}{48} ).Substitute into the equation:( frac{24 * frac{s^2 - 1}{48}}{s} - (s - 1) = 0 ).Simplify:( frac{(s^2 - 1)/2}{s} - s + 1 = 0 ).Which is:( frac{s^2 - 1}{2s} - s + 1 = 0 ).Multiply through by 2s:( s^2 - 1 - 2s^2 + 2s = 0 ).Combine like terms:( -s^2 + 2s - 1 = 0 ).Multiply by -1:( s^2 - 2s + 1 = 0 ).This factors as:( (s - 1)^2 = 0 ).So, ( s = 1 ).But ( s = sqrt{1 + 48c^2} = 1 ), so:( 1 + 48c^2 = 1 ).Thus,( 48c^2 = 0 ).So, ( c = 0 ).But if ( c = 0 ), then ( k = 0 ), which would mean the cost equation becomes:( 50n = 1,200 ).So, ( n = 24 ).But as I thought earlier, ( k = 0 ) might not be practical because complexity can't be zero. So, perhaps the maximum ( n ) is 24, but with ( k = 0 ), which might not be acceptable.Alternatively, maybe the problem expects us to find ( c ) such that ( n ) is as large as possible without ( k ) being zero. But since ( c ) can be any positive number, as ( c ) approaches zero, ( n ) approaches 24. So, practically, the maximum ( n ) is 24, but with ( k ) approaching zero.But in the context of the problem, maybe ( k ) must be at least some minimum value. Since the problem doesn't specify, perhaps we can accept ( c = 0 ) and ( n = 24 ).Wait, but if ( c = 0 ), then ( k = 0 ), which would make the cost equation ( 50n = 1,200 ), so ( n = 24 ). That seems to fit.But let me check if this makes sense. If each design uses 24 colors and has zero complexity, the cost per design is ( 50*24 + 25*0 = 1,200 ), which is correct. So, total cost for 10 designs is 12,000, which fits the budget.But is zero complexity acceptable? The problem says the graphic designer can produce a unique design for each album cover, but it doesn't specify that complexity must be positive. So, perhaps ( c = 0 ) is acceptable, and ( n = 24 ).But wait, the problem says "complexity ( k ) of each design is directly proportional to the number of colors ( n ) used". If ( k = 0 ), then ( c = 0 ), which is a proportionality constant. So, it's acceptable.Therefore, the proportionality constant ( c ) is 0, and the number of colors ( n ) is 24.But wait, that seems counterintuitive because complexity being zero would mean very simple designs, but perhaps it's allowed.Alternatively, maybe I made a mistake in the approach. Let me think again.We have:Total cost for 10 designs: 12,000.Cost per design: 1,200.Equation: ( 50n + 25k^2 = 1,200 ).With ( k = cn ).So, substituting:( 50n + 25c^2n^2 = 1,200 ).Let me rearrange:( 25c^2n^2 + 50n - 1,200 = 0 ).Divide by 25:( c^2n^2 + 2n - 48 = 0 ).This is a quadratic in ( n ):( c^2n^2 + 2n - 48 = 0 ).To solve for ( n ):( n = frac{-2 pm sqrt{4 + 192c^2}}{2c^2} ).Taking the positive root:( n = frac{-2 + sqrt{4 + 192c^2}}{2c^2} ).Now, to maximize ( n ), we can consider ( c ) as a variable and find the value of ( c ) that maximizes ( n ).But as ( c ) approaches zero, ( n ) approaches ( frac{-2 + sqrt{4}}{0} ), which is undefined. Wait, no, let me compute the limit as ( c ) approaches zero.As ( c to 0 ), ( sqrt{4 + 192c^2} approx 2 + frac{192c^2}{4} = 2 + 48c^2 ).So,( n approx frac{-2 + 2 + 48c^2}{2c^2} = frac{48c^2}{2c^2} = 24 ).So, as ( c ) approaches zero, ( n ) approaches 24.Therefore, the maximum ( n ) is 24, achieved when ( c = 0 ).But since ( c = 0 ) implies ( k = 0 ), which might not be practical, perhaps the problem expects ( c ) to be non-zero. But the problem doesn't specify any constraints on ( c ) or ( k ), so mathematically, ( c = 0 ) and ( n = 24 ) is the solution.Alternatively, maybe I misinterpreted the problem. Perhaps the goal is to find ( c ) such that the total cost is exactly 12,000, and then find ( n ). But since ( c ) is a proportionality constant, it's a fixed value, not a variable. So, perhaps we need to find ( c ) such that the equation ( 50n + 25c^2n^2 = 1,200 ) has a solution for ( n ), and then find ( n ).Wait, but without additional constraints, there are infinitely many solutions for ( c ) and ( n ). So, perhaps the problem is to find ( c ) such that the number of colors ( n ) is maximized, which we've determined is 24 when ( c = 0 ).Alternatively, maybe the problem is to find ( c ) such that the number of designs is maximized, but the number of designs is fixed at 10.Wait, perhaps I'm overcomplicating. Let me try to solve for ( c ) and ( n ) such that the total cost is 12,000 for 10 designs, with ( k = cn ).So, per design cost: 1,200.Equation: ( 50n + 25k^2 = 1,200 ).With ( k = cn ), so:( 50n + 25c^2n^2 = 1,200 ).Let me express this as:( 25c^2n^2 + 50n - 1,200 = 0 ).Divide by 25:( c^2n^2 + 2n - 48 = 0 ).This is a quadratic in ( n ):( c^2n^2 + 2n - 48 = 0 ).To solve for ( n ):( n = frac{-2 pm sqrt{4 + 192c^2}}{2c^2} ).Taking the positive root:( n = frac{-2 + sqrt{4 + 192c^2}}{2c^2} ).Now, to maximize ( n ), we can consider ( c ) as a variable and find the value of ( c ) that maximizes ( n ).But as ( c ) approaches zero, ( n ) approaches 24, as we saw earlier.But if ( c ) is non-zero, ( n ) will be less than 24.Therefore, the maximum ( n ) is 24, achieved when ( c = 0 ).So, the proportionality constant ( c ) is 0, and the number of colors ( n ) is 24.But let me check if this makes sense. If ( c = 0 ), then ( k = 0 ), so the cost per design is ( 50*24 + 25*0 = 1,200 ), which is correct. So, total cost for 10 designs is 12,000, which fits the budget.Therefore, the answer for Sub-problem 1 is ( c = 0 ) and ( n = 24 ).Sub-problem 2:Given the number of colors ( n ) found in Sub-problem 1, which is 24, determine the complexity ( k ) that maximizes the market value ( V ), which is modeled by ( V = 1000 + 75n - 10k^2 ).So, ( V = 1000 + 75*24 - 10k^2 ).Compute ( 75*24 ):75*24 = 1,800.So,( V = 1000 + 1,800 - 10k^2 = 2,800 - 10k^2 ).To maximize ( V ), we need to minimize ( k^2 ), which is achieved when ( k = 0 ).So, the complexity ( k ) that maximizes ( V ) is 0.But wait, in Sub-problem 1, we found ( c = 0 ), so ( k = 0 ). So, in this case, ( k = 0 ) is already the value that maximizes ( V ).Therefore, the complexity ( k ) that maximizes the market value ( V ) is 0.But let me double-check. The market value function is ( V = 1000 + 75n - 10k^2 ). Since ( n ) is fixed at 24, ( V ) is a quadratic function in ( k ) with a negative coefficient on ( k^2 ), so it opens downward, meaning the maximum occurs at the vertex.The vertex of ( V = -10k^2 + (1000 + 75*24) ) is at ( k = 0 ), since the linear term is zero. So, yes, ( k = 0 ) maximizes ( V ).Therefore, the answer for Sub-problem 2 is ( k = 0 ).But wait, in Sub-problem 1, we had ( k = 0 ) because ( c = 0 ). So, in this case, the complexity is already set to zero, which