answer:<think>Okay, so I have this problem where an author is self-publishing their books and wants to analyze the growth of their sales to optimize marketing strategies. They’re using a DIY platform that charges a fixed monthly fee and takes a percentage of the revenue. The revenue follows a logistic growth model, and I need to figure out the net profit over time and then find when it's maximized.Starting with part 1: I need to derive the expression for the net monthly profit ( P(t) ). The given formula for revenue is ( R(t) = frac{L}{1 + e^{-k(t - t_0)}} ). The net profit is given by ( P(t) = R(t) - F - pR(t) ). So, let me write that out:[ P(t) = R(t) - F - pR(t) ]I can factor out ( R(t) ) from the first and third terms:[ P(t) = R(t)(1 - p) - F ]Now, substituting the expression for ( R(t) ):[ P(t) = left( frac{L}{1 + e^{-k(t - t_0)}} right)(1 - p) - F ]So that simplifies to:[ P(t) = frac{L(1 - p)}{1 + e^{-k(t - t_0)}} - F ]I think that's the expression for ( P(t) ). It combines the logistic growth of revenue, subtracts the fixed fee and the percentage taken by the platform.Moving on to part 2: The author wants to maximize ( P(t) ) over 12 months. They give specific values: ( L = 10000 ), ( k = 0.5 ), ( t_0 = 6 ), ( F = 500 ), and ( p = 0.1 ). So, let's plug these into the expression we just derived.First, let me write the specific ( P(t) ):[ P(t) = frac{10000(1 - 0.1)}{1 + e^{-0.5(t - 6)}} - 500 ][ P(t) = frac{10000 times 0.9}{1 + e^{-0.5(t - 6)}} - 500 ][ P(t) = frac{9000}{1 + e^{-0.5(t - 6)}} - 500 ]So, to find the maximum, I need to find the value of ( t ) that maximizes this function. Since it's a continuous function, I can take the derivative of ( P(t) ) with respect to ( t ), set it equal to zero, and solve for ( t ).First, let me denote the denominator as ( D(t) = 1 + e^{-0.5(t - 6)} ). So, ( P(t) = frac{9000}{D(t)} - 500 ). The derivative ( P'(t) ) will be the derivative of ( frac{9000}{D(t)} ) minus the derivative of 500, which is zero.Using the quotient rule or chain rule, let's compute ( P'(t) ):Let me rewrite ( P(t) ) as ( 9000 times [D(t)]^{-1} - 500 ). So, the derivative is:[ P'(t) = 9000 times (-1) times [D(t)]^{-2} times D'(t) ]Compute ( D'(t) ):( D(t) = 1 + e^{-0.5(t - 6)} )So, ( D'(t) = 0 + e^{-0.5(t - 6)} times (-0.5) )[ D'(t) = -0.5 e^{-0.5(t - 6)} ]Plugging back into ( P'(t) ):[ P'(t) = -9000 times [D(t)]^{-2} times (-0.5 e^{-0.5(t - 6)}) ]Simplify the negatives:[ P'(t) = 9000 times 0.5 times e^{-0.5(t - 6)} times [D(t)]^{-2} ][ P'(t) = 4500 times e^{-0.5(t - 6)} times frac{1}{[1 + e^{-0.5(t - 6)}]^2} ]So, ( P'(t) = frac{4500 e^{-0.5(t - 6)}}{[1 + e^{-0.5(t - 6)}]^2} )To find the critical points, set ( P'(t) = 0 ):[ frac{4500 e^{-0.5(t - 6)}}{[1 + e^{-0.5(t - 6)}]^2} = 0 ]But ( e^{-0.5(t - 6)} ) is always positive, and the denominator is also always positive. So, the numerator is positive, and the denominator is positive, meaning ( P'(t) ) is always positive. Wait, that can't be right because if the derivative is always positive, the function is always increasing, which would mean the maximum occurs at the upper bound of the interval, which is ( t = 12 ).But wait, let me think again. The logistic growth model ( R(t) ) has an inflection point at ( t = t_0 ), which is 6 months. Before that, the growth rate is increasing, and after that, it's decreasing. So, the revenue is increasing at an increasing rate until ( t = 6 ), then increasing at a decreasing rate after that.But the net profit ( P(t) ) is revenue minus fixed costs and percentage. So, even though revenue is increasing, the net profit might have a maximum somewhere.Wait, but according to the derivative I calculated, ( P'(t) ) is always positive because all terms are positive. That suggests that ( P(t) ) is always increasing, so the maximum would be at ( t = 12 ). But that seems counterintuitive because the growth rate of revenue slows down after ( t = 6 ).Wait, maybe I made a mistake in computing the derivative. Let me double-check.Starting again:( P(t) = frac{9000}{1 + e^{-0.5(t - 6)}} - 500 )Let me denote ( u = -0.5(t - 6) ), so ( u = -0.5t + 3 ). Then, ( D(t) = 1 + e^{u} ).So, ( P(t) = frac{9000}{D(t)} - 500 )Then, ( dP/dt = -9000 times D'(t) / [D(t)]^2 )Compute ( D'(t) = derivative of ( 1 + e^{u} ) with respect to t:( D'(t) = e^{u} times du/dt = e^{-0.5t + 3} times (-0.5) )So, ( D'(t) = -0.5 e^{-0.5(t - 6)} )Thus, ( dP/dt = -9000 times (-0.5 e^{-0.5(t - 6)}) / [1 + e^{-0.5(t - 6)}]^2 )Which simplifies to:( dP/dt = (9000 times 0.5 e^{-0.5(t - 6)}) / [1 + e^{-0.5(t - 6)}]^2 )So, ( dP/dt = 4500 e^{-0.5(t - 6)} / [1 + e^{-0.5(t - 6)}]^2 )Which is the same as before. So, since ( e^{-0.5(t - 6)} ) is always positive, and denominator is positive, ( dP/dt ) is always positive. So, ( P(t) ) is strictly increasing over time.Therefore, the maximum net profit occurs at the maximum t, which is 12 months.But wait, let me think about the behavior of ( P(t) ). As ( t ) approaches infinity, ( R(t) ) approaches L, so ( P(t) ) approaches ( L(1 - p) - F ). So, in the long run, the net profit approaches a constant. So, in the interval from t=0 to t=12, since the function is increasing, the maximum occurs at t=12.But let me compute ( P(t) ) at t=6 and t=12 to see the difference.At t=6:( R(6) = 10000 / (1 + e^{-0.5(6 - 6)}) = 10000 / (1 + e^0) = 10000 / 2 = 5000 )So, ( P(6) = 5000 - 500 - 0.1*5000 = 5000 - 500 - 500 = 4000 )At t=12:( R(12) = 10000 / (1 + e^{-0.5(12 - 6)}) = 10000 / (1 + e^{-3}) )Compute ( e^{-3} approx 0.0498 ), so denominator ≈ 1.0498Thus, ( R(12) ≈ 10000 / 1.0498 ≈ 9525.74 )Then, ( P(12) = 9525.74 - 500 - 0.1*9525.74 ≈ 9525.74 - 500 - 952.57 ≈ 9525.74 - 1452.57 ≈ 8073.17 )So, P(t) increases from 4000 at t=6 to about 8073 at t=12. So, it's indeed increasing.But wait, what about at t=0?At t=0:( R(0) = 10000 / (1 + e^{-0.5(0 - 6)}) = 10000 / (1 + e^{3}) ≈ 10000 / (1 + 20.0855) ≈ 10000 / 21.0855 ≈ 474.34 )So, ( P(0) ≈ 474.34 - 500 - 0.1*474.34 ≈ 474.34 - 500 - 47.43 ≈ -73.09 )So, negative profit initially, but as t increases, it becomes positive and keeps increasing.Therefore, the maximum occurs at t=12.But wait, the problem says "over a 12-month period", so t ranges from 0 to 12. Since P(t) is increasing, the maximum is at t=12.But let me check if the derivative is always positive. Since ( e^{-0.5(t - 6)} ) is always positive, and denominator squared is positive, so yes, derivative is always positive. So, P(t) is always increasing, so maximum at t=12.But wait, maybe I should consider the second derivative to check concavity, but since the first derivative is always positive, it's sufficient.Alternatively, maybe the maximum occurs at t=6 because that's where the growth rate of R(t) is highest. But in terms of net profit, since it's R(t) minus fixed and percentage, the net profit's growth rate might be different.Wait, let me compute P(t) at t=6 and t=12 as I did before, and see the trend.At t=6: P=4000At t=12: P≈8073So, it's increasing. So, the maximum is at t=12.But wait, maybe the maximum profit occurs before t=12 because after a certain point, the increase in revenue might not compensate for the fixed cost? But in this case, fixed cost is fixed, so as revenue increases, net profit increases as well.Wait, but in the logistic model, revenue approaches L asymptotically. So, as t increases, R(t) approaches 10000, so P(t) approaches 10000*(1 - 0.1) - 500 = 9000 - 500 = 8500. So, at t=12, it's about 8073, which is approaching 8500.So, over 12 months, the maximum occurs at t=12.But let me check the derivative again. Since P'(t) is always positive, it's always increasing, so yes, maximum at t=12.Wait, but maybe I should consider that the derivative of P(t) is positive but decreasing. So, the rate of increase slows down, but the function is still increasing.Therefore, the maximum net profit occurs at t=12.But let me confirm by computing P(t) at t=12 and t=11, t=10, etc., to see if it's indeed increasing.At t=11:( R(11) = 10000 / (1 + e^{-0.5(11 - 6)}) = 10000 / (1 + e^{-2.5}) ≈ 10000 / (1 + 0.0821) ≈ 10000 / 1.0821 ≈ 9240.08 )Then, ( P(11) ≈ 9240.08 - 500 - 0.1*9240.08 ≈ 9240.08 - 500 - 924.01 ≈ 9240.08 - 1424.01 ≈ 7816.07 )At t=12: ~8073.17So, P(t) increases from ~7816 at t=11 to ~8073 at t=12.Similarly, at t=10:( R(10) = 10000 / (1 + e^{-0.5(10 - 6)}) = 10000 / (1 + e^{-2}) ≈ 10000 / (1 + 0.1353) ≈ 10000 / 1.1353 ≈ 8808.13 )( P(10) ≈ 8808.13 - 500 - 0.1*8808.13 ≈ 8808.13 - 500 - 880.81 ≈ 8808.13 - 1380.81 ≈ 7427.32 )So, P(t) increases from ~7427 at t=10 to ~7816 at t=11, and to ~8073 at t=12.Thus, it's consistently increasing, albeit at a decreasing rate, but still increasing.Therefore, the maximum net profit occurs at t=12.But wait, the problem says "over a 12-month period", so t is between 0 and 12. Since P(t) is increasing on this interval, the maximum is at t=12.But let me think again about the derivative. Since P'(t) is always positive, it's always increasing. So, yes, maximum at t=12.Wait, but maybe I should consider that the maximum profit might be at the point where the marginal revenue equals the marginal cost, but in this case, the fixed cost is fixed, and the percentage is a variable cost. So, the marginal cost is p, which is 0.1. The marginal revenue is the derivative of R(t), which is ( R'(t) = frac{dR}{dt} = frac{L k e^{-k(t - t_0)}}{(1 + e^{-k(t - t_0)})^2} ). So, setting ( R'(t) = p ), but I'm not sure if that's the right approach because the net profit is R(t) - F - pR(t). So, the derivative of net profit is R'(t)(1 - p). So, setting derivative to zero would require R'(t)(1 - p) = 0, but since 1 - p is positive, R'(t) must be zero. But R'(t) is zero only as t approaches infinity, which is beyond our 12-month period. So, within the interval, R'(t) is always positive, so P'(t) is always positive, hence maximum at t=12.Therefore, the value of t that maximizes P(t) is 12 months.</think>

answer:<think>Alright, so I have this problem where a retiree in Oxford is looking at voting patterns and wants to figure out how different age groups influence election outcomes. They've given me this function V(a) which is an integral from 0 to 1 of (x³ - 2a x² + a² x) dx. My task is to first find the critical age a₀ that maximizes V(a), and then calculate V(a) for three specific age groups: young (2.5 decades), middle-aged (4.5 decades), and older voters (6.5 decades). Then I need to discuss which group has the most influence.Okay, starting with part 1: finding the critical age a₀ where V(a) is maximized. Since V(a) is defined as an integral, I think I should first compute that integral to get V(a) in terms of a, and then find its maximum.So, let's compute the integral:V(a) = ∫₀¹ (x³ - 2a x² + a² x) dxI can split this integral into three separate terms:V(a) = ∫₀¹ x³ dx - 2a ∫₀¹ x² dx + a² ∫₀¹ x dxLet me compute each integral one by one.First integral: ∫ x³ dx from 0 to 1.The antiderivative of x³ is (x⁴)/4, so evaluating from 0 to 1:(1⁴)/4 - (0⁴)/4 = 1/4 - 0 = 1/4.Second integral: ∫ x² dx from 0 to 1.Antiderivative is (x³)/3, so evaluating:(1³)/3 - 0 = 1/3.Third integral: ∫ x dx from 0 to 1.Antiderivative is (x²)/2, so evaluating:(1²)/2 - 0 = 1/2.Putting it all together:V(a) = (1/4) - 2a*(1/3) + a²*(1/2)Simplify each term:V(a) = 1/4 - (2a)/3 + (a²)/2So now V(a) is a quadratic function in terms of a. To find the maximum, since it's a quadratic, I can look at the coefficient of a². The coefficient is 1/2, which is positive. Wait, but if the coefficient is positive, the parabola opens upwards, meaning it has a minimum, not a maximum. Hmm, that's confusing because the problem says to find the critical age where V(a) is maximized. Maybe I made a mistake?Wait, let me double-check the integral computation.First integral: ∫₀¹ x³ dx = 1/4. Correct.Second integral: ∫₀¹ x² dx = 1/3. Correct.Third integral: ∫₀¹ x dx = 1/2. Correct.So V(a) = 1/4 - (2a)/3 + (a²)/2. So yes, quadratic with a positive coefficient on a², so it opens upwards, meaning it has a minimum, not a maximum. So that would mean V(a) doesn't have a maximum—it goes to infinity as a increases. But that can't be right because a is the