answer：Okay, so I have this problem about urban development and population density. It involves partial differential equations, which I remember from my calculus classes, but I need to think carefully about how to approach it. Let me try to break it down step by step.First, the problem is divided into two sub-problems. The first one is to find the steady-state solution for the population density ( D_s(x,y) ) when the velocity field ( vec{v} ) is zero and the carrying capacity ( K ) is uniform. The second part is about analyzing the transportation network function ( T(x,y,t) ) given this steady-state density.Starting with sub-problem 1. The PDE given is:[frac{partial D}{partial t} = alpha nabla^2 D - beta nabla cdot (vec{v} D) + gamma D left(1 - frac{D}{K}right)]Since we're looking for the steady-state solution, that means the population density isn't changing with time anymore. So, the time derivative ( frac{partial D}{partial t} ) should be zero. Also, the velocity field ( vec{v} ) is zero, so the advection term ( -beta nabla cdot (vec{v} D) ) disappears. That simplifies the equation a lot.So, the equation becomes:[0 = alpha nabla^2 D_s + gamma D_s left(1 - frac{D_s}{K}right)]Hmm, this is an elliptic PDE because the time derivative is gone and we have the Laplacian term. The equation is:[alpha nabla^2 D_s + gamma D_s left(1 - frac{D_s}{K}right) = 0]I need to solve this for ( D_s(x,y) ). Let me rearrange the equation:[nabla^2 D_s = -frac{gamma}{alpha} D_s left(1 - frac{D_s}{K}right)]This looks like a nonlinear elliptic PDE because of the ( D_s^2 ) term. Nonlinear PDEs can be tricky, but maybe under certain assumptions, I can find a solution.Wait, the carrying capacity ( K ) is uniform, so it's a constant. Maybe I can assume that the steady-state density ( D_s ) is also uniform? If ( D_s ) is uniform, then its Laplacian would be zero because the second derivatives would be zero. Let me test this assumption.If ( D_s ) is constant, then ( nabla^2 D_s = 0 ). Plugging into the equation:[0 + gamma D_s left(1 - frac{D_s}{K}right) = 0]So, either ( gamma = 0 ) (which doesn't make sense because it's a growth rate), or ( D_s left(1 - frac{D_s}{K}right) = 0 ). Therefore, ( D_s = 0 ) or ( D_s = K ).But ( D_s = 0 ) would mean no population, which isn't realistic unless the city is uninhabited. So, the steady-state solution is ( D_s = K ) everywhere. That makes sense because the carrying capacity is the maximum population the area can sustain, so in the long run, the population should stabilize at ( K ).But wait, is this the only solution? What if ( D_s ) isn't uniform? Maybe there are non-uniform steady states? Hmm, the equation is nonlinear, so it's possible. But without any advection or sources/sinks, it's likely that the uniform solution is the only steady state, especially if the domain is the entire city with no boundaries or if we have periodic boundary conditions. But the problem doesn't specify boundary conditions, so maybe we can assume that the solution tends to zero at infinity or something. But since ( K ) is uniform, it's more plausible that the density reaches ( K ) everywhere.So, tentatively, I think the steady-state solution is ( D_s(x,y) = K ).Moving on to sub-problem 2. We need to analyze the long-term behavior of ( T(x,y,t) ) given ( D_s(x,y) = K ).The transportation network function is given by:[T(x,y,t) = T_0 + delta int_0^t frac{D(x,y,tau)}{1 + D(x,y,tau)} , dtau]As ( t to infty ), we need to find ( T(x,y,t) ).Given that in the steady state, ( D(x,y,tau) = K ) for all ( tau ) as ( t to infty ). So, substituting ( D = K ) into the integral:[T(x,y,t) = T_0 + delta int_0^t frac{K}{1 + K} , dtau]Simplify the integrand:[frac{K}{1 + K} = frac{K}{K + 1}]So, the integral becomes:[delta cdot frac{K}{K + 1} cdot t]Therefore, as ( t to infty ), ( T(x,y,t) ) becomes:[T(x,y,t) = T_0 + delta cdot frac{K}{K + 1} cdot t]Wait, but this suggests that ( T(x,y,t) ) grows linearly with time, which might not be realistic because travel time can't increase indefinitely. Maybe I made a wrong assumption.Hold on, the integral is from 0 to ( t ), so if ( D(x,y,tau) ) approaches ( K ) as ( tau ) increases, then for large ( t ), the integral would be approximately:[int_0^t frac{K}{1 + K} , dtau = frac{K}{1 + K} t]But that still leads to ( T(x,y,t) ) growing without bound. Maybe the model assumes that the population density doesn't change after reaching the steady state, so the integral is over the entire time, but the integrand becomes constant. Hmm, but that would still lead to linear growth.Alternatively, perhaps the transportation network function ( T(x,y,t) ) is being updated continuously, but in reality, travel time might reach a steady state as well. Maybe I need to think differently.Wait, the function ( T(x,y,t) ) is defined as:[T(x,y,t) = T_0 + delta int_0^t frac{D(x,y,tau)}{1 + D(x,y,tau)} , dtau]So, as ( t to infty ), unless ( frac{D}{1 + D} ) tends to zero, the integral will diverge. But in our case, ( D ) tends to ( K ), so ( frac{K}{1 + K} ) is a positive constant. Therefore, the integral grows without bound, which would mean ( T(x,y,t) ) tends to infinity.But in reality, travel time can't increase indefinitely. Maybe the model is only valid for finite times, or perhaps there's another factor I'm missing. Alternatively, maybe the integral is over a different variable or there's a different scaling.Wait, looking back at the problem statement, it says "the transportation network function ( T(x,y,t) ) is modeled by...". So, perhaps in the long term, the travel time increases linearly with time, which might indicate that the transportation network is degrading or becoming more congested as the population stabilizes. But that seems counterintuitive because if the population stabilizes, maybe the transportation network also stabilizes.Alternatively, perhaps the model is intended to show that as population density increases, the travel time increases, but once the population stabilizes, the rate of increase of travel time also stabilizes. So, in the long term, the derivative of ( T ) with respect to ( t ) becomes constant.Wait, let's compute the derivative of ( T ) with respect to ( t ):[frac{partial T}{partial t} = delta cdot frac{D(x,y,t)}{1 + D(x,y,t)}]In the steady state, ( D = K ), so:[frac{partial T}{partial t} = delta cdot frac{K}{1 + K}]Which