also maximizes the market value. So, it's consistent.But let me think again. If in Sub-problem 1, we set ( c = 0 ) to maximize ( n ), then in Sub-problem 2, with ( n = 24 ), the maximum ( V ) occurs at ( k = 0 ), which is already the case. So, no further adjustment is needed.Alternatively, if in Sub-problem 1, we had a different ( c ), then in Sub-problem 2, we might have a different ( k ). But in this case, since ( c = 0 ), ( k = 0 ) is already optimal for both cost and market value.So, to summarize:Sub-problem 1: ( c = 0 ), ( n = 24 ).Sub-problem 2: ( k = 0 ).But let me check if the problem expects ( k ) to be non-zero. The market value function is ( V = 1000 + 75n - 10k^2 ). So, higher ( n ) increases ( V ), but higher ( k ) decreases ( V ). So, to maximize ( V ), we need to maximize ( n ) and minimize ( k ).In Sub-problem 1, we maximized ( n ) by setting ( c = 0 ), which also set ( k = 0 ), which in turn maximizes ( V ). So, it's all consistent.Therefore, the answers are:Sub-problem 1: ( c = 0 ), ( n = 24 ).Sub-problem 2: ( k = 0 ).But let me make sure I didn't make a mistake in Sub-problem 1. If ( c = 0 ), then ( k = 0 ), which makes the cost equation ( 50n = 1,200 ), so ( n = 24 ). That's correct.Alternatively, if ( c ) is non-zero, then ( n ) would be less than 24, which would allow for a higher ( k ), but that would decrease ( V ). So, to maximize ( V ), we need to maximize ( n ) and minimize ( k ), which is achieved when ( c = 0 ) and ( k = 0 ).Therefore, the final answers are:Sub-problem 1: ( c = 0 ), ( n = 24 ).Sub-problem 2: ( k = 0 ).</think>

answer:<think>Okay, so I have this problem about analyzing the vulnerabilities of an AI model's neural network. It's divided into two parts. Let me start with the first one.Problem 1: Average Degree of the NetworkWe're told that the network has an exponential degree distribution, which is given by ( P(k) = C e^{-lambda k} ). Here, ( C ) is a normalization constant, and ( lambda ) is a positive constant. The task is to find the average degree ( langle k rangle ).Hmm, okay. I remember that for a probability distribution, the average or expected value ( langle k rangle ) is calculated as the sum over all possible values of ( k ) multiplied by their probabilities. So, in this case, it should be:[langle k rangle = sum_{k=0}^{infty} k P(k)]But wait, in a neural network, the degree ( k ) can't be zero because each neuron must have at least some connections, right? Or can it? Hmm, actually, in some models, nodes can have zero connections, but in a neural network, I think each neuron must have at least some inputs or outputs. Maybe in this case, ( k ) starts from 1? Or does it start from 0? The problem doesn't specify, so I might have to assume it starts from 0.But let's see. The exponential distribution is often used for continuous variables, but here we're dealing with discrete degrees. So, it's a discrete probability distribution where ( P(k) = C e^{-lambda k} ) for ( k = 0, 1, 2, ldots ). First, I need to find the normalization constant ( C ). Since it's a probability distribution, the sum over all ( k ) must equal 1:[sum_{k=0}^{infty} P(k) = 1 implies sum_{k=0}^{infty} C e^{-lambda k} = 1]This is a geometric series where each term is ( e^{-lambda} ) times the previous term. The sum of a geometric series ( sum_{k=0}^{infty} r^k = frac{1}{1 - r} ) when ( |r| < 1 ). Here, ( r = e^{-lambda} ), and since ( lambda > 0 ), ( e^{-lambda} < 1 ). So,[C sum_{k=0}^{infty} e^{-lambda k} = C cdot frac{1}{1 - e^{-lambda}} = 1]Therefore, solving for ( C ):[C = 1 - e^{-lambda}]Wait, let me check that. If ( sum_{k=0}^{infty} e^{-lambda k} = frac{1}{1 - e^{-lambda}} ), then ( C ) must be ( 1 - e^{-lambda} ) to make the sum equal to 1. Yeah, that seems right.Now, moving on to the average degree ( langle k rangle ):[langle k rangle = sum_{k=0}^{infty} k P(k) = sum_{k=0}^{infty} k C e^{-lambda k}]Substituting ( C = 1 - e^{-lambda} ):[langle k rangle = (1 - e^{-lambda}) sum_{k=0}^{infty} k e^{-lambda k}]Hmm, I need to compute this sum. I recall that the sum ( sum_{k=0}^{infty} k r^k = frac{r}{(1 - r)^2} ) for ( |r| < 1 ). Let me verify that.Yes, the formula is correct. So, in this case, ( r = e^{-lambda} ). Therefore,[sum_{k=0}^{infty} k e^{-lambda k} = frac{e^{-lambda}}{(1 - e^{-lambda})^2}]Therefore, substituting back into ( langle k rangle ):[langle k rangle = (1 - e^{-lambda}) cdot frac{e^{-lambda}}{(1 - e^{-lambda})^2} = frac{e^{-lambda}}{1 - e^{-lambda}}]Simplify that:[langle k rangle = frac{1}{e^{lambda} - 1}]Wait, let me double-check the algebra. Starting from:[langle k rangle = (1 - e^{-lambda}) cdot frac{e^{-lambda}}{(1 - e^{-lambda})^2}]Simplify numerator and denominator:The ( (1 - e^{-lambda}) ) in the numerator cancels with one in the denominator, leaving:[langle k rangle = frac{e^{-lambda}}{1 - e^{-lambda}} = frac{1}{e^{lambda} - 1}]Yes, that looks correct. So, the average degree is ( frac{1}{e^{lambda} - 1} ).But wait, another thought: in neural networks, nodes can't have zero connections because they need to receive inputs or send outputs. So, maybe the degree starts at 1 instead of 0. If that's the case, the sum would start at ( k = 1 ). Let me see if that affects the result.If ( k ) starts at 1, then the normalization constant would be:[sum_{k=1}^{infty} C e^{-lambda k} = C cdot frac{e^{-lambda}}{1 - e^{-lambda}}} = 1 implies C = frac{1 - e^{-lambda}}{e^{-lambda}}]Wait, that would complicate things. But the problem didn't specify whether ( k ) starts at 0 or 1. It just says the probability ( P(k) ) is given by that expression. So, unless specified otherwise, I think it's safer to assume that ( k ) starts at 0. Therefore, I think my original calculation is correct, and the average degree is ( frac{1}{e^{lambda} - 1} ).Problem 2: Expected Risk of FailureNow, the second part is about calculating the expected risk of failure ( R ) when nodes are targeted. The removal of a node with degree ( k ) results in failure with probability proportional to ( k^2 ). So, the failure probability is ( p(k) = c k^2 ), where ( c ) is a proportionality constant.But wait, actually, the problem says the failure probability is proportional to ( k^2 ). So, perhaps it's ( p(k) = c k^2 ), but we need to ensure that the total probability is 1? Or is it just proportional, so we don't need to normalize?Wait, no. The problem says the failure probability is proportional to ( k^2 ). So, the risk is the expectation of ( k^2 ) over the degree distribution. So, the expected risk ( R ) is:[R = sum_{k=0}^{infty} P(k) cdot k^2]Because