average age in decades, so it's a real number, but in reality, ages don't go to infinity. Wait, but in the problem, a is just a variable, so mathematically, V(a) is a quadratic that opens upwards, so it doesn't have a maximum—it tends to infinity as a increases.But the problem says to find the critical age a₀ where V(a) is maximized. Maybe I misunderstood the problem. Perhaps V(a) is supposed to be a function that does have a maximum? Or maybe I need to consider the derivative and find where it's zero, but since it's a quadratic, the critical point is the minimum.Wait, perhaps I need to double-check the integral again. Maybe I misread the integrand.The integrand is (x³ - 2a x² + a² x). So integrating term by term:∫x³ dx = 1/4,∫-2a x² dx = -2a*(1/3),∫a² x dx = a²*(1/2).So yes, that's correct. So V(a) = 1/4 - (2a)/3 + (a²)/2.So V(a) is indeed a quadratic in a, opening upwards. So it doesn't have a maximum—it only has a minimum. Therefore, unless there's a constraint on a, like a is between certain values, V(a) doesn't have a maximum. But the problem says to find the critical age a₀ where V(a) is maximized. Maybe I need to check if the function is concave or convex.Wait, the second derivative of V(a) with respect to a is 1, which is positive, so it's convex, meaning the critical point is a minimum. So perhaps the function doesn't have a maximum—it just increases as a moves away from the minimum. So unless a is bounded, V(a) can be made as large as desired by increasing a.But in the context of the problem, a is the average age in decades. So in reality, a can't be negative, but it can be any positive number. So as a increases, V(a) increases without bound. So in that case, the influence V(a) would be maximized as a approaches infinity, which doesn't make much sense in the real world.Wait, perhaps I made a mistake in interpreting the function. Maybe the integrand is supposed to be something else? Let me look again.The function is V(a) = ∫₀¹ (x³ - 2a x² + a² x) dx. Hmm, that seems correct. So integrating, we get 1/4 - (2a)/3 + (a²)/2.Wait, maybe the function is supposed to be V(a) = ∫₀¹ (x³ - 2a x² + a² x) dx, but perhaps it's a function that's being maximized over x, not a? No, the problem says V(a) is the integral, so it's a function of a.Alternatively, maybe the integrand is supposed to be a function that peaks somewhere, but when integrated over x, it becomes a quadratic in a.Wait, perhaps the problem is to maximize V(a) with respect to a, but since it's a quadratic opening upwards, the minimum is at a certain point, but the maximum is at the boundaries. But a can be any positive number, so unless there's a constraint, the maximum is unbounded.But the problem says "the critical age a₀ for which the influence of the respective age group on the voting outcome is maximized." So maybe I need to consider that V(a) is a function that is being maximized, but since it's a quadratic with a minimum, perhaps the maximum occurs at the endpoints? But a is an average age, so it can't be negative, but can be any positive number. So as a approaches infinity, V(a) approaches infinity as well.Wait, maybe I need to check my calculations again.Wait, let's compute V(a):V(a) = ∫₀¹ (x³ - 2a x² + a² x) dx= ∫₀¹ x³ dx - 2a ∫₀¹ x² dx + a² ∫₀¹ x dx= [x⁴/4]₀¹ - 2a [x³/3]₀¹ + a² [x²/2]₀¹= (1/4 - 0) - 2a (1/3 - 0) + a² (1/2 - 0)= 1/4 - (2a)/3 + (a²)/2Yes, that seems correct. So V(a) = (1/2)a² - (2/3)a + 1/4.So, as a quadratic in a, it's U-shaped, so it has a minimum at a = -b/(2a) where the quadratic is ax² + bx + c. So in this case, a (the coefficient) is 1/2, b is -2/3.So the critical point is at a = -b/(2a) = -(-2/3)/(2*(1/2)) = (2/3)/1 = 2/3.So the critical point is at a = 2/3 decades, which is approximately 6.666... years old. But since a is in decades, 2/3 of a decade is about 6.666 years. But in the context of the problem, age groups are in decades: 2.5, 4.5, 6.5. So 2/3 is about 0.666 decades, which is 6.666 years, which is a very young age, like a child. But in the problem, the age groups start at 18-34, which is 1.8 to 3.4 decades. So 2/3 is outside the range of the given age groups.But the problem is asking for the critical age a₀ where V(a) is maximized. But since V(a) is a quadratic opening upwards, it doesn't have a maximum—it only has a minimum. So the function decreases until a = 2/3 and then increases beyond that. So if we consider a > 2/3, V(a) increases as a increases. So in the context of the age groups given (2.5, 4.5, 6.5), which are all greater than 2/3, the function V(a) is increasing with a. Therefore, the older the age group, the higher the influence V(a). So the maximum influence would be at the highest a, which is 6.5 decades.Wait, but that contradicts the initial thought that the critical point is a minimum. So perhaps the problem is to find the critical point, which is a minimum, but the maximum would be at the boundaries. But since a can be any positive number, the maximum is unbounded. However, in the context of the problem, we have specific age groups, so perhaps among those, the one with the highest a has the highest V(a).But let me think again. The function V(a) is given as an integral, which results in a quadratic function of a. The quadratic has a minimum at a = 2/3, and since it's opening upwards, for a > 2/3, V(a) increases as a increases. Therefore, among the age groups given (2.5, 4.5, 6.5), which are all greater than 2/3, the influence V(a) increases with a. So the older the age group, the higher the influence.Therefore, the critical age a₀ where V(a) is maximized would be as a approaches infinity, but in the context of the problem, since we have specific age groups, the one with the highest a (6.5) would have the highest V(a).Wait, but the problem says "find the value of a such that V(a) is maximized." So if we consider a can be any positive number, then V(a) doesn't have a maximum—it goes to infinity as a increases. But if we consider a to be within a certain range, say, the age groups given, then the maximum would be at the highest a.But the problem doesn't specify any constraints on a, so mathematically, V(a) doesn't have a maximum—it can be made arbitrarily large by increasing a. Therefore, perhaps the problem is misworded, or I'm misunderstanding it.Alternatively, maybe the function V(a) is supposed to be a function that has a maximum, so perhaps the integrand was supposed to be something else, like x³ - 2a x² + a² x, but maybe with a negative sign somewhere else.Wait, let me check the integrand again: (x³ - 2a x² + a² x). Hmm, if I factor this, it becomes x(x² - 2a x + a²) = x(x - a)². So the integrand is x(x - a)². So integrating x(x - a)² from 0 to 1.Wait, maybe I can compute this integral differently by expanding (x - a)².(x - a)² = x² - 2a x + a², so x(x - a)² = x³ - 2a x² + a² x, which matches the integrand.So integrating x³ - 2a x² + a² x from 0 to 1 gives V(a) = 1/4 - (2a)/3 + (a²)/2.So, as established, it's a quadratic with a minimum at a = 2/3 and opens upwards.Therefore, unless a is constrained, V(a) doesn't have a maximum. So perhaps the problem is to find the critical point, which is a minimum, but the question says "maximized." Maybe it's a typo, and they meant to find the minimum? Or perhaps I'm missing something.Alternatively, maybe the function V(a) is supposed to be the influence, and perhaps it's being maximized over x, but no, the integral is over x, so V(a) is a function of a.Wait, perhaps the problem is to find the a that maximizes V(a), but since V(a) is a quadratic with a minimum, the maximum would be at the endpoints. But without endpoints, it's unbounded. So perhaps the problem is intended to have a maximum, so maybe I need to re-express V(a) differently.Wait, let me compute V(a) again:V(a) = 1/4 - (2a)/3 + (a²)/2.So, if I write this as V(a) = (1/2)a² - (2/3)a + 1/4.To find the critical point, take the derivative with respect to a:dV/da = 2*(1/2)a - 2/3 = a - 2/3.Set derivative to zero:a - 2/3 = 0 => a = 2/3.So the critical point is at a = 2/3, which is a minimum since the second derivative is positive.Therefore, V(a) has a minimum at a = 2/3 and increases as a moves away from that point in both directions. But since a represents age in decades, it can't be negative, so the function decreases from a=0 to a=2/3, and then increases from a=2/3 onwards.Therefore, if we consider a ≥ 0, the minimum is at a=2/3, and the function increases for a > 2/3. So, for a > 2/3, V(a) increases as a increases.Therefore, among the given age groups (2.5, 4.5, 6.5), which are all greater than 2/3, the influence V(a) increases with a. So the older the age group, the higher the influence.Therefore, the critical age a₀ where V(a) is maximized would be as a approaches infinity, but in the context of the problem, since we have specific age groups, the one with the highest a (6.5) would have the highest V(a).But the problem says "find the value of a such that V(a) is maximized." Since mathematically, V(a) doesn't have a maximum, but in the context of the problem, perhaps the maximum influence is at the highest a given, which is 6.5.Alternatively, maybe the problem is to find the a where V(a) is maximized, but since V(a) is a quadratic with a minimum, perhaps the maximum is at the boundaries of the domain of a. But since a can be any positive number, the maximum is unbounded.Wait, perhaps I'm overcomplicating this. Maybe the problem is intended to have a maximum, so perhaps I made a mistake in the integral.Wait, let me recompute the integral:∫₀¹ (x³ - 2a x² + a² x) dx= ∫₀¹ x³ dx - 2a ∫₀¹ x² dx + a² ∫₀¹ x dx= [x⁴/4]₀¹ - 2a [x³/3]₀¹ + a² [x²/2]₀¹= (1/4 - 0) - 2a*(1/3 - 0) + a²*(1/2 - 0)= 1/4 - (2a)/3 + (a²)/2.Yes, that's correct. So V(a) = (1/2)a² - (2/3)a + 1/4.So, as a quadratic, it's U-shaped, with a minimum at a=2/3.Therefore, the function V(a) is minimized at a=2/3, and increases as a moves away from that point. So, for a > 2/3, V(a) increases as a increases.Therefore, the influence V(a) is maximized as a increases without bound, but in reality, a is limited by the age groups. So among the given age groups, the older group (6.5) has the highest V(a).But the problem is asking for the critical age a₀ where V(a) is maximized. Since mathematically, it's unbounded, but in the context, perhaps the maximum is at the highest a given, which is 6.5.Alternatively, maybe the problem is intended to have a maximum, so perhaps the integrand was supposed to be something else, like x³ - 2a x² + a² x, but with a negative sign in front, making the quadratic open downwards.Wait, if the integrand was -(x³ - 2a x² + a² x), then V(a) would be -1/4 + (2a)/3 - (a²)/2, which is a quadratic opening downwards, with a maximum at a= (2/3)/(2*(1/2)) = 2/3. So that would make sense.But the problem states the integrand as (x³ - 2a x² + a² x), so I think that's correct.Therefore, perhaps the problem is intended to have a maximum at a=2/3, but since it's a minimum, maybe the maximum is at the boundaries. But without boundaries, it's unbounded.Wait, perhaps the problem is to find the a that maximizes V(a) for a in the range of the age groups given, i.e., between 2.5 and 6.5. But since V(a) is increasing in that interval, the maximum would be at a=6.5.Alternatively, maybe the problem is to find the a where V(a) is maximized, considering that a is the average age of a group, so perhaps a is constrained to be within the range of possible ages, but since a can be any positive number, it's still unbounded.Hmm, this is confusing. Maybe I should proceed with the calculations as per the given function, even if it seems counterintuitive.So, for part 1, the critical age a₀ where V(a) is maximized is at a=6.5, since among the given age groups, that's the highest a, and V(a) increases with a.But wait, that's not rigorous. The function V(a) is increasing for a > 2/3, so if we consider a can be any positive number, the maximum is unbounded. But if we consider only the given age groups, then 6.5 is the maximum.But the problem says "find the value of a such that V(a) is maximized," without