is a constant. Therefore, ( T(x,y,t) ) increases linearly with time in the long term. So, as ( t to infty ), ( T(x,y,t) ) tends to infinity. That seems to be the conclusion based on the model.But is this physically meaningful? In reality, transportation networks can adapt, so maybe the model doesn't account for that. Or perhaps the model is simplified and just shows that as population density increases, travel time increases indefinitely. So, in the context of this model, the answer is that ( T(x,y,t) ) tends to infinity as ( t to infty ).Alternatively, maybe the integral is over a different variable or there's a typo in the problem statement. But as given, I think this is the conclusion.So, to summarize:1. The steady-state population density is ( D_s(x,y) = K ).2. The transportation network function ( T(x,y,t) ) grows linearly with time as ( t to infty ), specifically:[T(x,y,t) = T_0 + delta cdot frac{K}{K + 1} cdot t]Which tends to infinity.But wait, maybe I should express it as ( T(x,y,t) ) approaching infinity, rather than writing it as a linear function. Because the question says "determine ( T(x,y,t) ) as ( t to infty )", so perhaps the answer is that ( T(x,y,t) ) tends to infinity.Alternatively, if the integral is improper, maybe it diverges, so ( T ) tends to infinity.Yeah, I think that's the case.So, final answers:1. ( D_s(x,y) = K )2. ( T(x,y,t) to infty ) as ( t to infty )But let me double-check.For sub-problem 1, the PDE reduces to:[alpha nabla^2 D_s + gamma D_s (1 - D_s/K) = 0]Assuming ( D_s ) is constant, then ( nabla^2 D_s = 0 ), so:[gamma D_s (1 - D_s/K) = 0]Which gives ( D_s = 0 ) or ( D_s = K ). Since ( D_s = 0 ) is trivial and not realistic, ( D_s = K ) is the steady state.For sub-problem 2, substituting ( D = K ) into the integral:[T = T_0 + delta int_0^t frac{K}{1 + K} dtau = T_0 + delta frac{K}{1 + K} t]As ( t to infty ), ( T to infty ).Yes, that seems correct.**Final Answer**1. The steady-state population density is (boxed{K}).2. The transportation network function ( T(x,y,t) ) tends to infinity as ( t to infty ), so (boxed{T(x,y,t) to infty}).

answer：Okay, so I have this problem about pollution levels in a river near some factories. The journalist wants to figure out which factory is contributing the most and the environmental impact. There are two parts to this problem. Let me tackle them one by one.Starting with part 1: The pollution level P(t) is modeled by the differential equation dP/dt = -kP + s(t), where s(t) is a combination of exponential, cosine, and constant terms. The initial condition is P(0) = P0. I need to find the expression for P(t).Alright, this looks like a linear first-order differential equation. The standard form is dP/dt + kP = s(t). So, the integrating factor method should work here. The integrating factor is e^(∫k dt) = e^(kt). Multiplying both sides by the integrating factor:e^(kt) dP/dt + k e^(kt) P = e^(kt) s(t)The left side is the derivative of (e^(kt) P). So, integrating both sides from 0 to t:∫₀ᵗ d/dτ (e^(kτ) P(τ)) dτ = ∫₀ᵗ e^(kτ) s(τ) dτWhich simplifies to:e^(kt) P(t) - e^(0) P(0) = ∫₀ᵗ e^(kτ) s(τ) dτSo,P(t) = e^(-kt) P0 + e^(-kt) ∫₀ᵗ e^(kτ) s(τ) dτThat's the general solution. Now, since s(t) is given as a e^(-λ t) + b cos(ω t) + c, I need to compute the integral ∫₀ᵗ e^(kτ) [a e^(-λ τ) + b cos(ω τ) + c] dτ.Let me break this integral into three parts:∫₀ᵗ a e^(kτ) e^(-λ τ) dτ + ∫₀ᵗ b e^(kτ) cos(ω τ) dτ + ∫₀ᵗ c e^(kτ) dτSimplify each integral:First integral: a ∫₀ᵗ e^{(k - λ)τ} dτSecond integral: b ∫₀ᵗ e^{kτ} cos(ω τ) dτThird integral: c ∫₀ᵗ e^{kτ} dτCompute each one:First integral: a [e^{(k - λ)τ} / (k - λ)] from 0 to t = a [e^{(k - λ)t} - 1] / (k - λ)Second integral: Hmm, integrating e^{kτ} cos(ω τ) dτ. I remember that the integral of e^{at} cos(bt) dt is e^{at} (a cos(bt) + b sin(bt)) / (a² + b²). So, applying that here:b [e^{kτ} (k cos(ω τ) + ω sin(ω τ)) / (k² + ω²)] from 0 to tWhich is b [e^{kt} (k cos(ω t) + ω sin(ω t)) - k] / (k² + ω²)Third integral: c [e^{kτ} / k] from 0 to t = c (e^{kt} - 1)/kPutting it all together:P(t) = e^(-kt) P0 + e^(-kt) [ a (e^{(k - λ)t} - 1)/(k - λ) + b (e^{kt} (k cos(ω t) + ω sin(ω t)) - k)/(k² + ω²) + c (e^{kt} - 1)/k ]Simplify each term:First term: e^(-kt) * a (e^{(k - λ)t} - 1)/(k - λ) = a (e^{-λ t} - e^{-kt}) / (k - λ)Second term: e^(-kt) * b [e^{kt} (k cos(ω t) + ω sin(ω t)) - k]/(k² + ω²) = b [ (k cos(ω t) + ω sin(ω t)) - k e^{-kt} ] / (k² + ω²)Third term: e^(-kt) * c (e^{kt} - 1)/k = c (1 - e^{-kt}) / kSo, combining all these:P(t) = e^(-kt) P0 + [a (e^{-λ t} - e^{-kt}) / (k - λ)] + [b (k cos(ω t) + ω sin(ω t) - k e^{-kt}) / (k² + ω²)] + [c (1 - e^{-kt}) / k]That seems a bit complicated, but I think that's the expression. Let me check if the dimensions make sense. Each term should have units of pollution level. The exponential terms are dimensionless, so multiplied by a, b, c which are presumably in pollution units, so that works.Wait, but when λ = k, the first term would have a division by zero. So, we need to consider that case separately. But since the problem didn't specify, maybe we can assume λ ≠ k. Otherwise, the integral would be different.So, assuming λ ≠ k, the expression is as above.Now, moving to part 2: The pollution level has a critical threshold T. If it exceeds T for more than 15 consecutive days, severe ecological damage occurs. Given P(0) = 0, find conditions on the parameters so that P(t) > T for at least 15 consecutive days.Hmm, so we need P(t) > T for t in [t0, t0 + 15] for some t0.Given that P(0) = 0, which is below T, so the pollution starts at 0 and then increases. We need to find when it crosses T and stays above for 15 days.First, let's recall the expression for P(t) when P(0) = 0. From part 1, P(t) is:P(t) = [a (e^{-λ t} - e^{-kt}) / (k - λ)] + [b (k cos(ω t) + ω sin(ω t) - k e^{-kt}) / (k² + ω²)] + [c (1 - e^{-kt}) / k]Since P(0) = 0, that expression holds.We need to analyze when P(t) > T for 15 consecutive days. So, the function P(t) must cross T and then remain above T for 15 days.To find the conditions, perhaps we can consider the steady-state behavior and the transient behavior.Looking at P(t), as t approaches infinity, the terms with e^{-kt} and e^{-λ t} go to zero (assuming λ > 0, which it is, since a, b, c > 0). So, the steady-state pollution level is:P_ss = [a (0 - 0) / (k - λ)] + [b (k * 1 + ω * 0 - 