the failure probability is proportional to ( k^2 ), so we can consider ( R ) as the expectation of ( k^2 ).So, we need to compute ( langle k^2 rangle ), which is:[langle k^2 rangle = sum_{k=0}^{infty} k^2 P(k) = sum_{k=0}^{infty} k^2 C e^{-lambda k}]Again, ( C = 1 - e^{-lambda} ). So,[langle k^2 rangle = (1 - e^{-lambda}) sum_{k=0}^{infty} k^2 e^{-lambda k}]I need to compute this sum. I remember that the sum ( sum_{k=0}^{infty} k^2 r^k ) has a formula. Let me recall.Yes, the formula is:[sum_{k=0}^{infty} k^2 r^k = frac{r(1 + r)}{(1 - r)^3}]Let me verify this. For ( |r| < 1 ), the sum can be derived by differentiating the generating function.Alternatively, I can derive it. Let me consider ( S = sum_{k=0}^{infty} r^k = frac{1}{1 - r} ).First derivative: ( S' = sum_{k=0}^{infty} k r^{k - 1} = frac{1}{(1 - r)^2} implies sum_{k=0}^{infty} k r^k = frac{r}{(1 - r)^2} ).Second derivative: Differentiate ( S' ):( S'' = sum_{k=0}^{infty} k(k - 1) r^{k - 2} = frac{2}{(1 - r)^3} ).Multiply both sides by ( r^2 ):( sum_{k=0}^{infty} k(k - 1) r^k = frac{2 r^2}{(1 - r)^3} ).But ( k(k - 1) = k^2 - k ), so:( sum_{k=0}^{infty} k^2 r^k - sum_{k=0}^{infty} k r^k = frac{2 r^2}{(1 - r)^3} ).We already know ( sum_{k=0}^{infty} k r^k = frac{r}{(1 - r)^2} ), so:( sum_{k=0}^{infty} k^2 r^k = frac{2 r^2}{(1 - r)^3} + frac{r}{(1 - r)^2} ).Factor out ( frac{r}{(1 - r)^2} ):( frac{r}{(1 - r)^2} left( frac{2 r}{1 - r} + 1 right ) = frac{r}{(1 - r)^2} cdot left( frac{2 r + 1 - r}{1 - r} right ) = frac{r (1 + r)}{(1 - r)^3} ).Yes, that matches the formula I remembered. So,[sum_{k=0}^{infty} k^2 e^{-lambda k} = frac{e^{-lambda} (1 + e^{-lambda})}{(1 - e^{-lambda})^3}]Therefore, substituting back into ( langle k^2 rangle ):[langle k^2 rangle = (1 - e^{-lambda}) cdot frac{e^{-lambda} (1 + e^{-lambda})}{(1 - e^{-lambda})^3} = frac{e^{-lambda} (1 + e^{-lambda})}{(1 - e^{-lambda})^2}]Simplify numerator and denominator:First, note that ( 1 - e^{-lambda} ) is in the denominator squared, so:[langle k^2 rangle = frac{e^{-lambda} (1 + e^{-lambda})}{(1 - e^{-lambda})^2}]Let me write this in terms of ( e^{lambda} ) to make it look cleaner.Multiply numerator and denominator by ( e^{2lambda} ):Numerator: ( e^{-lambda} (1 + e^{-lambda}) cdot e^{2lambda} = e^{lambda} (1 + e^{-lambda}) = e^{lambda} + 1 ).Denominator: ( (1 - e^{-lambda})^2 cdot e^{2lambda} = (e^{lambda} - 1)^2 ).So,[langle k^2 rangle = frac{e^{lambda} + 1}{(e^{lambda} - 1)^2}]Alternatively, we can factor the numerator:( e^{lambda} + 1 = e^{lambda} + 1 ), which doesn't factor further. So, that's the expression.But let me check if that makes sense. When ( lambda ) is small, meaning the degrees are more spread out, the average degree ( langle k rangle ) is roughly ( frac{1}{lambda} ) (since ( e^{lambda} approx 1 + lambda )), so ( langle k rangle approx frac{1}{lambda} ). Similarly, ( langle k^2 rangle ) would be roughly ( frac{2}{lambda^2} ), which is consistent with the formula above because:( e^{lambda} approx 1 + lambda + frac{lambda^2}{2} ), so ( e^{lambda} + 1 approx 2 + lambda + frac{lambda^2}{2} ), and ( (e^{lambda} - 1)^2 approx (lambda + frac{lambda^2}{2})^2 approx lambda^2 + lambda^3 + frac{lambda^4}{4} ). So, ( langle k^2 rangle approx frac{2 + lambda}{lambda^2} approx frac{2}{lambda^2} + frac{1}{lambda} ), which is roughly ( frac{2}{lambda^2} ), as expected for the second moment.So, that seems reasonable.Therefore, the expected risk of failure ( R ) is ( langle k^2 rangle ), which is ( frac{e^{lambda} + 1}{(e^{lambda} - 1)^2} ).But wait, let me make sure. The problem says the failure probability is proportional to ( k^2 ). So, if we have a risk proportional to ( k^2 ), then the expected risk would be the expectation of ( k^2 ), which is ( langle k^2 rangle ). So, yes, that's correct.Alternatively, if the failure probability is ( c k^2 ), then the expected risk would be ( sum P(k) c k^2 = c langle k^2 rangle ). But since the problem says it's proportional, we can just express ( R ) as ( langle k^2 rangle ), considering the constant of proportionality is absorbed into ( R ). So, I think expressing ( R ) as ( frac{e^{lambda} + 1}{(e^{lambda} - 1)^2} ) is acceptable.Wait, but let me think again. If the failure probability is proportional to ( k^2 ), then the expected risk would be the expectation of ( k^2 ), which is ( langle k^2 rangle ). So, yes, that's exactly what we calculated.Therefore, the expected risk ( R ) is ( frac{e^{lambda} + 1}{(e^{lambda} - 1)^2} ).But let me see if I can simplify this expression further. Let me write it as:[R = frac{e^{lambda} + 1}{(e^{lambda} - 1)^2}]Alternatively, factor numerator and denominator:Note that ( e^{lambda} + 1 = e^{lambda} + 1 ), and ( (e^{lambda} - 1)^2 ) is already factored. Alternatively, we can write ( e^{lambda} + 1 = e^{lambda/2}(e^{lambda/2} + e^{-lambda/2}) ), but I don't know if that helps.Alternatively, express in terms of hyperbolic functions. Since ( e^{lambda} + 1 = 2 e^{lambda/2} cosh(lambda/2) ), and ( e^{lambda} - 1 = 2 e^{lambda/2} sinh(lambda/2) ). So,[R = frac{2 e^{lambda/2} cosh(lambda/2)}{(2 e^{lambda/2} sinh(lambda/2))^2} = frac{cosh(lambda/2)}{4 e^{lambda} sinh^2(lambda/2)}]Wait, that might complicate it more. Alternatively, perhaps leave it as is.Alternatively, express in terms of ( langle k rangle ). Since ( langle k rangle = frac{1}{e^{lambda} - 1} ), then ( e^{lambda} - 1 = frac{1}{langle k rangle} ). So,[R = frac{e^{lambda} + 1}{(e^{lambda} - 1)^2} = frac{e^{lambda} + 1}{left( frac{1}{langle k rangle} right)^2} = (e^{lambda} + 1) langle k rangle^2]But ( e^{lambda} = 1 + frac{1}{langle k rangle} ), so substituting back:[R = left(1 + frac{1}{langle k rangle} + 1 right) langle k rangle^2 = left(2 + frac{1}{langle k rangle} right) langle k rangle^2 = 2 langle k rangle^2 + langle k rangle]But that might not necessarily be simpler. So, perhaps it's better to leave it in terms of ( lambda ).Therefore, the expected risk ( R ) is ( frac{e^{lambda} + 1}{(e^{lambda} - 1)^2} ).Final Answer1. The average degree is boxed{dfrac{1}{e^{lambda} - 1}}.2. The expected risk of failure is boxed{dfrac{e^{lambda} + 1}{(e^{lambda} - 1)^2}}.</think>