specifying constraints. So perhaps the answer is that V(a) doesn't have a maximum, but it has a minimum at a=2/3.But the problem says "maximized," so maybe I need to consider that the function is being maximized over x, but no, V(a) is defined as the integral over x, so it's a function of a.Wait, perhaps I need to consider that the integrand is x(x - a)², which is non-negative for x in [0,1], since x is between 0 and 1, and (x - a)² is always non-negative. Therefore, V(a) is the integral of a non-negative function, so V(a) is always non-negative.But since V(a) is a quadratic with a minimum at a=2/3, it's always positive, and increases as a moves away from 2/3.Therefore, the influence V(a) is minimized at a=2/3, and increases as a moves away from that point. So, for a > 2/3, V(a) increases as a increases.Therefore, the influence is higher for older age groups beyond 2/3 decades, which is about 6.666 years old. So, in the context of the given age groups (2.5, 4.5, 6.5), which are all above 2/3, the influence increases with a.Therefore, the older the age group, the higher the influence. So, the critical age a₀ where V(a) is maximized would be as a approaches infinity, but in the context of the problem, the highest given a is 6.5, so that group has the highest influence.But the problem is asking for a specific a₀, so perhaps I need to state that V(a) is maximized as a increases without bound, but in practical terms, among the given groups, the older group has the highest influence.Alternatively, maybe the problem is intended to have a maximum at a=2/3, but that's a minimum. So perhaps there's a mistake in the problem statement.Wait, maybe I need to consider that the influence is maximized at a=2/3, but that's the minimum. So perhaps the problem is misworded, and they meant to find the minimum. But the question says "maximized."Alternatively, maybe I need to consider that the influence is being maximized in terms of the function's behavior, but since it's a quadratic with a minimum, the maximum is at the endpoints.Wait, perhaps I should proceed to part 2 and see if that clarifies anything.So, part 2: calculate V(a) for each age group: 2.5, 4.5, 6.5 decades.Given V(a) = 1/4 - (2a)/3 + (a²)/2.Let's compute V(2.5), V(4.5), V(6.5).First, V(2.5):V(2.5) = 1/4 - (2*2.5)/3 + (2.5²)/2Compute each term:1/4 = 0.25(2*2.5)/3 = 5/3 ≈ 1.6667(2.5²)/2 = 6.25/2 = 3.125So V(2.5) = 0.25 - 1.6667 + 3.125Compute step by step:0.25 - 1.6667 = -1.4167-1.4167 + 3.125 = 1.7083So V(2.5) ≈ 1.7083Next, V(4.5):V(4.5) = 1/4 - (2*4.5)/3 + (4.5²)/2Compute each term:1/4 = 0.25(2*4.5)/3 = 9/3 = 3(4.5²)/2 = 20.25/2 = 10.125So V(4.5) = 0.25 - 3 + 10.125Compute step by step:0.25 - 3 = -2.75-2.75 + 10.125 = 7.375So V(4.5) = 7.375Next, V(6.5):V(6.5) = 1/4 - (2*6.5)/3 + (6.5²)/2Compute each term:1/4 = 0.25(2*6.5)/3 = 13/3 ≈ 4.3333(6.5²)/2 = 42.25/2 = 21.125So V(6.5) = 0.25 - 4.3333 + 21.125Compute step by step:0.25 - 4.3333 = -4.0833-4.0833 + 21.125 ≈ 17.0417So V(6.5) ≈ 17.0417So, summarizing:V(2.5) ≈ 1.7083V(4.5) = 7.375V(6.5) ≈ 17.0417Therefore, the influence increases with a, as expected since V(a) is increasing for a > 2/3.So, the older the age group, the higher the influence on the voting outcome.Therefore, the older voters (6.5 decades) have the most significant influence, followed by middle-aged voters (4.5), and then young voters (2.5).So, to answer part 1, the critical age a₀ where V(a) is maximized is as a approaches infinity, but in the context of the problem, the highest given a is 6.5, so that's the maximum influence.But since the problem asks for a specific a₀, perhaps I need to state that V(a) doesn't have a maximum, but it has a minimum at a=2/3. However, the problem says "maximized," so maybe I need to reconsider.Wait, perhaps I made a mistake in interpreting the function. Maybe V(a) is supposed to be the influence, and perhaps it's being maximized over x, but no, the integral is over x, so V(a) is a function of a.Alternatively, maybe the function is supposed to be V(a) = ∫₀¹ (x³ - 2a x² + a² x) dx, and perhaps the integrand is supposed to be a function that peaks at a certain a, but when integrated, it becomes a quadratic.Wait, let me think about the integrand: x(x - a)². So, for each x, the integrand is x times (x - a) squared. So, for a given x, the integrand is a quadratic in a, but when integrated over x, it becomes a quadratic in a.But regardless, the integral results in V(a) = (1/2)a² - (2/3)a + 1/4, which is a quadratic with a minimum at a=2/3.Therefore, the influence V(a) is minimized at a=2/3, and increases as a moves away from that point. So, for a > 2/3, V(a) increases as a increases.Therefore, the critical age a₀ where V(a) is maximized is as a approaches infinity, but in the context of the problem, the highest given a is 6.5, so that group has the highest influence.But the problem is asking for the critical age a₀ where V(a) is maximized, so perhaps the answer is that there is no maximum, but the influence increases without bound as a increases. However, in the context of the problem, the maximum influence is at the highest a given, which is 6.5.Alternatively, perhaps the problem is intended to have a maximum at a=2/3, but that's a minimum. So maybe the problem is misworded.Given that, I think the answer is that the influence V(a) is minimized at a=2/3 decades, and increases as a moves away from that point. Therefore, the older the age group, the higher the influence. So, among the given age groups, the older voters (6.5 decades) have the highest influence.Therefore, for part 1, the critical age a₀ where V(a) is maximized is as a approaches infinity, but in practical terms, the highest given a is 6.5.But since the problem asks for a specific value, perhaps I need to state that V(a) doesn't have a maximum, but it has a minimum at a=2/3. However, the problem says "maximized," so maybe I need to consider that the maximum is at the highest a given, which is 6.5.Alternatively, perhaps the problem is intended to have a maximum at a=2/3, but that's a minimum. So maybe the problem is misworded.Given the confusion, I think the best approach is to proceed with the calculations as per the given function, even if it seems counterintuitive.So, to summarize:1. The critical age a₀ where V(a) is maximized is as a approaches infinity, but in the context of the problem, the highest given a is 6.5 decades.2. Calculating V(a) for each age group:- Young voters (2.5): V ≈ 1.7083- Middle-aged voters (4.5): V = 7.375- Older voters (6.5): V ≈ 17.0417Therefore, the older voters have the most significant influence on the voting outcome in Oxford, followed by middle-aged voters, and then young voters.So, despite the mathematical function suggesting that V(a) can be made arbitrarily large by increasing a, in the context of the problem, the highest given a is 6.5, so that group has the highest influence.</think>

answer:<think>Alright, so I have this problem about a botanist studying sedges in different patches, represented as a graph. There are two parts to the problem. Let me try to tackle them one by one.Starting with part 1: We have a connected graph G with n vertices, each representing a patch. Some patches have sedge plants, and this subset is S. The function f(S) is defined as the sum of the degrees of the vertices in S divided by the number of vertices in S. I need to determine the conditions under which f(S) is maximized and prove that these conditions hold.Hmm, okay. So f(S) is the average degree of the vertices in S. To maximize this average, we need to select a subset S where the average degree is as high as possible. Since the graph is connected, all vertices have at least degree 1, but some might have higher degrees.I remember that in graph theory, the average degree of a graph is maximized when the subset S consists of the vertices with the highest degrees. So, intuitively, to maximize f(S), S should include the vertices with the highest degrees in G.But let me think more formally. Suppose we have a graph where vertices have degrees d1, d2, ..., dn, sorted in non-increasing order. So d1 ≥ d2 ≥ ... ≥ dn.If I take S as the first m vertices, then the average degree of S would be (d1 + d2 + ... + dm)/m. To maximize this, I should choose the m vertices with the highest degrees.Wait, but is this always the case? Suppose I have a graph where some high-degree vertices are connected to low-degree vertices. Would including a low-degree vertex in S decrease the average? Yes, it would. So to maximize the average, we should only include the highest-degree vertices.But is there a case where including a lower-degree vertex could somehow increase the average? For example, if adding a lower-degree vertex allows more high-degree vertices to be connected? Hmm, no, because the average is just the sum divided by the number. Adding a lower-degree vertex would only decrease the average.So, the maximum average degree subset S is the set of vertices with the highest degrees. Therefore, f(S) is maximized when S consists of the m vertices with the highest degrees in G, for some m.But the problem says "determine the conditions under which f(S) is maximized." So, perhaps more precisely, f(S) is maximized when S is the set of all vertices with degree at least some threshold. Or, in other words, S should be the subset of vertices with the highest degrees.Alternatively, maybe S should be a star graph, where one central vertex is connected to all others, but in that case, the average degree would be (n-1 + 1 + 1 + ... +1)/n, which is (n + (n-2))/n = (2n - 2)/n = 2 - 2/n. But if we take S as just the central vertex, the average degree is n-1, which is higher.Wait, so in that case, taking S as just the central vertex gives a higher average degree than taking all vertices. So, in general, the maximum average degree is achieved by the subset S consisting of the single vertex with the highest degree.But wait, is that always the case? Suppose we have two vertices with very high degrees, say both have degree n-1 in a complete graph. Then, taking both would give an average degree of (n-1 + n-1)/2 = n-1, which is the same as taking either one. So, in that case, the average doesn't decrease.But in a graph where the degrees are not all the same, taking the highest degree vertex will give a higher average than taking any other subset.Wait, let's test this with a simple example. Suppose we have a graph with 3 vertices: A connected to B and C, and B connected to C. So degrees are: A:2, B:2, C:2. So all have the same degree. Then, taking any subset S will have the same average degree. So, in this case, the maximum average is 2, and any subset will do.Another example: a star graph with center A connected to B, C, D. So degrees: A:3, B:1, C:1, D:1. If I take S as just A, average degree is 3. If I take S as A and B, average is (3+1)/2 = 2. So, 3 > 2, so taking just A is better.Another example: suppose a graph with two hubs. Let's say A connected to B, C, D, and E connected to B, C, D. So A and E have degree 3, and B, C, D have degree 2. So degrees: A:3, E:3, B:2, C:2, D:2. If I take S as A and E, the average is (3+3)/2 = 3. If I take S as A, E, B, the average is (3+3+2)/3 ≈ 2.666, which is less than 3. So, again, taking just the two highest degree vertices gives a higher average.Wait, but in this case, the average is the same as taking each individually. So, perhaps, when multiple vertices have the same maximum degree, including all of them in S will maintain the average.So, in general, to maximize f(S), S should include all vertices with degree at least as high as the minimum degree in S. So, S should be a set where all vertices have degree ≥ d, and d is as large as possible.Therefore, the maximum average degree is achieved when S is the set of all vertices with degree at least equal to the average degree of S. So, in other words, S should be a subset where all vertices have degree at least the average degree of S, and this average is maximized.Wait, that might be a bit circular. Alternatively, perhaps S should be a subset where adding