0) / (k² + ω²)] + [c (1 - 0) / k] = [b k / (k² + ω²)] + [c / k]So, P_ss = (b k)/(k² + ω²) + c/kIf this steady-state level is above T, then eventually, the pollution will stay above T indefinitely. So, one condition is that P_ss > T.But the problem says "exceeds T for at least 15 consecutive days," not necessarily indefinitely. So, even if P_ss <= T, if the transient response causes P(t) to go above T and stay there for 15 days, that's also a case.But since P_ss is the long-term level, if P_ss > T, then after some time, P(t) will stay above T forever, which certainly includes 15 consecutive days.If P_ss <= T, then P(t) might go above T temporarily, but eventually come back down. So, in that case, we need to ensure that the peak of P(t) is above T, and that the time it takes to come back below T is more than 15 days.But this seems complicated. Maybe we can consider two cases:Case 1: P_ss > T. Then, after some time, P(t) will stay above T forever, so certainly exceeds T for 15 days.Case 2: P_ss <= T. Then, we need to ensure that the maximum of P(t) is above T, and that the duration above T is at least 15 days.But perhaps the problem expects us to consider the steady-state condition, so the primary condition is P_ss > T.But let's think more carefully.Given that P(t) is a combination of decaying exponentials and a sinusoidal function, its behavior over time will depend on the parameters.If the steady-state is above T, then after transients die out, P(t) remains above T. So, if P_ss > T, then for sufficiently large t, P(t) > T, and since the exponential terms decay, the function approaches P_ss from below or above?Wait, let's see. The initial pollution is 0. The transient terms are:- a (e^{-λ t} - e^{-kt}) / (k - λ): Depending on whether λ > k or λ < k, this term can be increasing or decreasing.Similarly, the term with b involves a combination of exponential decay and oscillations.The term with c is (1 - e^{-kt}) / k, which increases from 0 to c/k as t increases.So, if P_ss > T, then as t increases, P(t) approaches P_ss. So, if P_ss > T, then after some time, P(t) will cross T and stay above it. So, the duration above T will be infinite, which certainly includes 15 days.If P_ss <= T, then P(t) might go above T temporarily, but eventually come back down. So, in that case, we need to ensure that the peak of P(t) is above T, and that the time between crossing T upwards and then downwards is at least 15 days.But this requires more detailed analysis.Alternatively, perhaps the problem assumes that the steady-state is the main factor, so the condition is simply P_ss > T.But let's compute P_ss:P_ss = (b k)/(k² + ω²) + c/kSo, for P_ss > T, we have:(b k)/(k² + ω²) + c/k > TWhich can be rewritten as:(b k² + c(k² + ω²)) / (k(k² + ω²)) > TSimplify numerator:b k² + c k² + c ω² = k²(b + c) + c ω²So,(k²(b + c) + c ω²) / (k(k² + ω²)) > TMultiply both sides by denominator (positive since k > 0):k²(b + c) + c ω² > T k(k² + ω²)Which is:k²(b + c) + c ω² - T k³ - T k ω² > 0Factor terms:k²(b + c - T k) + ω²(c - T k) > 0Hmm, not sure if that helps. Alternatively, maybe leave it as:(b k)/(k² + ω²) + c/k > TSo, that's one condition.But if P_ss <= T, then we need to ensure that the transient response causes P(t) to exceed T for at least 15 days.This would require analyzing the maximum of P(t) and ensuring that the time between crossing T upwards and downwards is at least 15.But this seems complicated because P(t) is a combination of exponential and sinusoidal terms.Alternatively, perhaps the dominant term is the sinusoidal one, so the pollution level oscillates around P_ss with some amplitude.If the amplitude is large enough, P(t) can exceed T even if P_ss is below T.So, the maximum pollution level would be P_ss plus the amplitude of the oscillation.Compute the amplitude of the oscillatory part.Looking back at P(t):The term involving b is [b (k cos(ω t) + ω sin(ω t) - k e^{-kt}) / (k² + ω²)]As t increases, the term with e^{-kt} decays, so the oscillatory part becomes [b (k cos(ω t) + ω sin(ω t)) / (k² + ω²)]The amplitude of this oscillation is b / sqrt(k² + ω²). Because k cos(ω t) + ω sin(ω t) can be written as sqrt(k² + ω²) cos(ω t - φ), where φ is some phase shift. So, the amplitude is b / sqrt(k² + ω²).Therefore, the maximum pollution level is P_ss + b / sqrt(k² + ω²).Similarly, the minimum is P_ss - b / sqrt(k² + ω²).So, if P_ss + b / sqrt(k² + ω²) > T, then the pollution level will exceed T periodically.But to have it exceed T for 15 consecutive days, we need that during the oscillation, it stays above T for at least 15 days.This is more complicated because it depends on the frequency ω.If the oscillation frequency is low, the time between peaks is longer, so it might stay above T for a longer duration.Alternatively, if the oscillation is such that the time between crossing T upwards and downwards is at least 15 days.But this seems too involved. Maybe the problem expects us to consider the steady-state condition, so that P_ss > T.Alternatively, perhaps the transient response is dominated by the exponential terms, so if the transient peak is above T and the decay is slow enough, it can stay above T for 15 days.But without knowing the exact parameters, it's hard to say.Wait, the problem says "assuming P(0) = 0", so the initial pollution is zero. So, the system starts from zero and then increases due to the sources.Given that, the pollution level will rise, possibly oscillate, and then approach P_ss.So, if P_ss > T, then eventually, it will stay above T forever. So, the time when it crosses T and stays above is when the transient dies out. So, the duration above T is infinite, which includes 15 days.If P_ss <= T, then the pollution level might go above T temporarily, but then come back down. So, in that case, we need the peak of P(t) to be above T, and the time between crossing T upwards and downwards to be at least 15 days.But this requires solving for when P(t) = T and ensuring the difference in t is at least 15.But this seems difficult because P(t) is a combination of exponentials and sinusoids.Alternatively, perhaps we can consider that the maximum pollution level is P_ss + b / sqrt(k² + ω²). So, if this maximum is above T, then P(t) will exceed T at some points.But to have it exceed T for 15 consecutive days, we need that the duration between two crossings of T is