answer:<think>Alright, so John has this old tractor he bought from his friend, and he wants to optimize his farming operations. There are two main problems here, both related to the tractor's efficiency and plowing time. Let me try to tackle them one by one.Starting with the first problem: calculating the total fuel consumed when using the tractor for 12 hours straight. The fuel efficiency is given by the function E(t) = 5 + sin(πt/4), where t is the number of hours the tractor has been running. Fuel efficiency here is in liters per hour, so I think that means the tractor consumes E(t) liters every hour. So, to find the total fuel consumed over 12 hours, I need to integrate E(t) from t=0 to t=12.Wait, is that right? Because fuel efficiency is usually liters per hour, so integrating E(t) over time would give the total fuel consumed. Yes, that makes sense. So, the integral of E(t) dt from 0 to 12 should give the total liters used.So, let me write that down:Total fuel = ∫₀¹² [5 + sin(πt/4)] dtI can split this integral into two parts:∫₀¹² 5 dt + ∫₀¹² sin(πt/4) dtCalculating the first integral:∫₀¹² 5 dt = 5t | from 0 to 12 = 5*12 - 5*0 = 60 litersNow, the second integral:∫₀¹² sin(πt/4) dtTo integrate sin(ax), the integral is -(1/a)cos(ax) + C. So here, a = π/4.So, integral becomes:- (4/π) cos(πt/4) | from 0 to 12Calculating at t=12:- (4/π) cos(π*12/4) = - (4/π) cos(3π) = - (4/π) * (-1) = 4/πAt t=0:- (4/π) cos(0) = - (4/π) * 1 = -4/πSo, subtracting:[4/π] - [-4/π] = 8/πSo, the second integral is 8/π liters.Therefore, total fuel consumed is 60 + 8/π liters.Calculating 8/π numerically: π is approximately 3.1416, so 8/3.1416 ≈ 2.5465 liters.So, total fuel ≈ 60 + 2.5465 ≈ 62.5465 liters.Hmm, so approximately 62.55 liters. But maybe I should keep it exact. So, 60 + 8/π liters. That would be the precise answer.Wait, let me double-check the integral:The integral of sin(πt/4) dt is indeed -4/π cos(πt/4). Plugging in 12 and 0:At t=12: cos(3π) = -1, so -4/π*(-1) = 4/πAt t=0: cos(0) = 1, so -4/π*(1) = -4/πSubtracting: 4/π - (-4/π) = 8/π. Yep, that's correct.So, total fuel is 60 + 8/π liters. If I need a decimal, it's about 62.55 liters.Alright, moving on to the second problem: calculating the total time to plow the field, including maintenance breaks.John's field is a rectangle, 300 meters by 200 meters. The tractor plows at 5 meters per minute. It must stop for maintenance every 3 hours, each maintenance taking 30 minutes.First, I need to find out how long it takes to plow the entire field, then add the maintenance time.But wait, the tractor plows at 5 meters per minute. Is that the speed? So, 5 meters per minute. So, in terms of area, how much area can it plow per minute?Wait, actually, the tractor's plowing speed is 5 meters per minute. But to find the total time, I need to know the total area and the rate at which the tractor plows.Wait, but the tractor's speed is 5 meters per minute. So, assuming it's plowing a row that's 5 meters long every minute? Or is it plowing width-wise?Wait, actually, the tractor's plowing speed is given as 5 meters per minute. So, if it's moving forward at 5 meters per minute, then in one minute, it can plow a strip of land 5 meters long (assuming the width is covered by the plow). But the field is 300 meters by 200 meters.Wait, maybe I need to think about the area. The field is 300m x 200m, so area is 60,000 square meters.If the tractor plows at 5 meters per minute, but how much area does it plow per minute? Hmm, perhaps I need more information. Wait, maybe the tractor plows a certain width as it moves forward.Wait, the problem doesn't specify the width of the plowed strip. Hmm, that's a problem. Maybe I need to assume that the tractor plows the entire width in one pass? Or perhaps it's moving along the length, plowing the width as it goes.Wait, the field is 300 meters by 200 meters. If the tractor is plowing along the length, which is 300 meters, then each pass would cover the entire 200 meters width? Or is it plowing 5 meters per minute along the length, so each minute it plows 5 meters of the 300-meter length, but the width is 200 meters.Wait, I think I need to clarify. The tractor's plowing speed is 5 meters per minute. So, if it's moving along the length, each minute it plows 5 meters of the 300-meter length, but the width plowed is the entire 200 meters? Or is the 5 meters per minute the forward speed, and the width is fixed?Wait, maybe the tractor plows a swath (a row) that is 5 meters wide as it moves forward. But the problem doesn't specify. Hmm.Wait, the problem says the tractor plows at a constant speed of 5 meters per minute. It doesn't specify the width, so perhaps we can assume that it's plowing the entire field in a single pass? That doesn't make sense because the field is 300m x 200m.Alternatively, maybe the tractor is moving along the length, which is 300 meters, and each pass covers the entire 200 meters width. But that would mean that each pass is 300 meters long, and the tractor plows 5 meters per minute, so each pass would take 300 / 5 = 60 minutes. But that seems like a lot.Wait, but if the tractor is moving along the length, plowing the entire width each time, then the number of passes needed would be the width divided by the swath width. But since the swath width isn't given, maybe we can assume that the tractor plows the entire width in one pass? That would mean only one pass is needed, but that seems unlikely because the field is 200 meters wide.Wait, perhaps the tractor plows a row that is 5 meters wide each time it goes forward 300 meters. So, each pass is 300 meters long, plowing 5 meters width. Then, the number of passes needed would be 200 / 5 = 40 passes.Each pass is 300 meters at 5 meters per minute, so time per pass is 300 / 5 = 60 minutes. So, total plowing time is 40 passes * 60 minutes = 2400 minutes.But wait, that seems like a lot. Let me check.Alternatively, maybe the tractor plows the entire 200 meters width in one pass, moving along the 300 meters length. So, each pass is 300 meters, plowing 200 meters width. So, the area plowed per pass is 300*200 = 60,000 square meters. But that's the entire field, so only one pass is needed. But that would mean the tractor plows the entire field in one pass, which is 300 meters at 5 meters per minute, so 60 minutes. But that seems too simplistic.Wait, perhaps the tractor plows a row that is 5 meters wide each time it goes along the 300 meters. So, each pass is 300 meters, plowing 5 meters width. So, total number of passes needed is 200 / 5 = 40 passes. Each pass takes 300 / 5 = 60 minutes. So, total plowing time is 40 * 60 = 2400 minutes.But that seems like a lot, and also, the tractor needs to stop for maintenance every 3 hours, which is 180 minutes. So, we need to calculate how many maintenance breaks are needed during the plowing.Wait, but if the total plowing time is 2400 minutes, and maintenance is every 3 hours (180 minutes), then the number of maintenance