any other vertex would decrease the average, and removing any vertex would also decrease the average. That is, S is a subset where all vertices have degree at least equal to the average degree of S, and all other vertices have degree less than this average.Yes, that makes sense. So, for a subset S, if every vertex in S has degree at least equal to the average degree of S, and every vertex not in S has degree less than the average degree of S, then S is a subset that maximizes the average degree.This is similar to the concept of a "dense" subgraph. In fact, the problem is related to finding a densest subgraph, which is a well-known problem in graph theory.The densest subgraph problem is to find a subset S of vertices that maximizes the average degree, which is exactly what f(S) is. It's known that the densest subgraph can be found using a max-flow algorithm, but in terms of conditions, it's when S is such that all vertices in S have degree at least the average degree of S, and all vertices outside S have degree less than that average.Therefore, the conditions under which f(S) is maximized are when S is a subset of vertices where each vertex in S has degree at least equal to the average degree of S, and each vertex not in S has degree less than the average degree of S.To prove this, suppose we have a subset S where some vertex in S has degree less than the average degree of S. Then, removing that vertex would increase the average degree of S, which contradicts the maximality of f(S). Similarly, if there is a vertex not in S with degree at least the average degree of S, adding it to S would not decrease the average degree, which again contradicts the maximality.Therefore, the maximum is achieved when S is such that all vertices in S have degree at least the average degree of S, and all vertices not in S have degree less than the average degree of S.So, that's part 1.Moving on to part 2: The number of sedge plants in a patch v is modeled by a Poisson distribution with parameter λ_v. The total number of sedge plants across all patches is a random variable X, which is the sum of these Poisson distributions. The botanist wants P(X ≥ k) ≥ 0.95. We need to find the relationship between k, the λ_v, and n, and find the minimum k such that this holds for the sum of λ_v being Λ.Okay, so X is the sum of independent Poisson random variables, each with parameter λ_v. The sum of independent Poisson variables is also Poisson with parameter equal to the sum of the individual parameters. So, X ~ Poisson(Λ), where Λ = sum_{v} λ_v.Wait, is that correct? Yes, because if X_i ~ Poisson(λ_i), then sum X_i ~ Poisson(sum λ_i). So, X ~ Poisson(Λ).Therefore, we need P(X ≥ k) ≥ 0.95. Since X is Poisson(Λ), we can write this as:P(X ≥ k) = 1 - P(X ≤ k - 1) ≥ 0.95So, P(X ≤ k - 1) ≤ 0.05We need to find the smallest k such that the cumulative distribution function of Poisson(Λ) evaluated at k - 1 is ≤ 0.05.In other words, k is the smallest integer such that P(X ≤ k - 1) ≤ 0.05.This is equivalent to finding the 95th percentile of the Poisson distribution with parameter Λ.For a Poisson distribution, the quantiles can be approximated, but there isn't a closed-form expression. However, we can relate k to Λ using the property that for large Λ, the Poisson distribution can be approximated by a normal distribution with mean Λ and variance Λ.Using the normal approximation, we can write:P(X ≤ k - 1) ≈ Φ((k - 1 - Λ)/sqrt(Λ)) ≤ 0.05Where Φ is the CDF of the standard normal distribution.We know that Φ(z) = 0.05 corresponds to z ≈ -1.6449 (the 5th percentile). Therefore:(k - 1 - Λ)/sqrt(Λ) ≤ -1.6449Solving for k:k - 1 ≤ Λ - 1.6449 * sqrt(Λ)So,k ≤ Λ - 1.6449 * sqrt(Λ) + 1But since k must be an integer, we take the ceiling of the right-hand side.However, this is an approximation. For exact values, we would need to compute the Poisson CDF until we find the smallest k such that P(X ≤ k - 1) ≤ 0.05.Alternatively, using the relationship between Poisson and chi-squared distributions, we can use the inequality:P(X ≥ k) ≥ 1 - e^{-Λ} sum_{i=0}^{k-1} frac{Λ^i}{i!}But this might not be helpful directly.Alternatively, using Markov's inequality, but that would give a weaker bound.Wait, but since we need P(X ≥ k) ≥ 0.95, which is a lower bound on the upper tail. Markov's inequality gives an upper bound on the upper tail, so it's not directly useful here.Alternatively, using the Chernoff bound for Poisson distributions.The Chernoff bound states that for a Poisson random variable X with parameter Λ,P(X ≥ k) ≤ e^{-Λ} (e^{(k - Λ)/2})^{2Λ/(k - Λ)}}But I'm not sure if that's directly applicable here.Alternatively, for a Poisson distribution, the mode is floor(Λ) or ceil(Λ). The distribution is unimodal and skewed. For large Λ, it's approximately normal.Given that, perhaps the normal approximation is sufficient for an approximate relationship.So, using the normal approximation, we can write:k ≈ Λ + z_{0.95} * sqrt(Λ)Wait, no. Wait, earlier I had:(k - 1 - Λ)/sqrt(Λ) ≤ -1.6449Which rearranges to:k ≤ Λ - 1.6449 * sqrt(Λ) + 1But since we want P(X ≥ k) ≥ 0.95, which corresponds to the upper tail, the z-score should be positive. Wait, maybe I messed up the direction.Wait, let's think again. If we approximate X as N(Λ, Λ), then:P(X ≤ k - 1) ≈ Φ((k - 1 - Λ)/sqrt(Λ)) ≤ 0.05So, (k - 1 - Λ)/sqrt(Λ) ≤ z_{0.05} = -1.6449Therefore,k - 1 ≤ Λ - 1.6449 * sqrt(Λ)So,k ≤ Λ - 1.6449 * sqrt(Λ) + 1But since k must be an integer, we take the floor of the right-hand side.Wait, but this gives an upper bound on k. However, we need the minimal k such that P(X ≥ k) ≥ 0.95, which is equivalent to P(X ≤ k - 1) ≤ 0.05.So, the minimal k is the smallest integer such that P(X ≤ k - 1) ≤ 0.05. Using the normal approximation, we can estimate k as:k ≈ Λ - 1.6449 * sqrt(Λ) + 1But since we need the minimal k, we might need to round up or down depending on the exact value.Alternatively, using the inverse of the Poisson CDF, but without exact computation, it's hard to get the precise k.However, the problem asks for the relationship between k, λ_v, and n, given that sum λ_v = Λ.So, the key relationship is that k is approximately equal to Λ minus roughly 1.645 times the square root of Λ, plus 1.But perhaps more precisely, using the normal approximation, the minimal k is approximately Λ + z_{0.95} * sqrt(Λ), but wait, no, because we're dealing with the lower tail.Wait, let me clarify. If we have X ~ Poisson(Λ), and we want P(X ≥ k) ≥ 0.95, then k is the 95th percentile of X. For a normal distribution, the 95th percentile is μ + z_{0.95} * σ, where μ = Λ and σ = sqrt(Λ). So, the 95th percentile is approximately Λ + 1.6449 * sqrt(Λ).But wait, in our earlier calculation, we had:(k - 1 - Λ)/sqrt(Λ) ≤ -1.6449Which led to k ≤ Λ - 1.6449 * sqrt(Λ) + 1But that seems contradictory because if we're looking for the 95th percentile, it should be higher than the mean, not lower.Wait, perhaps I made a mistake in the direction of the inequality.Let me re-examine.We have P(X ≥ k) ≥ 0.95, which is equivalent to P(X ≤ k - 1) ≤ 0.05.So, in terms of the normal approximation, we have:P(X ≤ k - 1) ≈ Φ((k - 1 - Λ)/sqrt(Λ)) ≤ 0.05So, (k - 1 - Λ)/sqrt(Λ) ≤ z_{0.05} = -1.6449Therefore,k - 1 ≤ Λ - 1.6449 * sqrt(Λ)So,k ≤ Λ - 1.6449 * sqrt(Λ) + 1But this suggests that k is less than Λ, which contradicts the intuition that the 95th percentile should be greater than the mean.Wait, no, actually, in the normal distribution, the 95th percentile is μ + z_{0.95} * σ, which is higher than μ. But in our case, we're dealing with the lower tail, so the 5th percentile is μ - z_{0.95} * σ.Wait, perhaps I confused the tails.Let me think again.We have X ~ Poisson(Λ). We want P(X ≥ k) ≥ 0.95, which is the same as P(X ≤ k - 1) ≤ 0.05.So, k - 1 is the 5th percentile of X.Therefore, using the normal approximation, the 5th percentile is μ - z_{0.95} * σ.So,k - 1 ≈ Λ - 1.6449 * sqrt(Λ)Therefore,k ≈ Λ - 1.6449 * sqrt(Λ) + 1But since k must be an integer, we take the ceiling of the right-hand side.Wait, but this gives k as approximately Λ minus something, which would be less than Λ, but the 5th percentile should be less than the mean, so k - 1 is less than Λ, so k is less than Λ + 1.But we need P(X ≥ k) ≥ 0.95, which would require k to be greater than the mean, right? Because the upper tail beyond k should be at least 95%.Wait, no, actually, no. If X is Poisson(Λ), the distribution is skewed to the right. The median is around Λ, but the 95th percentile is significantly higher.Wait, perhaps I'm confusing the direction.Wait, let's take an example. Suppose Λ = 10.Then, X ~ Poisson(10). The 95th percentile is around 16 or 17.But according to the formula above, k ≈ 10 - 1.6449 * sqrt(10) + 1 ≈ 10 - 5.22 + 1 ≈ 5.78, which is way too low.That can't be right because P(X ≥ 6) for Poisson(10) is much higher than 0.95.Wait, clearly, I have a mistake in my reasoning.Let me start over.We have X ~ Poisson(Λ). We need P(X ≥ k) ≥ 0.95.This is equivalent to P(X ≤ k - 1) ≤ 0.05.So, k - 1 is the 5th percentile of X.Therefore, k is the smallest integer such that P(X ≤ k - 1) ≤ 0.05.In terms of the normal approximation, the 5th percentile of X is approximately μ - z_{0.95} * σ.So,k - 1 ≈ Λ - 1.6449 * sqrt(Λ)Therefore,k ≈ Λ - 1.6449 * sqrt(Λ) + 1But in the example with Λ = 10, this gives k ≈ 10 - 5.22 + 1 ≈ 5.78, so k = 6.But for Poisson(10), P(X ≤ 5) ≈ 0.067, which is greater than 0.05, so we need k - 1 = 6, so k = 7.Wait, so actually, the approximation gives k ≈ 5.78, but the actual 5th percentile is around 6, so k = 7.Wait, maybe the formula is:k ≈ Λ + z_{0.95} * sqrt(Λ)But in that case, for Λ = 10, k ≈ 10 + 1.6449 * 3.16 ≈ 10 + 5.22 ≈ 15.22, so k = 16, which is more in line with the actual 95th percentile.Wait, now I'm confused.Wait, perhaps I need to consider that for the upper tail, we need to use the upper z-score.Wait, let's think carefully.We have X ~ Poisson(Λ). We want P(X ≥ k) ≥ 0.95.This is equivalent to P(X ≤ k - 1) ≤ 0.05.So, k - 1 is the 5th percentile of X.Therefore, using the normal approximation, the 5th percentile is μ - z_{0.95} * σ.So,k - 1 ≈ Λ - 1.6449 * sqrt(Λ)Thus,k ≈ Λ - 1.6449 * sqrt(Λ) + 1But in the example with Λ = 10, this gives k ≈ 10 - 5.22 + 1 ≈ 5.78, so k = 6.But for Poisson(10), P(X ≤ 5) ≈ 0.067, which is still greater than 0.05. So, we need k - 1 = 6, so k = 7.Wait, so the approximation underestimates k.Alternatively, perhaps the formula should be:k ≈ Λ + z_{0.95} * sqrt(Λ)But then, for Λ = 10, k ≈ 10 + 5.22 ≈ 15.22, so k = 16.But in reality, for Poisson(10), the 95th percentile is around 16 or 17.Wait, so perhaps the correct formula is:k ≈ Λ + z_{0.95} * sqrt(Λ)But then, why did we get that from the lower tail?Wait, maybe I need to think in terms of the upper tail.We have P(X ≥ k) ≥ 0.95.Using the normal approximation, P(X ≥ k) ≈ 1 - Φ((k - Λ)/sqrt(Λ)) ≥ 0.95So,Φ((k - Λ)/sqrt(Λ)) ≤ 0.05Therefore,(k - Λ)/sqrt(Λ) ≤ z_{0.05} = -1.6449So,k ≤ Λ - 1.6449 * sqrt(Λ)But this contradicts the earlier result.Wait, no, because if we're looking at the upper tail, P(X ≥ k) corresponds to the upper tail, so we need:1 - Φ((k - Λ)/sqrt(Λ)) ≥ 0.95Which implies:Φ((k - Λ)/sqrt(Λ)) ≤ 0.05Therefore,(k - Λ)/sqrt(Λ) ≤ z_{0.05} = -1.6449So,k ≤ Λ - 1.6449 * sqrt(Λ)But this suggests that k is less than Λ, which can't be right because the upper tail beyond k should be 95%, meaning k is much larger than Λ.Wait, this is confusing. Maybe the normal approximation isn't suitable for the upper tail because the Poisson distribution is skewed.Alternatively, perhaps we should use the inverse of the normal approximation.Wait, let's think differently.If we have X ~ Poisson(Λ), then for large Λ, X is approximately N(Λ, Λ). So, to find k such that P(X ≥ k) ≥ 0.95, we can write:P(X ≥ k) ≈ 1 - Φ((k - Λ)/sqrt(Λ)) ≥ 0.95So,Φ((k - Λ)/sqrt(Λ)) ≤ 0.05Therefore,(k - Λ)/sqrt(Λ) ≤ z_{0.05} = -1.6449So,k ≤ Λ - 1.6449 * sqrt(Λ)But this suggests that k is less than Λ, which is not correct because the upper tail beyond k should be 95%, meaning k should be much larger than Λ.Wait, perhaps I have the inequality direction wrong.Wait, if we want P(X ≥ k) ≥ 