at least 15 days.This would depend on the frequency ω. If ω is very low, the period is long, so the duration above T could be long.But without knowing ω, it's hard to specify.Alternatively, maybe the problem expects us to consider that the transient response causes P(t) to exceed T and stay above for 15 days if the decay rate k is small enough, so that the exponential terms decay slowly.But I'm not sure.Alternatively, perhaps the problem is simpler. If P_ss > T, then P(t) will eventually stay above T forever, so certainly exceeds T for 15 days. If P_ss <= T, then even if the peak is above T, it might not stay above for 15 days.So, the main condition is P_ss > T.Therefore, the condition is:(b k)/(k² + ω²) + c/k > TSo, that's the condition.But let me think again. If P_ss > T, then yes, after some time, P(t) will stay above T. But how long does it take to reach P_ss? If the decay rates are fast, it might reach P_ss quickly, but if decay rates are slow, it might take longer.But the problem says "exceeds T for at least 15 consecutive days". So, if P_ss > T, then after some t1, P(t) > T for all t > t1. So, the duration above T is infinite, which includes 15 days.Therefore, the condition is P_ss > T.So, summarizing:For part 1, the expression for P(t) is:P(t) = e^(-kt) P0 + [a (e^{-λ t} - e^{-kt}) / (k - λ)] + [b (k cos(ω t) + ω sin(ω t) - k e^{-kt}) / (k² + ω²)] + [c (1 - e^{-kt}) / k]For part 2, the condition is:(b k)/(k² + ω²) + c/k > TSo, that's the conclusion.**Final Answer**1. The expression for ( P(t) ) is:[boxed{P(t) = e^{-kt}P_0 + frac{a}{k - lambda}left(e^{-lambda t} - e^{-kt}right) + frac{b}{k^2 + omega^2}left(k cos(omega t) + omega sin(omega t) - k e^{-kt}right) + frac{c}{k}left(1 - e^{-kt}right)}]2. The condition for ( P(t) ) to exceed ( T ) for at least 15 consecutive days is:[boxed{frac{bk}{k^2 + omega^2} + frac{c}{k} > T}]

answer：Okay, so I have this problem about a horse trainer who wants to maximize the performance scores of his 12 horses by providing them with the best quality hay. The performance score is given by the function S(Q) = 50 + 0.8Q - 0.005Q², and the quality Q ranges from 0 to 100. The first part is to find the optimal Q that maximizes the average performance score. Then, in the second part, I need to check if the trainer can afford this optimal Q within a 2000 budget, considering the cost function C(Q) = 10Q + 0.1Q². If not, I have to find the maximum Q he can afford.Starting with the first part: maximizing the performance score. Since the performance score is a quadratic function in terms of Q, and the coefficient of Q² is negative (-0.005), the parabola opens downward, meaning the vertex is the maximum point. So, I need to find the vertex of this quadratic function.The general form of a quadratic function is f(Q) = aQ² + bQ + c, and the vertex occurs at Q = -b/(2a). In this case, a is -0.005 and b is 0.8. Plugging these into the formula:Q = -0.8 / (2 * -0.005) = -0.8 / (-0.01) = 80.So, the optimal Q is 80. That seems straightforward. Let me double-check by taking the derivative of S(Q) with respect to Q and setting it to zero. The derivative S'(Q) = 0.8 - 0.01Q. Setting this equal to zero:0.8 - 0.01Q = 0 0.01Q = 0.8 Q = 0.8 / 0.01 Q = 80.Yep, same result. So, the optimal Q is 80. That should maximize the performance score. Since the trainer has 12 horses, and each horse's performance is maximized at Q=80, the average performance will also be maximized at Q=80.Moving on to the second part: checking the budget. The cost function is C(Q) = 10Q + 0.1Q². I need to calculate the cost when Q=80 and see if it's within the 2000 limit.Calculating C(80):C(80) = 10*80 + 0.1*(80)² = 800 + 0.1*6400 = 800 + 640 = 1440.So, the cost at Q=80 is 1440, which is less than 2000. Therefore, the trainer can afford the optimal Q of 80 without any issues. But just to make sure, maybe I should check if there's a higher Q that might still be within the budget? Wait, the cost function is also a quadratic, opening upwards because the coefficient of Q² is positive (0.1). So, as Q increases beyond 80, the cost will increase as well. Since at Q=80, the cost is 1440, which is under 2000, the trainer could potentially buy higher quality hay if he wanted, but since 80 is the optimal point for performance, there's no need to go higher. But just for thoroughness, let's see what's the maximum Q the trainer can afford if he wanted to spend the entire budget.So, if the trainer wants to spend up to 2000, we can set C(Q) = 2000 and solve for Q:10Q + 0.1Q² = 2000 0.1Q² + 10Q - 2000 = 0 Multiply both sides by 10 to eliminate the decimal: Q² + 100Q - 20000 = 0.Now, solving this quadratic equation using the quadratic formula:Q = [-b ± sqrt(b² - 4ac)] / (2a) Where a=1, b=100, c=-20000.Discriminant D = 100² - 4*1*(-20000) = 10000 + 80000 = 90000.So, Q = [-100 ± sqrt(90000)] / 2 sqrt(90000) = 300.Thus, Q = (-100 + 300)/2 = 200/2 = 100 Or Q = (-100 - 300)/2 = -400/2 = -200.Since Q can't be negative, the maximum Q is 100. But wait, the quality Q only ranges up to 100, so the maximum possible Q is 100, which would cost:C(100) = 10*100 + 0.1*(100)² = 1000 + 1000 = 2000.So, the trainer can just afford Q=100 if he spends the entire budget. But since Q=80 is the optimal for performance, which is well within the budget, he doesn't need to go to 100. However, if he wanted to, he could, but it wouldn't improve performance beyond Q=80 because the performance function peaks at 80.Therefore, summarizing:1. The optimal Q is 80.2. The cost at Q=80 is 1440, which is within the 2000 budget. So, the trainer can afford it.But just to be thorough, let's make sure that Q=80 is indeed the maximum. Maybe I should check the second derivative to confirm it's a maximum.The second derivative of S(Q) is S''(Q) = -0.01, which is negative, confirming that Q=80 is indeed a maximum point.Also, checking the cost function, since it's a quadratic opening upwards, the cost increases as Q moves away from the vertex. The vertex of the cost function is at Q = -b/(2a) = -10/(2*0.1) = -10/0.2 = -50. But since Q can't be negative, the minimum cost occurs at Q=0, and cost increases as Q increases beyond that. So, the cost at Q=80 is 1440, and at Q=100, it's 2000. So, the trainer can choose any Q between 0 and 100, but the optimal performance is at 80, which is affordable.Wait, just thinking again: the cost function is C(Q) = 10Q + 0.1Q². So, for each horse, is this cost per horse or total cost? The problem says "the cost C of the hay is given by...", but it's not specified if it's per horse or total. Hmm, that's a bit ambiguous.Looking back at the problem statement: "the cost C of the hay is given by the function C(Q) = 10Q + 0.1Q²." It doesn't specify per horse or total. But since the trainer has 12 horses, maybe the cost is per horse? Or is it total cost for all 12?Wait, the first part talks about the average performance score of all 12 horses, so the function S(Q) is per horse. Then, the cost function C(Q) is probably per horse as well, because it's not specified otherwise. So, if C(Q) is per horse, then the total cost would be 12*C(Q). Let me check that.If that's the case, then the total cost would be 12*(10Q + 0.1Q²) = 120Q + 1.2Q². Then, we need to see if 120Q + 1.2Q² <= 2000.Wait, but the problem says "the cost C of the hay is given by...", so maybe it's total cost? Hmm, the wording is unclear. Let me read it again: "the cost C of the hay is given by the function C(Q) = 10Q + 0.1Q²." It doesn't specify per horse or total. But in the first part, S(Q) is the performance score of a horse, so it's per horse. So, perhaps C(Q) is also per horse.Therefore, total cost would be 12*C(Q). So, total cost = 12*(10Q + 0.1Q²) = 120Q + 1.2Q².So, if that's the case, then when Q=80, total cost is 120*80 + 1.2*(80)^2.Calculating that:120*80 = 9600 1.2*(6400) = 7680 Total cost = 9600 + 7680 = 17280.Wait, that's way over 2000. That can't be right. So, perhaps my initial assumption is wrong.Alternatively, maybe C(Q) is the total cost for all 12 horses. So, C(Q) = 10Q + 0.1Q² is the total cost. Then, at Q=80, total cost is 10*80 + 0.1*6400 = 800 + 640 = 1440, which is under 2000.But the problem is ambiguous. Hmm. Let me think.If C(Q) is total cost, then at Q=80, total cost is 1440, which is under 2000. So, the trainer can afford it.If C(Q) is per horse, then total cost is 12*(10Q + 0.1Q²) = 120Q + 1.2Q². At Q=80, that's 120*80 + 1.2*6400 = 9600 + 7680 = 17280, which is way over 2000. So, in that case, the trainer cannot afford Q=80.But the problem says "the cost C of the hay is given by...", without specifying per horse or total. So, perhaps it's total cost. Because if it were per horse, it would likely say "per horse" or "for each horse".Therefore, I think it's safer to assume that C(Q) is the total cost for all 12 horses. So, at Q=80, total cost is 1440, which is under 2000. So, the trainer can afford it.But just to make sure, let's see what the problem says: "the cost C of the hay is given by the function C(Q) = 10Q + 0.1Q²." It doesn't specify per horse, so it's probably total cost.Therefore, the answer is:1. Optimal Q is 80.2. The cost at Q=80 is 1440, which is within the 2000 budget. So, the trainer can afford it.But just to be thorough, let's also solve for the maximum Q if the cost were per horse. So, if C(Q) is per horse, then total cost is 12*C(Q) = 120Q + 1.2Q². Setting this equal to 2000:120Q + 1.2Q² = 2000 1.2Q² + 120Q - 2000 = 0 Divide all terms by 1.2 to simplify:Q² + 100Q - (2000/1.2) = 0 2000/1.2 = 1666.666...So, Q² + 100Q - 1666.666... = 0.Using quadratic formula:Q = [-100 ± sqrt(100² + 4*1*1666.666...)] / 2 = [-100 ± sqrt(10000 + 6666.666...)] / 2 = [-100 ± sqrt(16666.666...)] / 2 sqrt(16666.666...) ≈ 129.104.So, Q = (-100 + 129.104)/2 ≈ 29.104/2 ≈ 14.552.So, approximately Q=14.55. But since Q must be a whole number? Or can it be a decimal? The problem doesn't specify, so maybe it's okay to have a decimal.But in this case, since the cost is per horse, and the optimal Q is 80, which would require a total cost of 17280, which is way over 2000, the trainer cannot afford it. So, the maximum Q he can afford is approximately 14.55.But wait, the problem didn't specify whether C(Q) is per horse or total. So, this is a bit confusing. But given that in the first part, S(Q) is per horse, and the second part talks about the cost of the hay, which is likely total cost, because otherwise, the numbers don't make sense.So, I think the correct approach is to assume that C(Q) is total cost. Therefore, the trainer can afford Q=80.But just to make sure, let's see: if C(Q) is total cost, then 10Q + 0.1Q² is the total cost for all 12 horses. So, at Q=80, total cost is 1440, which is under 2000. So, the trainer can afford it.Therefore, the answers are:1. Optimal Q is 80.2. The trainer can afford it, as the cost is 1440, which is under 2000.But just to be absolutely thorough, let's calculate the cost if C(Q) is per horse, even though it seems unlikely.If C(Q) is per horse, then total cost is 12*(10Q + 0.1Q²) = 120Q + 1.2Q². At Q=80, that's 120*80 + 1.2*6400 = 9600 + 7680 = 17280, which is way over 2000. So, in that case, the trainer cannot afford Q=80. Then, we need to find the maximum Q such that 120Q + 1.2Q² <= 2000.So, 1.2Q² + 120Q - 2000 <= 0.Solving 1.2Q² + 120Q - 2000 = 0.Multiply all terms by 10 to eliminate decimals: 12Q² + 1200Q - 20000 = 0.Divide by 4: 3Q² + 300Q - 5000 = 0.Using quadratic formula:Q = [-300 ± sqrt(300² - 4*3*(-5000))]/(2*3) = [-300 ± sqrt(90000 + 60000)]/6 = [-300 ± sqrt(150000)]/6 sqrt(150000) ≈ 387.298.So, Q = (-300 + 387.298)/6 ≈ 87.298/6 ≈ 14.549.So, approximately 14.55. Since Q is a quality measure, it can be a decimal, so Q≈14.55.But again, this is under the assumption that C(Q) is per horse, which is not clearly stated. Given that, I think the more reasonable assumption is that C(Q) is total cost, so the trainer can afford Q=80.Therefore, final answers:1. Optimal Q is 80.2. The cost is 1440, which is within the 2000 budget. So, the trainer can afford it.But just to make sure, let's check the cost function again. If C(Q) is total cost, then 10Q + 0.1Q² is the total cost for all 12 horses. So, for Q=80, it's 10*80 + 0.1*80² = 800 + 640 = 1440. That's correct.Alternatively, if it's per horse, then 10Q + 0.1Q² per horse, times 12, is 120Q + 1.2Q². At Q=80, that's 120*80 + 1.2*6400 = 9600 + 7680 = 17280, which is way over 2000.Therefore, the problem likely assumes that C(Q) is total cost, not per horse, because otherwise, the numbers don't make sense. So, the trainer can afford the optimal Q=80.So, to wrap up:1. Optimal Q is 80.2. The cost is 1440, which is within the 2000 budget. Therefore, the trainer can afford it.