breaks is total plowing time divided by 180, but since you don't need a break after the last plowing session, it's (total plowing time / 180) - 1, but only if total plowing time is a multiple of 180. Otherwise, it's the integer division.Wait, let me think. If the tractor starts plowing, after every 3 hours, it stops for 30 minutes. So, the total time is plowing time plus the number of maintenance breaks times 30 minutes.But the number of maintenance breaks is equal to the number of times the tractor has been running for 3 hours. So, if the total plowing time is T, then the number of maintenance breaks is floor(T / 180). But actually, it's the number of 3-hour intervals completed, so if T is exactly divisible by 180, then it's T / 180 - 1, because the last interval doesn't require a break after. Hmm, this is getting a bit complicated.Alternatively, perhaps it's better to model it as: every 3 hours of plowing, there's a 30-minute break. So, the total time is plowing time + number_of_breaks * 30 minutes.Number of breaks is equal to the number of times the tractor has been running for 3 hours during the plowing. So, if the total plowing time is T, then the number of breaks is floor(T / 180). But if T is exactly divisible by 180, then it's T / 180 - 1, because the last 3 hours don't need a break after.Wait, let's take an example. If plowing time is 180 minutes (3 hours), then number of breaks is 0, because you plow for 3 hours, then you're done, no need for a break. If plowing time is 360 minutes (6 hours), then you have one break after the first 3 hours, so total breaks = 1. If plowing time is 540 minutes (9 hours), then breaks after 3 and 6 hours, so total breaks = 2.So, in general, number of breaks = floor(T / 180) if T is not a multiple of 180, else (T / 180) - 1.But in our case, if the total plowing time is 2400 minutes, let's see:2400 / 180 = 13.333... So, floor(13.333) = 13 breaks. But wait, that doesn't make sense because 13 breaks would mean 13*30 = 390 minutes of breaks, but the plowing time is 2400 minutes, which is 40 hours. 13 breaks would mean breaks after every 3 hours, but 13 breaks would cover 13*3 = 39 hours, but the total plowing time is 40 hours, so actually, the number of breaks should be 13, because after the 39th hour, you have a break, and then the last hour is plowing without a break.Wait, maybe it's better to think in terms of how many 3-hour intervals are there in the plowing time.If plowing time is T minutes, then the number of maintenance breaks is floor((T - 1) / 180). Because you don't need a break after the last interval.So, for T = 180, breaks = 0T = 360, breaks = 1T = 540, breaks = 2So, general formula: breaks = floor((T - 1) / 180)So, for T = 2400 minutes,breaks = floor((2400 - 1)/180) = floor(2399 / 180) ≈ floor(13.327) = 13 breaks.Each break is 30 minutes, so total break time = 13 * 30 = 390 minutes.Therefore, total time = plowing time + break time = 2400 + 390 = 2790 minutes.Convert that to hours: 2790 / 60 = 46.5 hours.Wait, but let me check if the plowing time is indeed 2400 minutes.Earlier, I assumed that the tractor plows 5 meters width each pass, so 200 / 5 = 40 passes, each taking 60 minutes, so 40 * 60 = 2400 minutes. But is that correct?Wait, the tractor's speed is 5 meters per minute. So, if it's moving along the 300-meter length, each pass is 300 meters, which at 5 meters per minute, takes 60 minutes. So, each pass is 60 minutes.But if each pass plows 5 meters width, then to cover the entire 200 meters width, you need 200 / 5 = 40 passes. So, total plowing time is 40 * 60 = 2400 minutes.Yes, that seems correct.But wait, maybe the tractor can plow the entire width in one pass? If so, then total plowing time would be 300 / 5 = 60 minutes, which seems too short, but let's see.If the tractor plows the entire 200 meters width in one pass, moving along the 300 meters length, then each pass is 300 meters, taking 60 minutes, and only one pass is needed. So, total plowing time is 60 minutes.But that seems unlikely because the field is 200 meters wide, which is quite large. So, it's more plausible that the tractor plows a swath of 5 meters width each time, requiring multiple passes.Therefore, I think the initial calculation of 2400 minutes of plowing time is correct.So, total time is 2400 + 390 = 2790 minutes, which is 46.5 hours.Wait, but let me think again. If the tractor plows 5 meters per minute, and each pass is 300 meters, taking 60 minutes, and each pass plows 5 meters width, then 40 passes are needed.But does the tractor have to turn around after each pass? Because turning around would take some time, but the problem doesn't mention that. So, perhaps we can ignore the turning time, as it's not specified.Alternatively, maybe the tractor can plow continuously, moving back and forth, but that would complicate the calculation. Since the problem doesn't specify, I think we can assume that each pass is 300 meters, taking 60 minutes, and turning around is instantaneous or already accounted for in the speed.So, with that, total plowing time is 2400 minutes, breaks are 13 * 30 = 390 minutes, total time is 2790 minutes, which is 46.5 hours.But let me check if 2400 minutes is indeed 40 hours, and 40 hours of plowing with breaks every 3 hours.Wait, 2400 minutes is 40 hours. So, every 3 hours, there's a 30-minute break. So, number of breaks is floor((40 - 1)/3) = floor(39/3) = 13 breaks.Yes, that matches. So, 13 breaks of 30 minutes each, totaling 390 minutes, which is 6.5 hours.So, total time is 40 + 6.5 = 46.5 hours.Therefore, the total time to plow the field, including maintenance breaks, is 46.5 hours.But wait, let me make sure about the plowing time. If the tractor plows 5 meters per minute, and each pass is 300 meters, then time per pass is 300 / 5 = 60 minutes. If each pass plows 5 meters width, then 200 / 5 = 40 passes. So, 40 * 60 = 2400 minutes, which is 40 hours. That seems correct.Alternatively, if the tractor plows the entire width in one pass, then plowing time is 60 minutes, but that seems unrealistic for a 200-meter width. So, I think the initial assumption is correct.Therefore, the total time is 46.5 hours.But let me check if the breaks are correctly calculated. If the tractor runs for 3 hours, then breaks for 0.5 hours, then runs another 3 hours, breaks, etc.So, in 40 hours of plowing, how many 3-hour intervals are there?40 / 3 ≈ 13.333. So, 13 full 3-hour intervals, each followed by a break, except the last interval which may be less than 3 hours.Wait, 13 breaks would mean 13 * 3.5 hours = 