0.95, then the z-score should be positive because we're in the upper tail.Wait, no, because:P(X ≥ k) = 1 - P(X ≤ k - 1)Using the normal approximation,1 - Φ((k - 1 - Λ)/sqrt(Λ)) ≥ 0.95So,Φ((k - 1 - Λ)/sqrt(Λ)) ≤ 0.05Which implies,(k - 1 - Λ)/sqrt(Λ) ≤ z_{0.05} = -1.6449So,k - 1 ≤ Λ - 1.6449 * sqrt(Λ)Therefore,k ≤ Λ - 1.6449 * sqrt(Λ) + 1But again, this suggests k is less than Λ, which is not correct.Wait, perhaps the normal approximation is not suitable for the upper tail because the Poisson distribution is skewed. Maybe we should use a different approximation, like the Wilson-Hilferty approximation or something else.Alternatively, perhaps we can use the relationship between Poisson and chi-squared distributions.It is known that if X ~ Poisson(Λ), then 2(X + 0.5) ~ approximately chi-squared with 2 degrees of freedom.But I'm not sure if that helps here.Alternatively, using the fact that for Poisson distribution, the quantile can be approximated by:k ≈ Λ + sqrt(Λ) * z_{0.95} + (z_{0.95})^2 / 2But I'm not sure about the exact form.Wait, perhaps using the saddlepoint approximation or other methods, but that might be too advanced.Alternatively, perhaps the relationship is simply that k is approximately Λ + z_{0.95} * sqrt(Λ). So, k ≈ Λ + 1.6449 * sqrt(Λ). This would make sense because for large Λ, the Poisson distribution is approximately normal, and the 95th percentile is μ + z * σ.But in our earlier example with Λ = 10, this would give k ≈ 10 + 5.22 ≈ 15.22, which is close to the actual 95th percentile of Poisson(10), which is around 16.So, perhaps the relationship is:k ≈ Λ + z_{0.95} * sqrt(Λ)Where z_{0.95} is approximately 1.6449.Therefore, the minimal k is approximately Λ + 1.6449 * sqrt(Λ).But since k must be an integer, we take the ceiling of this value.So, the relationship is k ≈ Λ + 1.6449 * sqrt(Λ).But the problem states that sum_{v} λ_v = Λ. So, the total parameter is Λ.Therefore, the minimal k is approximately Λ + 1.6449 * sqrt(Λ).But to express it more formally, we can write:k ≈ Λ + z_{0.95} * sqrt(Λ)Where z_{0.95} is the 95th percentile of the standard normal distribution, approximately 1.6449.Therefore, the minimal k is roughly Λ plus 1.645 times the square root of Λ.However, since the problem asks for the relationship, not necessarily the approximation, perhaps we can express it in terms of the Poisson quantile function.But without exact computation, the best we can do is relate k to Λ and the square root of Λ.Therefore, the minimal k is approximately Λ + 1.645 * sqrt(Λ).So, summarizing:For part 1, f(S) is maximized when S is the subset of vertices with the highest degrees, specifically where all vertices in S have degree at least the average degree of S, and all vertices not in S have degree less than this average.For part 2, the minimal k is approximately Λ + 1.645 * sqrt(Λ), where Λ is the sum of all λ_v.But let me check the exact wording of the problem for part 2:"If the botanist wants to ensure that the probability P(X ≥ k) is at least 0.95 for a given threshold k, find the relationship between k, the parameters λ_v, and the number of patches n. What is the minimum k such that this inequality holds for sum_{v} λ_v = Λ?"So, they want the relationship, not necessarily an approximation. But since X is Poisson(Λ), the exact k is the smallest integer such that P(X ≤ k - 1) ≤ 0.05.But without exact computation, we can express k in terms of Λ and the Poisson quantile function.However, since the problem mentions the sum of Poisson distributions, which is Poisson(Λ), and asks for the relationship, perhaps the answer is that k is approximately Λ + z_{0.95} * sqrt(Λ), where z_{0.95} is the 95th percentile of the standard normal.Alternatively, if we use the exact Poisson quantile, it's the smallest k such that the cumulative distribution function up to k - 1 is ≤ 0.05.But since the problem asks for the relationship, perhaps the answer is:k ≈ Λ + 1.645 * sqrt(Λ)So, the minimal k is roughly Λ plus 1.645 times the square root of Λ.Therefore, the relationship is k ≈ Λ + 1.645 * sqrt(Λ).So, putting it all together:1. f(S) is maximized when S is the subset of vertices with the highest degrees, specifically where all vertices in S have degree at least the average degree of S, and all vertices not in S have degree less than this average.2. The minimal k is approximately Λ + 1.645 * sqrt(Λ), where Λ is the sum of all λ_v.But let me write the final answers more formally.For part 1, the conditions are that S is a subset where all vertices in S have degree at least the average degree of S, and all vertices not in S have degree less than the average degree of S.For part 2, the minimal k is approximately Λ + 1.645 * sqrt(Λ).But to express it more precisely, since the problem might expect an exact expression, perhaps using the inverse of the Poisson CDF, but without exact computation, the normal approximation is the way to go.Therefore, the relationship is k ≈ Λ + z_{0.95} * sqrt(Λ), where z_{0.95} ≈ 1.6449.So, the minimal k is approximately Λ + 1.645 * sqrt(Λ).Therefore, the final answers are:1. f(S) is maximized when S is the subset of vertices with the highest degrees, specifically where all vertices in S have degree at least the average degree of S, and all vertices not in S have degree less than this average.2. The minimal k is approximately Λ + 1.645 * sqrt(Λ).But let me check if there's a more precise way to express this without approximation.Alternatively, perhaps using Markov's inequality, but that would give a weaker bound.Markov's inequality states that P(X ≥ k) ≤ Λ / k.But we want P(X ≥ k) ≥ 0.95, so:Λ / k ≤ 0.05Therefore,k ≥ Λ / 0.05 = 20ΛBut this is a very loose bound and not tight.Therefore, the normal approximation is better.So, I think the answer is k ≈ Λ + 1.645 * sqrt(Λ).Therefore, the minimal k is approximately Λ + 1.645 * sqrt(Λ).So, summarizing:1. The function f(S) is maximized when S is the subset of vertices with the highest degrees, specifically where all vertices in S have degree at least the average degree of S, and all vertices not in S have degree less than this average.2. The minimal k is approximately Λ + 1.645 * sqrt(Λ), where Λ is the total sum of λ_v.Therefore, the final answers are:1. S should be the subset of vertices with the highest degrees, where each vertex in S has degree at least the average degree of S, and each vertex not in S has degree less than this average.2. The minimal k is approximately Λ + 1.645 * sqrt(Λ).But to write it more formally, perhaps using the inverse of the standard normal distribution.Alternatively, since the problem might expect an exact expression, perhaps it's better to write it in terms of the Poisson quantile function, but without an exact formula, the normal approximation is the way to go.Therefore, the final answers are:1. The conditions for maximizing f(S) are that S is a subset where all vertices in S have degree at least the average degree of S, and all vertices not in S have degree less than this average.2. The minimal k is approximately Λ + 1.645 * sqrt(Λ).So, in boxed form:1. boxed{S text{ is the subset of vertices with the highest degrees, where each vertex in } S text{ has degree at least the average degree of } S}2. boxed{k approx Lambda + 1.645 sqrt{Lambda}}</think>

answer:<think>Alright, so I have this problem about a geologist studying a pollution spill in a river. The concentration of pollutants is modeled by the function ( P(x, y, t) ). The first part asks me to find the general solution for ( P(x, y, t) ) given a specific partial differential equation (PDE) and an initial Gaussian distribution. The second part is about placing sensors at two points and figuring out when the concentration drops below a certain threshold.Let me start with the first part. The PDE given is:[frac{partial P}{partial t} = D left( frac{partial^2 P}{partial x^2} + frac{partial^2 P}{partial y^2} right) - kP]This looks like a diffusion equation with an additional decay term. So, it's a combination of diffusion and exponential decay. The initial condition is a Gaussian distribution:[P(x, y, 0) = P_0 e^{-frac{(x - x_0)^2 + (y - y_0)^2}{4sigma^2}}]Hmm, okay. So, I need to solve this PDE with the given initial condition. I remember that for linear PDEs, especially those that are parabolic like the diffusion equation, methods like separation of variables or Fourier transforms can be useful. Since the equation is in two spatial dimensions, maybe I can use the method of eigenfunction expansion or perhaps look for a solution in terms of Green's functions.Wait, the equation is linear, so maybe I can write the solution as a convolution of the initial condition with the Green's function of the operator. That might be a good approach. Let me recall that the Green's function for the diffusion equation in two dimensions is known, and with the decay term, it might just be a product of the standard Green's function and an exponential decay factor.Alternatively, maybe I can use the Fourier transform method. Let me try that. If I take the Fourier transform in both x and y directions, the PDE becomes an ordinary differential equation (ODE) in time. That might be manageable.So, let's denote the Fourier transform of ( P(x, y, t) ) as ( tilde{P}(k_x, k_y, t) ). Then, the Fourier transform of the PDE would be:[frac{partial tilde{P}}{partial t} = -D (k_x^2 + k_y^2) tilde{P} - k tilde{P}]This simplifies to:[frac{partial tilde{P}}{partial t} = - (D (k_x^2 + k_y^2) + k) tilde{P}]This is a simple ODE whose solution is:[tilde{P}(k_x, k_y, t) = tilde{P}(k_x, k_y, 0) e^{- (D (k_x^2 + k_y^2) + k) t}]Now, I need to find ( tilde{P}(k_x, k_y, 0) ), which is the Fourier transform of the initial condition ( P(x, y, 0) ). The initial condition is a Gaussian, so its Fourier transform should be another Gaussian. Let me compute that.The initial condition is:[P(x, y, 0) = P_0 e^{-frac{(x - x_0)^2 + (y - y_0)^2}{4sigma^2}}]The Fourier transform in two dimensions is:[tilde{P}(k_x, k_y, 0) = int_{-infty}^{infty} int_{-infty}^{infty} P(x, y, 0) e^{-i (k_x x + k_y y)} dx dy]Since the Gaussian is separable in x and y, I can write this as the product of two one-dimensional Fourier transforms:[tilde{P}(k_x, k_y, 0) = tilde{P}_x(k_x) tilde{P}_y(k_y)]Where each one-dimensional transform is:[tilde{P}_x(k_x) = int_{-infty}^{infty} P_0 e^{-frac{(x - x_0)^2}{4sigma^2}} e^{-i k_x x} dx][tilde{P}_y(k_y) = int_{-infty}^{infty} e^{-frac{(y - y_0)^2}{4sigma^2}} e^{-i k_y y} dy]I remember that the Fourier transform of a Gaussian ( e^{-a x^2} ) is another Gaussian ( sqrt{frac{pi}{a}} e^{-frac{k^2}{4a}} ). But here, we have a shifted Gaussian. So, I can use the shift property of Fourier transforms, which states that:[mathcal{F}{e^{-a(x - x_0)^2}} = sqrt{frac{pi}{a}} e^{-frac{k^2}{4a}} e^{-i k x_0}]So, applying this to ( tilde{P}_x(k_x) ):[tilde{P}_x(k_x) = P_0 sqrt{2pi sigma^2} e^{-frac{k_x^2 sigma^2}{2}} e^{-i k_x x_0}]Similarly, for ( tilde{P}_y(k_y) ):[tilde{P}_y(k_y) = sqrt{2pi sigma^2} e^{-frac{k_y^2 sigma^2}{2}} e^{-i k_y y_0}]Wait, hold on. The initial condition is ( P_0 e^{-frac{(x - x_0)^2 + (y - y_0)^2}{4sigma^2}} ). So, actually, the exponent is ( -frac{(x - x_0)^2}{4sigma^2} - frac{(y - y_0)^2}{4sigma^2} ). Therefore, in the Fourier transform, each integral will have ( a = frac{1}{4sigma^2} ), so ( sqrt{frac{pi}{a}} = sqrt{4pi sigma^2} = 2sigma sqrt{pi} ).Wait, let me recast that. Let me define ( a = frac{1}{4sigma^2} ), so ( sqrt{frac{pi}{a}} = sqrt{4pi sigma^2} = 2sigma sqrt{pi} ). So, each Fourier transform would be:[tilde{P}_x(k_x) = P_0 cdot 2sigma sqrt{pi} e^{-frac{k_x^2 sigma^2}{2}} e^{-i k_x x_0}][tilde{P}_y(k_y) = 2sigma sqrt{pi} e^{-frac{k_y^2 sigma^2}{2}} e^{-i k_y y_0}]Wait, no, actually, the initial condition is ( P_0 e^{-frac{(x - x_0)^2 + (y - y_0)^2}{4sigma^2}} ), so the integral over x is:[int_{-infty}^{infty} P_0 e^{-frac{(x - x_0)^2}{4sigma^2}} e^{-i k_x x} dx = P_0 cdot sqrt{4pi sigma^2} e^{-frac{k_x^2 sigma^2}{2}} e^{-i k_x x_0}]Similarly for y. So, actually, the Fourier transform of the initial condition is:[tilde{P}(k_x, k_y, 0) = P_0 cdot (2sigma sqrt{pi})^2 e^{-frac{(k_x^2 + k_y^2) sigma^2}{2}} e^{-i (k_x x_0 + k_y y_0)}]Simplifying, that's:[tilde{P}(k_x, k_y, 0) = P_0 cdot 4pi sigma^2 e^{-frac{(k_x^2 + k_y^2) sigma^2}{2}} e^{-i (k_x x_0 + k_y y_0)}]So, going back to the expression for ( tilde{P}(k_x, k_y, t) ):[tilde{P}(k_x, k_y, t) = P_0 cdot 4pi sigma^2 e^{-frac{(k_x^2 + k_y^2) sigma^2}{2}} e^{-i (k_x x_0 + k_y y_0)} e^{- (D (k_x^2 + k_y^2) + k) t}]Now, to find ( P(x, y, t) ), I need to take the inverse Fourier transform of ( tilde{P}(k_x, k_y, t) ). That is:[P(x, y, t) = frac{1}{(2pi)^2} int_{-infty}^{infty} int_{-infty}^{infty} tilde{P}(k_x, k_y, t) e^{i (k_x x + k_y y)} dk_x dk_y]Substituting ( tilde{P} ):[P(x, y, t) = frac{P_0 cdot 4pi sigma^2}{(2pi)^2} int_{-infty}^{infty} int_{-infty}^{infty} e^{-frac{(k_x^2 + k_y^2) sigma^2}{2}} e^{-i (k_x x_0 + k_y y_0)} e^{- (D (k_x^2 + k_y^2) + k) t} e^{i (k_x x + k_y y)} dk_x dk_y]Simplify the constants:[frac{4pi sigma^2}{(2pi)^2} = frac{4pi sigma^2}{4pi^2} = frac{sigma^2}{pi}]So,[P(x, y, t) = frac{P_0 sigma^2}{pi} int_{-infty}^{infty} int_{-infty}^{infty} e^{-frac{(k_x^2 + k_y^2) sigma^2}{2}} e^{-i (k_x x_0 + k_y y_0)} e^{- (D (k_x^2 + k_y^2) + k) t} e^{i (k_x x + k_y y)} dk_x dk_y]Let me combine the exponents:The exponentials involving ( k_x ) and ( k_y ) can be combined as:[e^{-frac{(k_x^2 + k_y^2) sigma^2}{2}} e^{- (D (k_x^2 + k_y^2) + k) t} e^{i k_x (x - x_0)} e^{i k_y (y - y_0)}]So, let me write the exponent as:[- left( frac{sigma^2}{2} + D t right) (k_x^2 + k_y^2) + i k_x (x - x_0) + i k_y (y - y_0) - k t]So, the integral becomes:[P(x, y, t) = frac{P_0 sigma^2}{pi} e^{-k t} int_{-infty}^{infty} int_{-infty}^{infty} e^{- left( frac{sigma^2}{2} + D t right) (k_x^2 + k_y^2)} e^{i k_x (x - x_0)} e^{i k_y (y - y_0)} dk_x dk_y]Now, this integral is separable into x and y components:[int_{-infty}^{infty} e^{- a k_x^2 + i k_x b} dk_x times int_{-infty}^{infty} e^{- a k_y^2 + i k_y c} dk_y]Where ( a = frac{sigma^2}{2} + D t ), ( b = x - x_0 ), and ( c = y - y_0 ).I remember that the integral ( int_{-infty}^{infty} e^{-a k^2 + i k b} dk = sqrt{frac{pi}{a}} e^{- frac{b^2}{4a}} ).So, applying this to both x and y integrals:Each integral becomes ( sqrt{frac{pi}{a}} e^{- frac{b^2}{4a}} ), so the product is ( frac{pi}{a} e^{- frac{b^2 + c^2}{4a}} ).Therefore, substituting back:[P(x, y, t) = frac{P_0 sigma^2}{pi} e^{-k t} times frac{pi}{a} e^{- frac{(x - x_0)^2 + (y - y_0)^2}{4a}}]Simplify:[P(x, y, t) = frac{P_0 sigma^2}{a} e^{-k t} e^{- frac{(x - x_0)^2 + (y - y_0)^2}{4a}}]But ( a = frac{sigma^2}{2} + D t ), so:[P(x, y, t) = frac{P_0 sigma^2}{frac{sigma^2}{2} + D t} e^{-k t} e^{- frac{(x - x_0)^2 + (y - y_0)^2}{4 left( frac{sigma^2}{2} + D t right)}}]Simplify the denominator:[frac{sigma^2}{frac{sigma^2}{2} + D t} = frac{2 sigma^2}{sigma^2 + 2 D t}]So,[P(x, y, t) = P_0 cdot frac{2 sigma^2}{sigma^2 + 2 D t} e^{-k t} e^{- frac{(x - x_0)^2 + (y - y_0)^2}{4 left( frac{sigma^2}{2} + D t right)}}]Let me rewrite the exponent in the Gaussian term:[frac{(x - x_0)^2 + (y - y_0)^2}{4 left( frac{sigma^2}{2} + D t right)} = frac{(x - x_0)^2 + (y - y_0)^2}{2 sigma^2 + 4 D t}]So, the entire expression becomes:[P(x, y, t) = frac{2 P_0 sigma^2}{sigma^2 + 2 D t} e^{-k t} e^{- frac{(x - x_0)^2 + (y - y_0)^2}{2 sigma^2 + 4 D t}}]Hmm, that seems a bit messy, but I think it's correct. Let me check the dimensions to see if everything makes sense. The exponent should be dimensionless, and the prefactor should have the same dimensions as P.Given that D has units of [length]^2/[time], k has units of [1/time], and σ has units of [length]. So, in the exponent:- ( frac{(x - x_0)^2}{2 sigma^2 + 4 D t} ): numerator is [length]^2, denominator is [length]^2, so dimensionless.- Similarly for the y term.- ( -k t ): dimensionless.The prefactor:- ( frac{2 P_0 sigma^2}{sigma^2 + 2 D t} ): numerator is [concentration][length]^2, denominator is [length]^2, so overall [concentration], which matches P.So, that seems consistent.Therefore, the general solution is:[P(x, y, t) = frac{2 P_0 sigma^2}{sigma^2 + 2 D t} e^{-k t} e^{- frac{(x - x_0)^2 + (y - y_0)^2}{2 sigma^2 + 4 D t}}]Alternatively, we can write this as:[P(x, y, t) = frac{2 P_0 sigma^2}{sigma^2 + 2 D t} e^{-k t - frac{(x - x_0)^2 + (y - y_0)^2}{2 sigma^2 + 4 D t}}]That should be the general solution.Now, moving on to part 2. The geologist places sensors at points ( (x_1, y_1) ) and ( (x_2, y_2) ). I need to derive the expressions for the pollutant concentration at these points over time and calculate the time ( t ) at which the concentration first drops below a threshold ( P_{text{threshold}} ).So, for each sensor location, plug in ( x = x_i ), ( y = y_i ) into the general solution.Let me denote ( r_i^2 = (x_i - x_0)^2 + (y_i - y_0)^2 ). So, for each sensor, the concentration is:[P(x_i, y_i, t) = frac{2 P_0 sigma^2}{sigma^2 + 2 D t} e^{-k t - frac{r_i^2}{2 sigma^2 + 4 D t}}]Simplify the exponent:Let me write the exponent as:[- k t - frac{r_i^2}{2 sigma^2 + 4 D t} = - left( k t + frac{r_i^2}{2 sigma^2 + 4 D t} right )]So, the concentration is:[P(x_i, y_i, t) = frac{2 P_0 sigma^2}{sigma^2 + 2 D t} e^{- left( k t + frac{r_i^2}{2 sigma^2 + 4 D t} right )}]Now, to find the time ( t ) when ( P(x_i, y_i, t) = P_{text{threshold}} ), we set up the equation:[frac{2 P_0 sigma^2}{sigma^2 + 2 D t} e^{- left( k t + frac{r_i^2}{2 sigma^2 + 4 D t} right )} = P_{text{threshold}}]This equation looks transcendental, meaning it's unlikely to have a closed-form solution. So, we might need to solve it numerically. However, perhaps we can manipulate it a bit to make it more manageable.Let me denote ( s = D t ). Then, ( t = s / D ). Let me substitute this into the equation:First, rewrite the equation:[frac{2 P_0 sigma^2}{sigma^2 + 2 s} e^{- left( frac{k s}{D} + frac{r_i^2}{2 sigma^2 + 4 s} right )} = P_{text{threshold}}]Hmm, maybe not particularly helpful. Alternatively, let me take natural logarithms on both sides to linearize the equation:[ln left( frac{2 P_0 sigma^2}{sigma^2 + 2 D t} right ) - k t - frac{r_i^2}{2 sigma^2 + 4 D t} = ln P_{text{threshold}}]Let me denote ( A = 2 P_0 sigma^2 ), ( B = sigma^2 ), ( C = k ), ( D ) is already used, ( E = r_i^2 ). So, the equation becomes:[ln left( frac{A}{B + 2 D t} right ) - C t - frac{E}{B + 4 D t} = ln P_{text{threshold}}]But this still seems complicated. Maybe we can consider that for small times, the exponential decay is dominated by the ( -k t ) term, and for larger times, the diffusion term becomes more significant. But without specific values, it's hard to approximate.Alternatively, perhaps we can make a substitution to simplify the expression. Let me set ( u = D t ), so ( t = u / D ). Then, the equation becomes:[frac{2 P_0 sigma^2}{sigma^2 + 2 u} e^{- left( frac{k u}{D} + frac{r_i^2}{2 sigma^2 + 4 u} right )} = P_{text{threshold}}]Again, not particularly helpful. Maybe another substitution? Let me think.Alternatively, perhaps we can write the equation as:[frac{2 P_0 sigma^2}{sigma^2 + 2 D t} e^{-k t} e^{- frac{r_i^2}{2 sigma^2 + 4 D t}} = P_{text{threshold}}]Let me denote ( tau = t ), then:[frac{2 P_0 sigma^2}{sigma^2 + 2 D tau} e^{-k tau} e^{- frac{r_i^2}{2 sigma^2 + 4 D tau}} = P_{text{threshold}}]This still seems difficult to solve analytically. Maybe we can consider that for the time when the concentration drops below the threshold, the exponential term is the dominant factor. But I'm not sure.Alternatively, perhaps we can make an approximation for small times or large times.Wait, for small times, the term ( 2 D t ) is much smaller than ( sigma^2 ), so the denominator ( sigma^2 + 2 D t approx sigma^2 ). Similarly, the term ( 2 sigma^2 + 4 D t approx 2 sigma^2 ). So, in that case, the concentration simplifies to:[P(x_i, y_i, t) approx frac{2 P_0 sigma^2}{sigma^2} e^{-k t} e^{- frac{r_i^2}{2 sigma^2}} = 2 P_0 e^{-k t} e^{- frac{r_i^2}{2 sigma^2}}]But wait, that's only if ( 2 D t ll sigma^2 ). So, for small t, the concentration decreases exponentially due to decay and the initial Gaussian spread.For larger times, the terms involving t dominate, so the denominator ( sigma^2 + 2 D t ) and the exponent ( frac{r_i^2}{2 sigma^2 + 4 D t} ) become more influenced by t.But without specific values, it's hard to make a general approximation. Therefore, perhaps the best approach is to recognize that the equation is transcendental and needs to be solved numerically.Alternatively, we can write the equation in terms of dimensionless variables. Let me define:Let ( tau = D t ), so ( t = tau / D ). Let me also define ( mu = k / D ), which is a dimensionless constant. Then, the equation becomes:[frac{2 P_0 sigma^2}{sigma^2 + 2 tau} e^{- mu tau - frac{r_i^2}{2 sigma^2 + 4 tau}} = P_{text{threshold}}]This substitution might help in some cases, but it's still not straightforward.Alternatively, perhaps we can write the equation as:[frac{2 P_0 sigma^2}{sigma^2 + 2 D t} e^{-k t} e^{- frac{r_i^2}{2 sigma^2 + 4 D t}} = P_{text{threshold}}]Let me take the natural logarithm:[ln left( frac{2 P_0 sigma^2}{sigma^2 + 2 D t} right ) - k t - frac{r_i^2}{2 sigma^2 + 4 D t} = ln P_{text{threshold}}]Let me denote ( f(t) = ln left( frac{2 P_0 sigma^2}{sigma^2 + 2 D t} right ) - k t - frac{r_i^2}{2 sigma^2 + 4 D t} - ln P_{text{threshold}} = 0 )We can solve this equation numerically using methods like Newton-Raphson. However, since we don't have specific values, we can't compute it here. But the expression is set up for numerical computation.Alternatively, if we assume that the decay term ( k t ) is negligible compared to the diffusion term, or vice versa, we might approximate.But without knowing the relative magnitudes of D, k, and the distances ( r_i ), it's hard to make such assumptions.Therefore, the conclusion is that the time ( t ) when the concentration drops below ( P_{text{threshold}} ) can be found by solving the equation:[frac{2 P_0 sigma^2}{sigma^2 + 2 D t} e^{-k t - frac{r_i^2}{2 sigma^2 + 4 D t}} = P_{text{threshold}}]Numerically, for each sensor location ( (x_i, y_i) ), compute ( r_i ), then solve for ( t ) using numerical methods.So, summarizing:1. The general solution is:[P(x, y, t) = frac{2 P_0 sigma^2}{sigma^2 + 2 D t} e^{-k t} e^{- frac{(x - x_0)^2 + (y - y_0)^2}{2 sigma^2 + 4 D t}}]2. For each sensor at ( (x_i, y_i) ), the concentration is:[P(x_i, y_i, t) = frac{2 P_0 sigma^2}{sigma^2 + 2 D t} e^{-k t - frac{r_i^2}{2 sigma^2 + 4 D t}}]And the time ( t ) when this drops below ( P_{text{threshold}} ) is found by solving the equation above numerically.I think that's as far as I can go analytically. The key takeaway is that the solution involves a Gaussian that spreads over time due to diffusion, while also decaying exponentially due to the decay term ( k ). The concentration at any point depends on both the spreading of the Gaussian and the exponential decay.Final Answer1. The general solution for the pollutant concentration is:[boxed{P(x, y, t) = frac{2 P_0 sigma^2}{sigma^2 + 2 D t} e^{-k t} e^{- frac{(x - x_0)^2 + (y - y_0)^2}{2 sigma^2 + 4 D t}}}]2. The time ( t ) at which the concentration at each sensor first drops below ( P_{text{threshold}} ) is found by solving:[boxed{frac{2 P_0 sigma^2}{sigma^2 + 2 D t} e^{-k t - frac{(x_i - x_0)^2 + (y_i - y_0)^2}{2 sigma^2 + 4 D t}} = P_{text{threshold}}}]for each sensor location ( (x_i, y_i) ).</think>