answer：Alright, so I have this problem about designing an airplane wing to minimize drag and maximize lift. It involves some calculus and parametric equations. Let me try to break it down step by step.First, the wing's cross-section is modeled by the parametric equations:[ x(t) = 1.5(1 - t)^2 ][ y(t) = 0.5t(2 - t) ]where ( t ) is in the interval [0, 1]. The problem has two parts: minimizing the drag coefficient ( C_d ) and maximizing the lift coefficient ( C_l ). Let me tackle them one by one.**1. Minimize Drag Coefficient ( C_d ):**The drag coefficient is given by:[ C_d = int_0^1 sqrt{left(frac{dx}{dt}right)^2 + left(frac{dy}{dt}right)^2} left( kappa(t)^2 + frac{dkappa(t)}{dt} right) dt ]where ( kappa(t) ) is the curvature of the airfoil.First, I need to find the expressions for ( frac{dx}{dt} ) and ( frac{dy}{dt} ).Starting with ( x(t) = 1.5(1 - t)^2 ):[ frac{dx}{dt} = 1.5 times 2(1 - t)(-1) = -3(1 - t) ]And for ( y(t) = 0.5t(2 - t) ):First, expand ( y(t) ):[ y(t) = 0.5(2t - t^2) = t - 0.5t^2 ]So,[ frac{dy}{dt} = 1 - t ]Next, I need the second derivatives for the curvature formula.Compute ( frac{d^2x}{dt^2} ):[ frac{d^2x}{dt^2} = frac{d}{dt}(-3(1 - t)) = 3 ]Compute ( frac{d^2y}{dt^2} ):[ frac{d^2y}{dt^2} = frac{d}{dt}(1 - t) = -1 ]Now, let's compute the curvature ( kappa(t) ):[ kappa(t) = frac{ left| frac{dx}{dt} frac{d^2y}{dt^2} - frac{dy}{dt} frac{d^2x}{dt^2} right| }{ left( left( frac{dx}{dt} right)^2 + left( frac{dy}{dt} right)^2 right)^{3/2} } ]Plugging in the derivatives:Numerator:[ left| (-3(1 - t))(-1) - (1 - t)(3) right| ]Simplify:First term: ( (-3(1 - t))(-1) = 3(1 - t) )Second term: ( (1 - t)(3) = 3(1 - t) )So, numerator:[ |3(1 - t) - 3(1 - t)| = |0| = 0 ]Wait, that can't be right. If the numerator is zero, the curvature is zero everywhere? That would mean the airfoil is a straight line, which doesn't make sense because the parametric equations are quadratic, so they should form a curve.Let me double-check the calculations.Compute numerator:[ frac{dx}{dt} frac{d^2y}{dt^2} - frac{dy}{dt} frac{d^2x}{dt^2} ]Which is:[ (-3(1 - t))(-1) - (1 - t)(3) ]Calculate each term:First term: ( (-3(1 - t))(-1) = 3(1 - t) )Second term: ( (1 - t)(3) = 3(1 - t) )So, subtracting:[ 3(1 - t) - 3(1 - t) = 0 ]So, yes, the numerator is zero. That means the curvature ( kappa(t) ) is zero for all t in [0,1]. Hmm, that seems odd.Wait, maybe I made a mistake in computing the derivatives.Let me recompute ( frac{dx}{dt} ):[ x(t) = 1.5(1 - t)^2 ]So,[ frac{dx}{dt} = 1.5 times 2(1 - t)(-1) = -3(1 - t) ]That seems correct.( frac{dy}{dt} ):[ y(t) = 0.5t(2 - t) = t - 0.5t^2 ]So,[ frac{dy}{dt} = 1 - t ]Correct.Second derivatives:( frac{d^2x}{dt^2} = 3 )( frac{d^2y}{dt^2} = -1 )Correct.So, plugging into the curvature formula:Numerator:[ (-3(1 - t))(-1) - (1 - t)(3) = 3(1 - t) - 3(1 - t) = 0 ]So, yes, curvature is zero. That means the airfoil is a straight line? But the parametric equations are quadratic, so it should be a parabola or something.Wait, let me plot the parametric equations to see.x(t) = 1.5(1 - t)^2y(t) = 0.5t(2 - t) = t - 0.5t^2So, when t=0: x=1.5, y=0t=1: x=0, y=0.5(2 -1)=0.5t=0.5: x=1.5*(0.5)^2=1.5*0.25=0.375, y=0.5*0.5*(2 -0.5)=0.25*1.5=0.375So, plotting these points, it's a curve from (1.5,0) to (0,0.5), passing through (0.375,0.375). It's a parabola opening to the left.But according to the curvature calculation, it's zero everywhere. That can't be right because a parabola has non-zero curvature.Wait, maybe I messed up the formula for curvature.The formula is:[ kappa = frac{ |x'y'' - y'x''| }{ (x'^2 + y'^2)^{3/2} } ]Wait, in our case, x' = -3(1 - t), y' = 1 - tx'' = 3, y'' = -1So, numerator:x'y'' - y'x'' = (-3(1 - t))*(-1) - (1 - t)*(3) = 3(1 - t) - 3(1 - t) = 0So, yes, numerator is zero. So curvature is zero.But that contradicts the fact that it's a parabola.Wait, maybe because the parametric equations are not in the standard form. Let me think.Wait, in Cartesian coordinates, a parabola has non-zero curvature. But in parametric form, if the parametric equations are such that the numerator is zero, then curvature is zero. That seems contradictory.Wait, perhaps the parametric equations are actually representing a straight line? But when I plug in t=0, t=0.5, t=1, it's a curve, not a straight line.Wait, let's see: x(t) = 1.5(1 - t)^2, y(t) = t - 0.5t^2Let me express y in terms of x.From x(t) = 1.5(1 - t)^2, solve for t:x = 1.5(1 - 2t + t^2)x = 1.5 - 3t + 1.5t^2Bring all terms to one side:1.5t^2 - 3t + (1.5 - x) = 0Multiply both sides by 2 to eliminate decimals:3t^2 - 6t + 3 - 2x = 0Divide by 3:t^2 - 2t + 1 - (2/3)x = 0Which is:(t - 1)^2 = (2/3)xSo, y(t) = t - 0.5t^2Express y in terms of t:y = t - 0.5t^2But from (t - 1)^2 = (2/3)x, so t - 1 = sqrt((2/3)x), but that's only for t >=1, which isn't the case here.Wait, maybe express t in terms of x:From x = 1.5(1 - t)^2, so (1 - t)^2 = x / 1.5, so 1 - t = sqrt(x / 1.5), so t = 1 - sqrt(x / 1.5)Then, plug into y:y = [1 - sqrt(x / 1.5)] - 0.5[1 - sqrt(x / 1.5)]^2This seems complicated, but let's see if it's a straight line.Wait, if curvature is zero, it should be a straight line, but the parametric equations don't look like a straight line.Wait, maybe I made a mistake in the curvature formula. Let me double-check.The formula for curvature of a parametric curve x(t), y(t) is:[ kappa = frac{ |x'y'' - y'x''| }{ (x'^2 + y'^2)^{3/2} } ]Yes, that's correct.So, plugging in:x' = -3(1 - t), y' = 1 - tx'' = 3, y'' = -1So,x'y'' - y'x'' = (-3(1 - t))*(-1) - (1 - t)*(3) = 3(1 - t) - 3(1 - t) = 0So, numerator is zero, hence curvature is zero.But that would mean the parametric curve is a straight line, but when I plot it, it's a parabola.Wait, maybe I messed up the derivatives.Wait, x(t) = 1.5(1 - t)^2x'(t) = 1.5 * 2(1 - t)(-1) = -3(1 - t)x''(t) = 3y(t) = 0.5t(2 - t) = t - 0.5t^2y'(t) = 1 - ty''(t) = -1So, x'y'' - y'x'' = (-3(1 - t))*(-1) - (1 - t)*(3) = 3(1 - t) - 3(1 - t) = 0So, it's correct. The curvature is zero everywhere on the curve. That means the curve is a straight line.But when I plot x(t) and y(t), it's a parabola. How is that possible?Wait, maybe I'm confusing the parametric equations with Cartesian coordinates.Wait, in parametric form, even if the Cartesian equation is a parabola, the curvature can be zero if the parametric equations are such that the numerator is zero.But that seems contradictory because a parabola has non-zero curvature.Wait, perhaps the parametric equations are actually representing a straight line in disguise.Wait, let me check if the parametric equations satisfy a linear relationship.From x(t) = 1.5(1 - t)^2 and y(t) = t - 0.5t^2.Let me see if y is a linear function of x.Express t in terms of x:x = 1.5(1 - t)^2 => (1 - t)^2 = x / 1.5 => 1 - t = sqrt(x / 1.5) => t = 1 - sqrt(x / 1.5)Then, y = t - 0.5t^2 = [1 - sqrt(x / 1.5)] - 0.5[1 - sqrt(x / 1.5)]^2This is definitely not a linear function of x, so the curve is not a straight line. Therefore, the curvature shouldn't be zero.But according to the calculations, the curvature is zero. There must be a mistake in the curvature formula.Wait, maybe I mixed up the formula. Let me check again.Curvature for parametric equations is:[ kappa = frac{ |x'y'' - y'x''| }{ (x'^2 + y'^2)^{3/2} } ]Yes, that's correct.Wait, let me compute the numerator again:x' = -3(1 - t)y' = 1 - tx'' = 3y'' = -1So,x'y'' = (-3(1 - t))*(-1) = 3(1 - t)y'x'' = (1 - t)*3 = 3(1 - t)So, x'y'' - y'x'' = 3(1 - t) - 3(1 - t) = 0So, numerator is zero. Therefore, curvature is zero.But that's impossible because the curve is a parabola, which has non-zero curvature.Wait, maybe the parametric equations are such that the curve is a straight line, but when I plot it, it's a parabola. That doesn't make sense.Wait, let me compute some points:t=0: x=1.5, y=0t=0.5: x=1.5*(0.5)^2=1.5*0.25=0.375, y=0.5*0.5*(2 -0.5)=0.25*1.5=0.375t=1: x=0, y=0.5*(2 -1)=0.5So, the points are (1.5,0), (0.375,0.375), (0,0.5)Plotting these, it's a curve, not a straight line.Wait, maybe the parametric equations are actually a straight line in disguise. Let me check if the points lie on a straight line.Compute the slope between (1.5,0) and (0.375,0.375):Slope = (0.375 - 0)/(0.375 - 1.5) = 0.375 / (-1.125) = -0.333...Slope between (0.375,0.375) and (0,0.5):Slope = (0.5 - 0.375)/(0 - 0.375) = 0.125 / (-0.375) = -0.333...So, same slope. Therefore, the three points lie on a straight line. Wait, so the parametric equations are actually representing a straight line?But when I plug in t=0, t=0.5, t=1, they lie on a straight line. So, the entire curve is a straight line? That's why the curvature is zero.Wait, but x(t) and y(t) are quadratic functions, so shouldn't they form a parabola? But according to the points, it's a straight line.Wait, let me check if the parametric equations satisfy a linear relationship.From x(t) = 1.5(1 - t)^2 and y(t) = t - 0.5t^2.Let me see if y can be expressed as a linear function of x.From x(t) = 1.5(1 - t)^2, solve for t:(1 - t)^2 = x / 1.5Take square root:1 - t = sqrt(x / 1.5)So,t = 1 - sqrt(x / 1.5)Then, plug into y(t):y = t - 0.5t^2 = [1 - sqrt(x / 1.5)] - 0.5[1 - sqrt(x / 1.5)]^2Let me expand this:Let me denote s = sqrt(x / 1.5)Then,y = (1 - s) - 0.5(1 - 2s + s^2) = 1 - s - 0.5 + s - 0.5s^2 = 0.5 - 0.5s^2So,y = 0.5 - 0.5s^2 = 0.5 - 0.5*(x / 1.5) = 0.5 - (x / 3)So,y = 0.5 - (x / 3)Which is a linear equation. So, the parametric equations actually represent a straight line.Wait, so the parametric equations are quadratic, but they trace a straight line? That's interesting.So, in that case, the curvature is zero everywhere, as we calculated.Therefore, the drag coefficient ( C_d ) is:[ C_d = int_0^1 sqrt{left(frac{dx}{dt}right)^2 + left(frac{dy}{dt}right)^2} left( 0^2 + 0 right) dt = 0 ]Because both ( kappa(t) ) and its derivative are zero.But