45.5 hours, but that doesn't add up.Wait, no, because each break is after 3 hours of plowing. So, for each 3 hours of plowing, you have a 0.5-hour break. So, the total time is plowing time + number_of_breaks * break_time.Number of breaks is the number of times the tractor has completed 3 hours of plowing.So, in 40 hours of plowing, how many 3-hour segments are there?40 / 3 = 13.333, so 13 full 3-hour segments, each followed by a break, and then 0.333 hours (20 minutes) of plowing without a break.Therefore, number of breaks is 13.Each break is 0.5 hours, so total break time is 13 * 0.5 = 6.5 hours.Therefore, total time is 40 + 6.5 = 46.5 hours.Yes, that matches.So, to summarize:1. Total fuel consumed: 60 + 8/π liters ≈ 62.55 liters.2. Total time to plow the field: 46.5 hours.But let me write the exact value for the fuel consumption: 60 + 8/π liters.And for the time, 46.5 hours, which can also be written as 46 hours and 30 minutes.But the problem asks for the total time, so 46.5 hours is fine.Wait, but let me check if the tractor plows continuously for 3 hours, then breaks, etc. So, the first 3 hours of plowing, then 0.5 hours break, then next 3 hours, etc.So, in 40 hours of plowing, the number of breaks is 13, as calculated.Therefore, the total time is 40 + 6.5 = 46.5 hours.Yes, that seems correct.So, final answers:1. Total fuel consumed: 60 + 8/π liters.2. Total time: 46.5 hours.But let me make sure I didn't make any calculation mistakes.For the fuel:Integral of E(t) from 0 to 12:∫₀¹² [5 + sin(πt/4)] dt = 5t + (-4/π)cos(πt/4) from 0 to 12.At t=12:5*12 = 60cos(π*12/4) = cos(3π) = -1So, (-4/π)*(-1) = 4/πAt t=0:5*0 = 0cos(0) = 1So, (-4/π)*1 = -4/πSubtracting:[60 + 4/π] - [0 - 4/π] = 60 + 4/π + 4/π = 60 + 8/πYes, correct.For the time:Total plowing time: 40 hours.Number of breaks: 13.Total break time: 13 * 0.5 = 6.5 hours.Total time: 40 + 6.5 = 46.5 hours.Yes, correct.So, I think that's the solution.</think>

answer:<think>Okay, so I've got this problem about a clothing line collaboration where an entrepreneur is trying to figure out the best marketing and sales strategy. There are two main factors influencing success: online engagement (E) and conversion rate (C). Both of these depend on marketing spend (M) and time spent engaging with the audience (T). The functions given are:1. ( E(M, T) = 3M^{0.6}T^{0.4} + 5sinleft(frac{M}{T}right) )2. ( C(M, T) = frac{e^{0.05(M + T)}}{1 + e^{0.05(M + T)}} )And there are two sub-problems to solve.Starting with part a). We need to calculate the total engagement and conversion rate when M is 10,000 and T is 50 hours. Then, find the partial derivatives of E and C with respect to M and T at these values.Alright, let's tackle the first part: calculating E and C.First, plug in M = 10,000 and T = 50 into E(M, T):( E(10000, 50) = 3*(10000)^{0.6}*(50)^{0.4} + 5sinleft(frac{10000}{50}right) )Let me compute each part step by step.Calculating ( (10000)^{0.6} ). Since 10000 is 10^4, so ( (10^4)^{0.6} = 10^{2.4} ). 10^2 is 100, 10^0.4 is approximately 2.51188643150958. So 100 * 2.511886 ≈ 251.1886.Similarly, ( (50)^{0.4} ). 50 is 5*10, so maybe compute 50^0.4. Let's see, 50^0.4 is equal to e^(0.4*ln50). ln50 is approximately 3.91202. So 0.4*3.91202 ≈ 1.5648. Then e^1.5648 ≈ 4.783.So, 3 * 251.1886 * 4.783. Let me compute 251.1886 * 4.783 first.251.1886 * 4 = 1004.7544251.1886 * 0.783 ≈ 251.1886 * 0.7 = 175.832, 251.1886 * 0.083 ≈ 20.855. So total ≈ 175.832 + 20.855 ≈ 196.687.So total 1004.7544 + 196.687 ≈ 1201.4414.Then multiply by 3: 3 * 1201.4414 ≈ 3604.3242.Next, the sine term: 5*sin(10000/50) = 5*sin(200). Since 200 radians is a large angle, but sine is periodic with period 2π ≈ 6.283. So 200 / 6.283 ≈ 31.83, so 31 full periods and 0.83*6.283 ≈ 5.21 radians. So sin(200) = sin(5.21). 5.21 radians is more than π (≈3.14) but less than 2π. Let's compute sin(5.21). Since 5.21 - π ≈ 5.21 - 3.14 ≈ 2.07 radians. So sin(5.21) = -sin(2.07). Because sin(π + x) = -sin(x). So sin(2.07) is approximately sin(2.07) ≈ 0.808. So sin(5.21) ≈ -0.808. Therefore, 5*sin(200) ≈ 5*(-0.808) ≈ -4.04.So total E ≈ 3604.3242 - 4.04 ≈ 3600.2842.So E ≈ 3600.28.Now, calculating C(M, T):( C(10000, 50) = frac{e^{0.05*(10000 + 50)}}{1 + e^{0.05*(10000 + 50)}} )Compute exponent: 0.05*(10050) = 502.5So ( C = frac{e^{502.5}}{1 + e^{502.5}} ). Hmm, e^502.5 is an astronomically large number, so the denominator is approximately e^502.5, so C ≈ 1. Because e^502.5 / (1 + e^502.5) ≈ 1.But wait, that can't be right because 0.05*(M + T) is 0.05*(10000 + 50) = 0.05*10050 = 502.5. So e^502.5 is a huge number, so C ≈ 1. So the conversion rate is almost 100%.But that seems too high. Maybe I made a mistake. Wait, 0.05*(M + T) is 0.05*(10000 + 50) = 502.5. So yes, e^502.5 is a gigantic number, so the conversion rate is practically 1. So C ≈ 1.Wait, but in reality, conversion rates can't be 100%, so maybe the function is intended to model a sigmoid curve where as M + T increases, C approaches 1. So at M + T = 10050, C is 1. So that's correct.So, E ≈ 3600.28 and C ≈ 1.Now, moving on to the partial derivatives.First, for E(M, T):Compute ( frac{partial E}{partial M} ) and ( frac{partial E}{partial T} ).Given ( E(M, T) = 3M^{0.6}T^{0.4} + 5sinleft(frac{M}{T}right) )So, partial derivative with respect to M:( frac{partial E}{partial M} = 3*0.6*M^{-0.4}T^{0.4} + 5*cosleft(frac{M}{T}right)*frac{1}{T} )Similarly, partial derivative with respect to T:( frac{partial E}{partial T} = 3*0.4*M^{0.6}T^{-0.6} + 5*cosleft(frac{M}{T}right)*left(-frac{M}{T^2}right) )Now, plug in M = 10000 and T = 50.First, compute ( frac{partial E}{partial M} ):Compute each term:First term: 3*0.6 = 1.8. Then, M^{-0.4} = (10000)^{-0.4} = 1/(10000^{0.4}) = 1/(10^{4*0.4}) = 1/(10^{1.6}) ≈ 1/39.8107 ≈ 0.02512.T^{0.4} = 50^{0.4} ≈ 4.783 as before.So first term: 1.8 * 0.02512 * 4.783 ≈ 1.8 * (0.02512 * 4.783). Let's compute 0.02512 * 4.783 ≈ 0.120. So 1.8 * 0.120 ≈ 0.216.Second term: 5 * cos(M/T) * (1/T). M/T = 10000/50 = 200. So cos(200). As before, 200 radians is equivalent to 200 - 31*2π ≈ 200 - 194.774 ≈ 5.226 radians. So cos(5.226). 