answer:<think>Okay, so Jane is trying to figure out which insulation material is better for her house. She has two materials, A and B, each with different thermal conductivities and thicknesses. She wants to minimize heat loss, so I need to help her calculate which one is better and maybe even a combination of both.First, let's tackle part 1. She wants the total heat loss per hour through the walls for each material. The formula given is Q = (k * A * ΔT * t) / d. I need to plug in the values for each material.Starting with Material A. The thermal conductivity k_A is 0.035 W/(m·K). The area A is 200 m². The temperature difference ΔT is 20°C, which is the same as 20 K since the size of the degree is the same in Celsius and Kelvin. The time t is 1 hour, which I need to convert into seconds because the units of thermal conductivity are in watts, which is joules per second. So 1 hour is 3600 seconds. The thickness d_A is 0.1 meters.Plugging into the formula: Q_A = (0.035 * 200 * 20 * 3600) / 0.1. Let me compute that step by step.First, multiply 0.035 by 200: 0.035 * 200 = 7.Then multiply by 20: 7 * 20 = 140.Multiply by 3600: 140 * 3600. Hmm, 140 * 3600. Let's see, 100 * 3600 = 360,000, 40 * 3600 = 144,000. So total is 360,000 + 144,000 = 504,000.Then divide by 0.1: 504,000 / 0.1 = 5,040,000 Joules per hour. So Q_A is 5,040,000 J/hour.Now for Material B. k_B is 0.025 W/(m·K). The area A is still 200 m². ΔT is still 20 K, time is 3600 seconds, and thickness d_B is 0.08 meters.So Q_B = (0.025 * 200 * 20 * 3600) / 0.08.Calculating step by step:0.025 * 200 = 5.5 * 20 = 100.100 * 3600 = 360,000.Divide by 0.08: 360,000 / 0.08. Let's see, 360,000 divided by 0.08 is the same as 360,000 * 12.5, because 1/0.08 is 12.5. So 360,000 * 12.5.Breaking that down: 360,000 * 10 = 3,600,000; 360,000 * 2.5 = 900,000. So total is 3,600,000 + 900,000 = 4,500,000 J/hour.So Q_B is 4,500,000 J/hour.Comparing the two, Material B results in less heat loss than Material A. So if she uses Material B, she'll lose less heat per hour.Now, moving on to part 2. Jane wants to use a combination of both materials to minimize heat loss. The proportion of Material A is x, and Material B is 1 - x. The total thickness remains constant at 0.1 meters.I need to derive the effective thermal conductivity k_eff for the combined setup. Then find the value of x that minimizes heat loss.First, since the total thickness is constant, the thickness of each material in the combination will be proportional to their proportions. So, the thickness of Material A will be x * 0.1 meters, and the thickness of Material B will be (1 - x) * 0.1 meters.Wait, actually, is that correct? Or is the total thickness the sum of the individual thicknesses? Hmm, the problem says the total thickness remains constant at 0.1 meters. So, if she uses a combination, the thickness of A plus the thickness of B equals 0.1 meters. So, if the proportion of A is x, then the thickness of A is x * d_total, which is x * 0.1, and thickness of B is (1 - x) * 0.1.But actually, wait, in insulation, when you have multiple layers, the total thermal resistance is the sum of each layer's thermal resistance. Thermal resistance R is given by R = d / k. So, for each material, R_A = d_A / k_A, R_B = d_B / k_B. The total R_total is R_A + R_B.But in this case, the total thickness is fixed, so d_A + d_B = 0.1 m. So, if we let d_A = x * 0.1, then d_B = (1 - x) * 0.1.Therefore, the total thermal resistance R_total = d_A / k_A + d_B / k_B = (x * 0.1) / k_A + ((1 - x) * 0.1) / k_B.But we need the effective thermal conductivity k_eff. How do we relate R_total to k_eff?Wait, the effective thermal conductivity for a composite material with layers in series is not straightforward. It's actually the total thermal resistance that matters, not a single k_eff. But perhaps the question is considering the combination as a single material with an effective k, so maybe it's treating the combined material as a parallel or series combination.Wait, no, in building insulation, materials are typically in series, meaning one layer after another. So the total thermal resistance is the sum of each layer's resistance.But the formula for heat transfer is Q = (A * ΔT * t) / R_total.So, if we have two materials in series, the total R is R_A + R_B. So, the effective thermal conductivity k_eff would be such that R_total = d_total / k_eff. But wait, that's only if the materials were in parallel, but in series, it's additive.Wait, maybe I need to think differently. Let me recall the formula for combined thermal conductivities.When materials are in series (i.e., layered), the total thermal resistance is the sum of individual resistances. So, R_total = R_A + R_B = d_A / k_A + d_B / k_B.If we want to express this as an effective thermal conductivity, assuming the total thickness is d_total = d_A + d_B, then k_eff would satisfy R_total = d_total / k_eff.So, k_eff = d_total / R_total.Therefore, k_eff = d_total / (d_A / k_A + d_B / k_B).Given that d_total is 0.1 m, d_A = x * 0.1, d_B = (1 - x) * 0.1.Plugging in:k_eff = 0.1 / [ (x * 0.1) / k_A + ((1 - x) * 0.1) / k_B ]Simplify:k_eff = 0.1 / [ 0.1 (x / k_A + (1 - x) / k_B ) ]The 0.1 cancels out:k_eff = 1 / (x / k_A + (1 - x) / k_B )So that's the expression for k_eff.Now, to minimize heat loss, we need to minimize Q. Since Q = (k_eff * A * ΔT * t) / d_total, and d_total is fixed, minimizing Q is equivalent to minimizing k_eff. Because A, ΔT, t, and d_total are constants.So, to minimize Q, we need to minimize k_eff. Therefore, we need to find the value of x that minimizes k_eff.So, let's write k_eff as:k_eff(x) = 1 / [ (x / k_A) + ( (1 - x) / k_B ) ]We can write this as:k_eff(x) = 1 / [ (x / 0.035) + ( (1 - x) / 0.025 ) ]We need to find the x that minimizes k_eff(x). Since k_eff is a function of x, we can take the derivative of k_eff with respect to x, set it equal to zero, and solve for x.But before taking derivatives, let me note that k_eff is a reciprocal function. So, to minimize k_eff, we need to maximize the denominator, because k_eff = 1 / denominator. So, maximizing the denominator will minimize k_eff.Therefore, instead of minimizing k_eff, we can equivalently maximize the denominator function:f(x) = (x / 0.035) + ( (1 - x) / 0.025 )So, let's define f(x) = x / 0.035 + (1 - x) / 0.025.We need to find the x that maximizes f(x).Compute f(x):f(x) = (x / 0.035) + ( (1 - x) / 0.025 )Let me compute the coefficients:1 / 0.035 ≈ 28.57141 / 0.025 = 40So, f(x) = 28.5714 x + 40 (1 - x) = 28.5714 x + 40 - 40x = (28.5714 - 40) x + 40 = (-11.4286) x + 40.So, f(x) is a linear function in x with a negative slope. Therefore, it is maximized when x is minimized.Since x is the proportion of Material A, which has a higher thermal conductivity (0.035) compared to Material B (0.025). So, Material B is a better insulator.Therefore, to maximize f(x), which is equivalent to minimizing k_eff, we should minimize x, meaning use as much Material B as possible.But x is the proportion of Material A, so the minimal x is 0, meaning use all Material B.Wait, but the total thickness is fixed at 0.1 m. If x = 0, then d_A = 0, and d_B = 0.1 m. That would give the maximum f(x) because f(x) = 40 when x=0.Similarly, if x=1, f(x) = 28.5714, which is less than 40.Therefore, the maximum of f(x) is at x=0, so the minimal k_eff is when x=0.But wait, that seems counterintuitive because if we use all Material B, which is better, we get less heat loss. So, indeed, the minimal heat loss is achieved when x=0, meaning using only Material B.But let me double-check. If we use a combination, maybe we can get a better effective k? Wait, no, because Material B has lower k, so using more of it would give a lower k_eff.Wait, but in our expression, k_eff = 1 / [ (x / k_A) + ( (1 - x) / k_B ) ]Since k_B < k_A, (1 - x)/k_B is larger than (1 - x)/k_A. So, as x decreases, the denominator increases, so k_eff decreases.Therefore, to minimize k_eff, we need to minimize x, i.e., maximize the proportion of Material B.Hence, the minimal heat loss occurs when x=0, meaning using only Material B.But wait, let me think again. If we use only Material B, the thickness is 0.1 m, but Material B's original thickness was 0.08 m. Wait, in the first part, Material B was 0.08 m, but in the combination, we are allowing the thickness to be up to 0.1 m.Wait, hold on. In the first part, Material B had a thickness of 0.08 m, but in the combination, we're considering the total thickness as 0.1 m, so if we use only Material B, its thickness would be 0.1 m, which is more than its original 0.08 m. So, actually, in the first part, the thickness was fixed for each material, but in the combination, the thickness is variable as per the proportion.So, in the first part, for Material B, the thickness was 0.08 m, but in the combination, if we set x=0, the thickness of B becomes 0.1 m, which is thicker, so it should provide better insulation, meaning lower heat loss.Wait, but in the first part, the heat loss was calculated with the given thickness. So, if we can increase the thickness of Material B, we can get even lower heat loss.But in the combination, the total thickness is fixed at 0.1 m, so if we use more of Material B, we can have a lower k_eff.Wait, but in the first part, Material B was 0.08 m, which is less than 0.1 m. So, if we use all Material B with 0.1 m, it's better than using 0.08 m.But in the first part, the heat loss for Material B was calculated with its original thickness, which was 0.08 m. So, if we use 0.1 m of Material B, the heat loss would be even less.But in the combination, we are considering varying the proportion x, but keeping the total thickness at 0.1 m. So, if x=0, d_B=0.1 m, which is better than d_B=0.08 m. So, indeed, using all Material B in the combination setup (with 0.1 m thickness) would give lower heat loss than using Material B alone with 0.08 m.Therefore, the conclusion is that to minimize heat loss, Jane should use as much Material B as possible, i.e., x=0.But let me verify this with calculus to be thorough.We have k_eff(x) = 1 / [ (x / 0.035) + ( (1 - x) / 0.025 ) ]Let me write f(x) = (x / 0.035) + ( (1 - x) / 0.025 )We can compute f(x) = (x / 0.035) + ( (1 - x) / 0.025 ) = (x / 0.035) + (1 / 0.025) - (x / 0.025) = (x)(1/0.035 - 1/0.025) + 1/0.025Compute 1/0.035 ≈ 28.5714, 1/0.025 = 40.So, f(x) = x(28.5714 - 40) + 40 = x(-11.4286) + 40.So, f(x) is a linear function decreasing with x. Therefore, it's maximum at x=0, minimum at x=1.Therefore, k_eff(x) = 1 / f(x) is minimized when f(x) is maximized, which is at x=0.Hence, the minimal k_eff occurs at x=0, so Jane should use all Material B.Therefore, the value of x that minimizes heat loss is 0.But wait, let me think again. If x=0, we're using all Material B with thickness 0.1 m. In the first part, Material B had a thickness of 0.08 m, so using 0.1 m would be better. So, indeed, the minimal heat loss is achieved when x=0.Alternatively, if we set x=1, we're using all Material A with 0.1 m, which is worse than using Material B.Therefore, the optimal x is 0.But let me compute the heat loss for x=0 and x=1 to confirm.For x=0:k_eff = 1 / [0 / 0.035 + (1 - 0)/0.025] = 1 / (0 + 40) = 1/40 = 0.025 W/(m·K). Wait, that's the same as Material B's k. But wait, Material B's k is 0.025, but in the combination, we're using 0.1 m of it, whereas in the first part, it was 0.08 m.Wait, but in the first part, the heat loss was calculated with the given thickness. So, if we use 0.1 m of Material B, the heat loss would be:Q = (0.025 * 200 * 20 * 3600) / 0.1 = (0.025 * 200 * 20 * 3600) / 0.1.Compute numerator: 0.025 * 200 = 5; 5 * 20 = 100; 100 * 3600 = 360,000.Divide by 0.1: 360,000 / 0.1 = 3,600,000 J/hour.Wait, but in the first part, with 0.08 m, it was 4,500,000 J/hour. So, using 0.1 m of Material B gives lower heat loss, as expected.Similarly, if we use x=1, k_eff = 1 / [1 / 0.035 + 0 / 0.025] = 1 / (28.5714) ≈ 0.035 W/(m·K), which is Material A's k. Then, heat loss would be:Q = (0.035 * 200 * 20 * 3600) / 0.1 = same as Material A in part 1, which was 5,040,000 J/hour.So, indeed, using x=0 gives lower heat loss than x=1.Therefore, the optimal x is 0, meaning use all Material B with 0.1 m thickness.But wait, let me think if there's a way to get even lower k_eff by combining both materials. Since k_eff is 1 / [ (x / k_A) + ( (1 - x) / k_B ) ], and since k_B < k_A, the term (1 - x)/k_B is larger than (1 - x)/k_A. So, as x decreases, the denominator increases, making k_eff smaller. Therefore, the minimal k_eff is achieved when x is as small as possible, which is x=0.Hence, the conclusion is that x=0 minimizes heat loss.But just to be thorough, let's compute k_eff for x=0 and x=1.For x=0:k_eff = 1 / [0 + 1/0.025] = 1 / 40 = 0.025 W/(m·K).For x=1:k_eff = 1 / [1/0.035 + 0] ≈ 1 / 28.5714 ≈ 0.035 W/(m·K).So, indeed, k_eff is minimized at x=0.Therefore, the value of x that minimizes heat loss is 0.But wait, let me think again. If we use a combination, maybe we can get a lower k_eff than 0.025? No, because k_eff is the effective conductivity, and since Material B has k=0.025, which is lower than Material A's 0.035, using more of B will give a lower k_eff. But since B's k is already the lower one, using all B gives the minimal k_eff.Therefore, the optimal x is 0.But let me also consider the derivative approach to confirm.We have k_eff(x) = 1 / [ (x / 0.035) + ( (1 - x) / 0.025 ) ]Let me denote f(x) = (x / 0.035) + ( (1 - x) / 0.025 )So, k_eff(x) = 1 / f(x)To minimize k_eff, we need to maximize f(x).Compute derivative of f(x):f'(x) = 1/0.035 - 1/0.025 ≈ 28.5714 - 40 = -11.4286Since f'(x) is negative, f(x) is decreasing in x. Therefore, maximum at x=0.Hence, x=0 is the point where f(x) is maximized, leading to k_eff minimized.Therefore, the answer is x=0.But wait, let me think about the physical meaning. If we have two materials in series, the total thermal resistance is the sum. So, using more of the material with lower k (better insulator) will give higher total R, hence lower heat loss.Therefore, to minimize heat loss, we should maximize the thermal resistance, which is achieved by using as much as possible of the material with lower k, which is Material B.Hence, x=0.So, summarizing:1. Heat loss for A: 5,040,000 J/hour Heat loss for B: 4,500,000 J/hour2. The effective thermal conductivity k_eff = 1 / [ (x / 0.035) + ( (1 - x) / 0.025 ) ] The value of x that minimizes heat loss is 0.Therefore, Jane should use only Material B with a thickness of 0.1 meters to minimize heat loss.</think>

answer:<think>Okay, so I'm a student at Lee University, and I need to analyze the effectiveness of three communication channels: social media, email newsletters, and webinars. The task has two parts: reach analysis and engagement analysis. Let me try to break this down step by step.Starting with the reach analysis. I need to calculate the total reach for each channel over the first 12 months. The reach functions are given as functions of time ( t ) in months. For each channel, I have a specific function:- Social Media: ( R_{SM}(t) = 500t^2 - 1500t + 2000 )- Email Newsletters: ( R_{EN}(t) = 800ln(t + 1) + 1000 )- Webinars: ( R_{WB}(t) = 1000e^{-0.1t} + 1500 )To find the total reach over the first 12 months, I need to integrate each of these functions from ( t = 0 ) to ( t = 12 ). Integration will give me the area under the curve, which represents the cumulative reach over that time period.Let me tackle each channel one by one.1. Social Media Reach:The function is ( R_{SM}(t) = 500t^2 - 1500t + 2000 ). This is a quadratic function, and integrating it should be straightforward.The integral of ( R_{SM}(t) ) from 0 to 12 is:[int_{0}^{12} (500t^2 - 1500t + 2000) , dt]Let me compute the integral term by term.- Integral of ( 500t^2 ) is ( frac{500}{3}t^3 )- Integral of ( -1500t ) is ( -frac{1500}{2}t^2 = -750t^2 )- Integral of ( 2000 ) is ( 2000t )So, putting it all together:[left[ frac{500}{3}t^3 - 750t^2 + 2000t right]_{0}^{12}]Now, plugging in ( t = 12 ):First term: ( frac{500}{3} times 12^3 )12 cubed is 1728, so ( frac{500}{3} times 1728 = 500 times 576 = 288,000 )Second term: ( -750 times 12^2 )12 squared is 144, so ( -750 times 144 = -108,000 )Third term: ( 2000 times 12 = 24,000 )Adding these together: 288,000 - 108,000 + 24,000 = 204,000Now, plugging in ( t = 0 ) gives all terms as zero, so the total reach is 204,000.Wait, that seems a bit high. Let me double-check my calculations.First term: ( frac{500}{3} times 1728 )1728 divided by 3 is 576, so 500 * 576 is indeed 288,000.Second term: 750 * 144. Let me compute 750 * 100 = 75,000; 750 * 44 = 33,000. So total is 75,000 + 33,000 = 108,000. Since it's negative, it's -108,000.Third term: 2000 * 12 = 24,000.So 288,000 - 108,000 = 180,000; 180,000 + 24,000 = 204,000. Yeah, that seems correct.So total reach for Social Media is 204,000.2. Email Newsletters Reach:The function is ( R_{EN}(t) = 800ln(t + 1) + 1000 ). This involves integrating a natural logarithm function.The integral from 0 to 12 is:[int_{0}^{12} (800ln(t + 1) + 1000) , dt]I can split this into two integrals:[800 int_{0}^{12} ln(t + 1) , dt + int_{0}^{12} 1000 , dt]First, let's compute ( int ln(t + 1) , dt ). Integration by parts is needed here. Let me recall that ( int ln(u) du = uln(u) - u + C ).Let ( u = t + 1 ), so ( du = dt ). Then,[int ln(t + 1) dt = (t + 1)ln(t + 1) - (t + 1) + C]So, evaluating from 0 to 12:At t = 12: ( (13)ln(13) - 13 )At t = 0: ( (1)ln(1) - 1 = 0 - 1 = -1 )So the definite integral is:( [13ln(13) - 13] - [-1] = 13ln(13) - 13 + 1 = 13ln(13) - 12 )Multiplying by 800:( 800 times (13ln(13) - 12) )Let me compute this numerically.First, compute ( ln(13) ). I know that ( ln(10) approx 2.3026 ), ( ln(13) ) is a bit more. Let me use calculator approximation: ( ln(13) approx 2.5649 ).So, 13 * 2.5649 ≈ 33.3437Then, 33.3437 - 12 = 21.3437Multiply by 800: 21.3437 * 800 ≈ 17,075Wait, 21 * 800 = 16,800; 0.3437 * 800 ≈ 275. So total ≈ 16,800 + 275 = 17,075.Now, the second integral is ( int_{0}^{12} 1000 dt = 1000t ) evaluated from 0 to 12, which is 1000*12 - 1000*0 = 12,000.So total reach for Email Newsletters is 17,075 + 12,000 = 29,075.Wait, that seems low compared to Social Media. Let me check my calculations again.Wait, 800*(13 ln13 -12). Let me compute 13 ln13:13 * 2.5649 ≈ 33.343733.3437 - 12 = 21.343721.3437 * 800 = 17,075. That's correct.Then, adding 12,000 gives 29,075. Hmm, okay, maybe it's correct.3. Webinars Reach:The function is ( R_{WB}(t) = 1000e^{-0.1t} + 1500 ). Integrating this from 0 to 12.So,[int_{0}^{12} (1000e^{-0.1t} + 1500) , dt]Again, split into two integrals:[1000 int_{0}^{12} e^{-0.1t} dt + int_{0}^{12} 1500 dt]First integral: ( int e^{-0.1t} dt ). The integral of ( e^{kt} ) is ( frac{1}{k}e^{kt} ). So here, k = -0.1.Thus,[int e^{-0.1t} dt = frac{1}{-0.1} e^{-0.1t} + C = -10 e^{-0.1t} + C]Evaluated from 0 to 12:At t = 12: ( -10 e^{-1.2} )At t = 0: ( -10 e^{0} = -10 )So the definite integral is:( (-10 e^{-1.2}) - (-10) = -10 e^{-1.2} + 10 = 10(1 - e^{-1.2}) )Multiply by 1000:( 1000 * 10(1 - e^{-1.2}) = 10,000(1 - e^{-1.2}) )Compute ( e^{-1.2} ). I know that ( e^{-1} approx 0.3679 ), ( e^{-1.2} ) is a bit less. Let me compute it:Using Taylor series or calculator approximation: ( e^{-1.2} ≈ 0.3012 )So, 1 - 0.3012 = 0.6988Multiply by 10,000: 0.6988 * 10,000 = 6,988Second integral: ( int_{0}^{12} 1500 dt = 1500*12 = 18,000 )So total reach for Webinars is 6,988 + 18,000 = 24,988Wait, let me verify:10,000*(1 - e^{-1.2}) ≈ 10,000*(1 - 0.3012) ≈ 10,000*0.6988 ≈ 6,988. Correct.1500*12 = 18,000. Correct.Total: 6,988 + 18,000 = 24,988. Okay.So, summarizing the reach:- Social Media: 204,000- Email Newsletters: 29,075- Webinars: 24,988Wait, that seems like Social Media has a significantly higher reach than the other two. That makes sense because the quadratic function grows faster, while the others have logarithmic and exponential decay.Now, moving on to the engagement analysis. Engagement is given as a percentage of reach for each channel, with functions:- Social Media: ( E_{SM}(t) = 10sinleft(frac{pi t}{6}right) + 20 )- Email Newsletters: ( E_{EN}(t) = 15cosleft(frac{pi t}{6}right) + 30 )- Webinars: ( E_{WB}(t) = 5e^{-0.05t} + 25 )I need to find the average engagement percentage over the first 12 months. The average value of a function over an interval [a, b] is given by:[frac{1}{b - a} int_{a}^{b} E(t) dt]So, for each channel, compute the integral of E(t) from 0 to 12, then divide by 12.Let me compute each one.1. Social Media Engagement:Function: ( E_{SM}(t) = 10sinleft(frac{pi t}{6}right) + 20 )Average engagement:[frac{1}{12} int_{0}^{12} left(10sinleft(frac{pi t}{6}right) + 20right) dt]Let me compute the integral:[int_{0}^{12} 10sinleft(frac{pi t}{6}right) dt + int_{0}^{12} 20 dt]First integral:Let ( u = frac{pi t}{6} ), so ( du = frac{pi}{6} dt ), which implies ( dt = frac{6}{pi} du )So,[10 int sin(u) cdot frac{6}{pi} du = frac{60}{pi} int sin(u) du = frac{60}{pi} (-cos(u)) + C = -frac{60}{pi} cosleft(frac{pi t}{6}right) + C]Evaluated from 0 to 12:At t = 12: ( -frac{60}{pi} cos(2pi) = -frac{60}{pi} (1) = -frac{60}{pi} )At t = 0: ( -frac{60}{pi} cos(0) = -frac{60}{pi} (1) = -frac{60}{pi} )So, the definite integral is:( (-frac{60}{pi}) - (-frac{60}{pi}) = 0 )Wait, that's interesting. The integral of the sine function over a full period is zero. Since the period of ( sin(frac{pi t}{6}) ) is ( frac{2pi}{pi/6} = 12 ) months, so over 0 to 12, it's exactly one full period. Hence, the integral is zero.Second integral:[int_{0}^{12} 20 dt = 20t bigg|_{0}^{12} = 240 - 0 = 240]So, total integral is 0 + 240 = 240Average engagement:( frac{240}{12} = 20 )%Wait, that seems straightforward. The sine function averages out over the period, so only the constant term contributes. So, average engagement is 20%.2. Email Newsletters Engagement:Function: ( E_{EN}(t) = 15cosleft(frac{pi t}{6}right) + 30 )Average engagement:[frac{1}{12} int_{0}^{12} left(15cosleft(frac{pi t}{6}right) + 30right) dt]Compute the integral:[int_{0}^{12} 15cosleft(frac{pi t}{6}right) dt + int_{0}^{12} 30 dt]First integral:Again, let ( u = frac{pi t}{6} ), so ( du = frac{pi}{6} dt ), ( dt = frac{6}{pi} du )So,[15 int cos(u) cdot frac{6}{pi} du = frac{90}{pi} int cos(u) du = frac{90}{pi} sin(u) + C = frac{90}{pi} sinleft(frac{pi t}{6}right) + C]Evaluated from 0 to 12:At t = 12: ( frac{90}{pi} sin(2pi) = 0 )At t = 0: ( frac{90}{pi} sin(0) = 0 )So, the definite integral is 0 - 0 = 0Second integral:[int_{0}^{12} 30 dt = 30t bigg|_{0}^{12} = 360 - 0 = 360]Total integral: 0 + 360 = 360Average engagement:( frac{360}{12} = 30 )%Same logic as Social Media: the cosine function over its full period integrates to zero, so only the constant term contributes. So, average engagement is 30%.3. Webinars Engagement:Function: ( E_{WB}(t) = 5e^{-0.05t} + 25 )Average engagement:[frac{1}{12} int_{0}^{12} left(5e^{-0.05t} + 25right) dt]Compute the integral:[5 int_{0}^{12} e^{-0.05t} dt + int_{0}^{12} 25 dt]First integral:Let me compute ( int e^{-0.05t} dt ). The integral is ( frac{1}{-0.05} e^{-0.05t} + C = -20 e^{-0.05t} + C )Evaluated from 0 to 12:At t = 12: ( -20 e^{-0.6} )At t = 0: ( -20 e^{0} = -20 )So, definite integral:( (-20 e^{-0.6}) - (-20) = -20 e^{-0.6} + 20 = 20(1 - e^{-0.6}) )Multiply by 5:( 5 * 20(1 - e^{-0.6}) = 100(1 - e^{-0.6}) )Compute ( e^{-0.6} ). I know ( e^{-0.5} ≈ 0.6065 ), ( e^{-0.6} ≈ 0.5488 )So, 1 - 0.5488 = 0.4512Multiply by 100: 45.12Second integral:[int_{0}^{12} 25 dt = 25t bigg|_{0}^{12} = 300 - 0 = 300]Total integral: 45.12 + 300 = 345.12Average engagement:( frac{345.12}{12} ≈ 28.76 )%So, approximately 28.76%, which we can round to 28.8%.Wait, let me compute more accurately:First, ( e^{-0.6} ) is approximately 0.5488116.So, 1 - 0.5488116 = 0.4511884Multiply by 100: 45.11884Second integral: 25*12 = 300Total integral: 45.11884 + 300 = 345.11884Divide by 12: 345.11884 / 12 ≈ 28.7599 ≈ 28.76%So, approximately 28.76%.So, summarizing the average engagement percentages:- Social Media: 20%- Email Newsletters: 30%- Webinars: ~28.76%Therefore, Email Newsletters have the highest average engagement, followed by Webinars, then Social Media.Wait, let me just make sure I didn't make any calculation errors.For Webinars:Integral of 5e^{-0.05t} from 0 to12:Computed as 5 * [ (-20)(e^{-0.6} - 1) ] = 5 * [ -20 e^{-0.6} + 20 ] = 100(1 - e^{-0.6})Which is 100*(1 - 0.5488) = 100*0.4512 = 45.12Then, 25*12=300. Total 345.12Average: 345.12 /12=28.76. Correct.Yes, that seems right.So, in conclusion, the reach analysis shows Social Media has the highest total reach, followed by Email Newsletters, then Webinars. The engagement analysis shows Email Newsletters have the highest average engagement, followed by Webinars, then Social Media.Final Answer1. Total Reach: - Social Media: boxed{204000} - Email Newsletters: boxed{29075} - Webinars: boxed{24988}2. Average Engagement: - Social Media: boxed{20%} - Email Newsletters: boxed{30%} - Webinars: boxed{28.8%}</think>