5.226 - π ≈ 2.086 radians. So cos(5.226) = -cos(2.086). cos(2.086) ≈ -0.416. So cos(5.226) ≈ 0.416. Wait, no: cos(π + x) = -cos(x). So cos(5.226) = -cos(2.086). cos(2.086) ≈ -0.416, so cos(5.226) ≈ 0.416.Wait, let me double-check. cos(5.226): 5.226 radians is approximately 300 degrees (since π radians ≈ 180 degrees, so 5.226 ≈ 300 degrees). Cos(300 degrees) is 0.5. But 5.226 radians is actually more than π (3.14) but less than 2π (6.28). Let me compute cos(5.226):Using calculator: cos(5.226) ≈ cos(5.226) ≈ 0.416. So positive.So 5 * 0.416 * (1/50) ≈ 5 * 0.416 * 0.02 ≈ 5 * 0.00832 ≈ 0.0416.So total ( frac{partial E}{partial M} ≈ 0.216 + 0.0416 ≈ 0.2576 ).Now, ( frac{partial E}{partial T} ):First term: 3*0.4 = 1.2. Then, M^{0.6} = (10000)^{0.6} ≈ 251.1886 as before. T^{-0.6} = 50^{-0.6} ≈ 1/(50^{0.6}) ≈ 1/(50^{0.6}).Compute 50^{0.6}: 50^{0.6} = e^{0.6*ln50} ≈ e^{0.6*3.91202} ≈ e^{2.3472} ≈ 10.5. So T^{-0.6} ≈ 1/10.5 ≈ 0.095238.So first term: 1.2 * 251.1886 * 0.095238 ≈ 1.2 * (251.1886 * 0.095238). Compute 251.1886 * 0.095238 ≈ 23.9. So 1.2 * 23.9 ≈ 28.68.Second term: 5 * cos(M/T) * (-M/T^2). M/T = 200, as before. cos(200) ≈ 0.416. So 5 * 0.416 * (-10000)/(50^2) = 5 * 0.416 * (-10000)/2500 = 5 * 0.416 * (-4) = 5 * (-1.664) ≈ -8.32.So total ( frac{partial E}{partial T} ≈ 28.68 - 8.32 ≈ 20.36 ).Wait, that seems quite large. Let me double-check the calculations.First term: 3*0.4*M^{0.6}T^{-0.6} = 1.2*(10000)^{0.6}*(50)^{-0.6}.We had (10000)^{0.6} ≈ 251.1886 and (50)^{-0.6} ≈ 0.095238. So 1.2 * 251.1886 * 0.095238 ≈ 1.2 * (251.1886 * 0.095238).251.1886 * 0.095238 ≈ 23.9. So 1.2 * 23.9 ≈ 28.68. That seems correct.Second term: 5 * cos(M/T) * (-M/T^2). M=10000, T=50, so M/T^2 = 10000/2500 = 4. So 5 * cos(200) * (-4). cos(200) ≈ 0.416. So 5 * 0.416 * (-4) ≈ 5 * (-1.664) ≈ -8.32. So total is 28.68 - 8.32 ≈ 20.36. That seems correct.Now, for the partial derivatives of C(M, T):Given ( C(M, T) = frac{e^{0.05(M + T)}}{1 + e^{0.05(M + T)}} ). Let's denote ( u = 0.05(M + T) ), so ( C = frac{e^u}{1 + e^u} = sigma(u) ), where σ is the sigmoid function.The derivative of σ(u) with respect to u is σ(u)(1 - σ(u)).So, ( frac{partial C}{partial M} = sigma'(u) * 0.05 = 0.05 * C * (1 - C) ).Similarly, ( frac{partial C}{partial T} = 0.05 * C * (1 - C) ).Given that at M=10000, T=50, u = 0.05*(10050) = 502.5, so C ≈ 1 as before. So 1 - C ≈ 0.Therefore, both partial derivatives ( frac{partial C}{partial M} ) and ( frac{partial C}{partial T} ) are approximately 0.05 * 1 * 0 = 0.So, summarizing:E ≈ 3600.28C ≈ 1Partial derivatives:( frac{partial E}{partial M} ≈ 0.2576 )( frac{partial E}{partial T} ≈ 20.36 )( frac{partial C}{partial M} ≈ 0 )( frac{partial C}{partial T} ≈ 0 )So, that's part a).Now, moving on to part b). The entrepreneur wants to maximize the overall effectiveness F(M, T) = E(M, T) * C(M, T). We need to find the critical points of F(M, T) and analyze their nature. Given the constraint M + 2T = 10000, find the optimal M and T to maximize F.First, without constraints, we can find critical points by setting the partial derivatives of F to zero. But since there's a constraint, we'll use Lagrange multipliers.But let's first write F(M, T) = E(M, T) * C(M, T).Given E(M, T) = 3M^{0.6}T^{0.4} + 5 sin(M/T)C(M, T) = e^{0.05(M + T)} / (1 + e^{0.05(M + T)})So F = [3M^{0.6}T^{0.4} + 5 sin(M/T)] * [e^{0.05(M + T)} / (1 + e^{0.05(M + T)})]This is a complicated function. Given the constraint M + 2T = 10000, we can express M = 10000 - 2T, and substitute into F to make it a function of T alone.So let's do that.Let M = 10000 - 2T. Then, substitute into F:F(T) = [3*(10000 - 2T)^{0.6} * T^{0.4} + 5 sin((10000 - 2T)/T)] * [e^{0.05*(10000 - 2T + T)} / (1 + e^{0.05*(10000 - 2T + T)})]Simplify the exponent in C:0.05*(10000 - T) = 500 - 0.05TSo C becomes e^{500 - 0.05T} / (1 + e^{500 - 0.05T})But e^{500} is an enormous number, so unless T is very large, C will be approximately 1. However, since M = 10000 - 2T, T can't be more than 5000 (since M must be non-negative). So T ranges from 0 to 5000.But even at T=5000, M=0, so M/T = 0, so sin(0)=0. So E becomes 3*0^{0.6}*5000^{0.4} + 5*0 = 0 + 0 = 0. So F=0.At T=0, M=10000, but T=0 is problematic because M/T is undefined. So T must be greater than 0.But given that C is almost 1 for all practical T, because 0.05*(10000 - T) is still a large number. For example, at T=5000, 0.05*(10000 - 5000)=250, so e^{250} is still huge, so C≈1.Therefore, F ≈ E(M, T) * 1 ≈ E(M, T). So maximizing F is approximately equivalent to maximizing E(M, T) under the constraint M + 2T = 10000.So, we can simplify the problem to maximizing E(M, T) with M = 10000 - 2T.So, E(T) = 3*(10000 - 2T)^{0.6} * T^{0.4} + 5 sin((10000 - 2T)/T)Simplify (10000 - 2T)/T = 10000/T - 2.So, E(T) = 3*(10000 - 2T)^{0.6} * T^{0.4} + 5 sin(10000/T - 2)Now, to find the maximum of E(T), we can take the derivative of E with respect to T and set it to zero.But this seems quite complex due to the sine term. However, given that 10000/T - 2 is a large number when T is small, the sine function oscillates rapidly. But since we're looking for a maximum, perhaps the sine term contributes a small oscillation around the main term.But let's consider that for T in a reasonable range, say T=50 as in part a), 10000/50=200, so 200 - 2=198, sin(198) as before. But for larger T, say T=1000, 10000/1000=10, so 10 - 2=8, sin(8)≈0.989. So the sine term can contribute up to 5*0.989≈4.945.But the first term, 3*(10000 - 2T)^{0.6}*T^{0.4}, is likely the dominant term.So, perhaps we can approximate F ≈ E, and focus on maximizing E.So, let's define E(T) = 3*(10000 - 2T)^{0.6} * T^{0.4} + 5 sin(10000/T - 2)To find the maximum, we can take the derivative dE/dT and set it to zero.Compute dE/dT:First term: d/dT [3*(10000 - 2T)^{0.6} * T^{0.4}]Let’s denote f(T) = (10000 - 2T)^{0.6} * T^{0.4}Using product rule:f’(T) = 0.6*(10000 - 2T)^{-0.4}*(-2)*T^{0.4} + (10000 - 2T)^{0.6}*0.4*T^{-0.6}Simplify:= -1.2*(10000 - 2T)^{-0.4} * T^{0.4} + 0.4*(10000 - 2T)^{0.6} * T^{-0.6}Factor out common terms:= (10000 - 2T)^{-0.4} * T^{-0.6} [ -1.2*T^{0.8} + 0.4*(10000 - 2T)^{1.0} ]Wait, let me see:Wait, f’(T) = -1.2*(10000 - 2T)^{-0.4} * T^{0.4} + 0.4*(10000 - 2T)^{0.6} * T^{-0.6}We can factor out (10000 - 2T)^{-0.4} * T^{-0.6}:= (10000 - 2T)^{-0.4} * T^{-0.6} [ -1.2*T^{1.0} + 0.4*(10000 - 2T)^{1.0} ]Because:-1.2*(10000 - 2T)^{-0.4} * T^{0.4} = -1.2*(10000 - 2T)^{-0.4} * T^{0.4} = -1.2*T^{0.4}*(10000 - 2T)^{-0.4}Similarly, 0.4*(10000 - 2T)^{0.6} * T^{-0.6} = 0.4*(10000 - 2T)^{0.6} * T^{-0.6}So, factoring out (10000 - 2T)^{-0.4} * T^{-0.6}:= (10000 - 2T)^{-0.4} * T^{-0.6} [ -1.2*T^{1.0} + 0.4*(10000 - 2T)^{1.0} ]So, f’(T) = (10000 - 2T)^{-0.4} * T^{-0.6} [ -1.2T + 0.4*(10000 - 2T) ]Simplify inside the brackets:-1.2T + 0.4*10000 - 0.8T = (-1.2T - 0.8T) + 4000 = -2T + 4000So, f’(T) = (10000 - 2T)^{-0.4} * T^{-0.6} * (-2T + 4000)Therefore, the derivative of the first term is 3*f’(T) = 3*(10000 - 2T)^{-0.4} * T^{-0.6} * (-2T + 4000)Now, the derivative of the second term: 5 sin(10000/T - 2)Let’s denote g(T) = sin(10000/T - 2)g’(T) = cos(10000/T - 2) * (-10000/T^2)So, the derivative of the second term is 5 * g’(T) = -50000 cos(10000/T - 2) / T^2Therefore, total derivative dE/dT:= 3*(10000 - 2T)^{-0.4} * T^{-0.6} * (-2T + 4000) - 50000 cos(10000/T - 2) / T^2Set this equal to zero for critical points:3*(10000 - 2T)^{-0.4} * T^{-0.6} * (-2T + 4000) - 50000 cos(10000/T - 2) / T^2 = 0This equation is quite complex and likely doesn't have an analytical solution. So, we'll need to solve it numerically.Given that, perhaps we can make an approximation by considering that the sine term's derivative is small compared to the first term, especially since the first term is likely dominant. So, perhaps we can approximate by ignoring the sine term's derivative and solve for when the first term's derivative is zero.So, set 3*(10000 - 2T)^{-0.4} * T^{-0.6} * (-2T + 4000) ≈ 0Since (10000 - 2T)^{-0.4} and T^{-0.6} are always positive for T in (0, 5000), the term (-2T + 4000) must be zero.So, -2T + 4000 = 0 → T = 2000.So, T ≈ 2000, then M = 10000 - 2*2000 = 6000.But we need to check if this is indeed a maximum. Also, we need to consider the effect of the sine term.Let me compute E(T) at T=2000:E(2000) = 3*(6000)^{0.6}*(2000)^{0.4} + 5 sin(6000/2000 - 2) = 3*(6000)^{0.6}*(2000)^{0.4} + 5 sin(3 - 2) = 3*(6000)^{0.6}*(2000)^{0.4} + 5 sin(1) ≈ 3*(6000)^{0.6}*(2000)^{0.4} + 5*0.8415 ≈ 3*(6000)^{0.6}*(2000)^{0.4} + 4.2075Compute (6000)^{0.6}:6000 = 6*1000 = 6*10^3, so (6000)^{0.6} = (6)^{0.6}*(10^3)^{0.6} = 6^{0.6}*10^{1.8} ≈ 3.3019 * 63.0957 ≈ 208.12Similarly, (2000)^{0.4} = (2*10^3)^{0.4} = 2^{0.4}*10^{1.2} ≈ 1.3195 * 15.8489 ≈ 20.85So, 3*208.12*20.85 ≈ 3*4340 ≈ 13020So, E ≈ 13020 + 4.2075 ≈ 13024.21Now, let's check T=2000 ± some value to see if E is higher.But given the complexity, perhaps T=2000 is the optimal point.However, we need to consider the derivative including the sine term. Let's compute the derivative at T=2000:First term: 3*(6000)^{-0.4}*(2000)^{-0.6}*(-2*2000 + 4000) = 3*(6000)^{-0.4}*(2000)^{-0.6}*(0) = 0Second term: -50000 cos(6000/2000 - 2)/ (2000)^2 = -50000 cos(3 - 2)/4,000,000 = -50000 cos(1)/4,000,000 ≈ -50000*0.5403/4,000,000 ≈ -27015/4,000,000 ≈ -0.00675So, the derivative at T=2000 is approximately -0.00675, which is slightly negative. So, E is decreasing at T=2000. Therefore, the maximum might be slightly before T=2000.Alternatively, perhaps the critical point is near T=2000 but slightly less.Alternatively, maybe the maximum is indeed around T=2000, considering the approximation.But given the complexity, perhaps the optimal point is around T=2000, M=6000.Alternatively, let's consider that the derivative of the first term is zero at T=2000, but the sine term's derivative is negative, so the total derivative is negative, meaning E is decreasing at T=2000. So, the maximum might be slightly before T=2000.But without solving numerically, it's hard to find the exact point. However, given the problem's context, perhaps T=2000 is the optimal point.Alternatively, let's consider that the sine term is periodic and might have local maxima nearby, but given that the first term is much larger, the overall maximum is likely near T=2000.Therefore, the optimal values are approximately M=6000 and T=2000.But let's check at T=1900:M=10000 - 2*1900=6200Compute E(T)=3*(6200)^{0.6}*(1900)^{0.4} +5 sin(6200/1900 -2)=3*(6200)^{0.6}*(1900)^{0.4} +5 sin(3.263 -2)=3*(6200)^{0.6}*(1900)^{0.4} +5 sin(1.263)≈3*(6200)^{0.6}*(1900)^{0.4} +5*0.951≈3*(6200)^{0.6}*(1900)^{0.4}+4.755Compute (6200)^{0.6}: 6200=6.2*10^3, so (6.2)^{0.6}=approx 6.2^0.6≈3.32, and (10^3)^{0.6}=10^{1.8}=63.0957, so total≈3.32*63.0957≈209.3(1900)^{0.4}=1.9*10^3^{0.4}=1.9^{0.4}*10^{1.2}≈1.3195*15.8489≈20.85So, 3*209.3*20.85≈3*4360≈13080So, E≈13080 +4.755≈13084.755Compare to T=2000: E≈13024.21So, E is higher at T=1900. So, perhaps the maximum is around T=1900.Similarly, let's try T=1800:M=10000 - 2*1800=6400E(T)=3*(6400)^{0.6}*(1800)^{0.4} +5 sin(6400/1800 -2)=3*(6400)^{0.6}*(1800)^{0.4} +5 sin(3.555 -2)=3*(6400)^{0.6}*(1800)^{0.4} +5 sin(1.555)≈3*(6400)^{0.6}*(1800)^{0.4} +5*0.999≈3*(6400)^{0.6}*(1800)^{0.4}+4.995Compute (6400)^{0.6}=64^{0.6}*100^{0.6}= (8^2)^{0.6}*10^{1.2}=8^{1.2}*10^{1.2}= (8^{1.2})*(10^{1.2})= approx 8^1=8, 8^0.2≈1.5157, so 8*1.5157≈12.1256; 10^{1.2}≈15.8489. So, 12.1256*15.8489≈192.5(1800)^{0.4}=1.8*10^3^{0.4}=1.8^{0.4}*10^{1.2}≈1.3195*15.8489≈20.85So, 3*192.5*20.85≈3*4016≈12048So, E≈12048 +4.995≈12052.995Which is less than at T=1900.So, E is higher at T=1900 than at T=1800 and T=2000.Wait, but at T=1900, E≈13084.755, which is higher than at T=2000 (13024.21). So, perhaps the maximum is around T=1900.But let's try T=1950:M=10000 - 2*1950=10000 - 3900=6100E(T)=3*(6100)^{0.6}*(1950)^{0.4} +5 sin(6100/1950 -2)=3*(6100)^{0.6}*(1950)^{0.4} +5 sin(3.128 -2)=3*(6100)^{0.6}*(1950)^{0.4} +5 sin(1.128)≈3*(6100)^{0.6}*(1950)^{0.4} +5*0.901≈3*(6100)^{0.6}*(1950)^{0.4}+4.505Compute (6100)^{0.6}=6.1*10^3^{0.6}=6.1^{0.6}*10^{1.8}≈3.32*63.0957≈209.3(1950)^{0.4}=1.95*10^3^{0.4}=1.95^{0.4}*10^{1.2}≈1.3195*15.8489≈20.85So, 3*209.3*20.85≈3*4360≈13080So, E≈13080 +4.505≈13084.505Which is slightly less than at T=1900.Wait, but at T=1900, E≈13084.755, which is higher than at T=1950.So, perhaps the maximum is around T=1900.But let's try T=1850:M=10000 - 2*1850=10000 - 3700=6300E(T)=3*(6300)^{0.6}*(1850)^{0.4} +5 sin(6300/1850 -2)=3*(6300)^{0.6}*(1850)^{0.4} +5 sin(3.405 -2)=3*(6300)^{0.6}*(1850)^{0.4} +5 sin(1.405)≈3*(6300)^{0.6}*(1850)^{0.4} +5*0.986≈3*(6300)^{0.6}*(1850)^{0.4}+4.93Compute (6300)^{0.6}=6.3*10^3^{0.6}=6.3^{0.6}*10^{1.8}≈3.32*63.0957≈209.3(1850)^{0.4}=1.85*10^3^{0.4}=1.85^{0.4}*10^{1.2}≈1.3195*15.8489≈20.85So, 3*209.3*20.85≈3*4360≈13080So, E≈13080 +4.93≈13084.93Which is slightly higher than at T=1900.Wait, but at T=1900, E≈13084.755, which is slightly less than at T=1850.Wait, this is confusing. Maybe the maximum is around T=1850-1900.But given the time constraints, perhaps the optimal T is around 1900, M=6200.But let's consider that the derivative at T=2000 is slightly negative, so the function is decreasing after T=2000, but increasing before T=2000. So, the maximum is at T=2000.Alternatively, perhaps the maximum is indeed at T=2000, considering the first term's derivative is zero there, and the sine term's derivative is small.Given the complexity, perhaps the optimal point is M=6000, T=2000.But let's check the second derivative to confirm if it's a maximum.Alternatively, given the problem's context, perhaps the optimal point is M=6000, T=2000.So, in conclusion, the optimal values are M=6000 and T=2000.</think>