that seems too easy. The problem says to derive the expression and determine the value. So, maybe I did something wrong.Wait, but according to the calculations, the curvature is zero, so ( C_d = 0 ). But in reality, even a straight wing would have some drag, but maybe in this model, it's zero.Alternatively, perhaps the parametric equations are not correctly representing a straight line, but my calculations show that they do.Wait, let me check with another point. Let's take t=0.25:x(t) = 1.5*(1 - 0.25)^2 = 1.5*(0.75)^2 = 1.5*0.5625 = 0.84375y(t) = 0.5*0.25*(2 - 0.25) = 0.125*(1.75) = 0.21875Now, check if this lies on the line y = 0.5 - x/3:y = 0.5 - 0.84375/3 ≈ 0.5 - 0.28125 = 0.21875Yes, it does. So, all points lie on the straight line y = 0.5 - x/3.Therefore, the parametric equations represent a straight line, so curvature is zero, and hence ( C_d = 0 ).But that seems counterintuitive because a straight wing would have some drag, but maybe in this model, it's zero.Alternatively, perhaps I made a mistake in interpreting the parametric equations. Maybe they are not supposed to be straight lines, but I think the math shows they are.So, moving on, perhaps the problem is designed this way, so ( C_d = 0 ).**2. Maximize Lift Coefficient ( C_l ):**The lift coefficient is given by:[ C_l = int_0^1 left( frac{dy}{dt} right) left(1 + left(frac{dy}{dx}right)^2 right)^{1/2} dt ]First, I need to find ( frac{dy}{dx} ). Since we have parametric equations, ( frac{dy}{dx} = frac{dy/dt}{dx/dt} ).From earlier, we have:[ frac{dy}{dt} = 1 - t ][ frac{dx}{dt} = -3(1 - t) ]So,[ frac{dy}{dx} = frac{1 - t}{-3(1 - t)} = -frac{1}{3} ]So, ( frac{dy}{dx} ) is constant, -1/3.Therefore, ( 1 + left(frac{dy}{dx}right)^2 = 1 + (1/3)^2 = 1 + 1/9 = 10/9 )So, ( sqrt{10/9} = sqrt{10}/3 )Therefore, the integrand becomes:[ (1 - t) times sqrt{10}/3 ]Thus,[ C_l = int_0^1 (1 - t) times sqrt{10}/3 , dt ]Factor out constants:[ C_l = frac{sqrt{10}}{3} int_0^1 (1 - t) dt ]Compute the integral:[ int_0^1 (1 - t) dt = left[ t - frac{t^2}{2} right]_0^1 = (1 - 0.5) - (0 - 0) = 0.5 ]So,[ C_l = frac{sqrt{10}}{3} times 0.5 = frac{sqrt{10}}{6} ]Simplify:[ C_l = frac{sqrt{10}}{6} ]So, the lift coefficient is ( sqrt{10}/6 ).But wait, let me double-check the steps.First, ( frac{dy}{dx} = frac{dy/dt}{dx/dt} = frac{1 - t}{-3(1 - t)} = -1/3 ). Correct.So, ( 1 + (dy/dx)^2 = 1 + 1/9 = 10/9 ). Correct.Square root is ( sqrt{10}/3 ). Correct.Then, integrand is ( (1 - t) times sqrt{10}/3 ). Correct.Integral of (1 - t) from 0 to1 is 0.5. Correct.So, ( C_l = sqrt{10}/6 ). Correct.Therefore, the answers are:1. ( C_d = 0 )2. ( C_l = sqrt{10}/6 )But wait, the problem says "minimize drag coefficient" and "maximize lift coefficient". So, if ( C_d = 0 ), that's already minimal. And ( C_l = sqrt{10}/6 ) is the value.But maybe I should express ( C_d ) as zero, and ( C_l ) as ( sqrt{10}/6 ).Alternatively, perhaps I made a mistake in interpreting the parametric equations as a straight line, but according to the calculations, they are.So, final answers:1. ( C_d = 0 )2. ( C_l = sqrt{10}/6 )

answer：Alright, so I have this problem about a social worker referring individuals to a human rights lawyer. The problem has two parts, and I need to figure out both. Let me take them one at a time.**Problem 1:** I need to calculate the probability that exactly ( k ) individuals are referred in a month, given that ( N ) follows a Poisson distribution with parameter ( lambda ).Hmm, okay. So, the number of referrals ( N ) is Poisson distributed. The Poisson probability mass function is given by:[P(N = k) = frac{e^{-lambda} lambda^k}{k!}]So, for part 1, it seems straightforward. The probability that exactly ( k ) individuals are referred in a month is just the Poisson probability formula. I don't think I need to do anything more complicated here because ( N ) is already given as Poisson. So, I can just write that down.But wait, let me make sure. The question says "Given that ( N ) follows a Poisson distribution with parameter ( lambda ), calculate the probability that exactly ( k ) individuals are referred in a month." Yeah, that's exactly the Poisson PMF. So, I think that's it for part 1.**Problem 2:** Now, ( X ) represents the number of individuals who win their cases in a given month. ( X ) is a binomial random variable given ( N ). I need to derive the expected value and variance of ( X ).Okay, so ( X ) is binomial given ( N ). That means, for a given ( N = n ), ( X ) has a binomial distribution with parameters ( n ) and ( p = 0.6 ) (since 60% win their cases). So, ( X | N = n sim text{Binomial}(n, 0.6) ).To find the expected value and variance of ( X ), I can use the law of total expectation and the law of total variance.First, the expected value:[E[X] = E[E[X | N]]]Since ( X | N = n ) is binomial, ( E[X | N = n] = n times 0.6 ). Therefore,[E[X] = E[0.6 N] = 0.6 E[N]]But ( N ) is Poisson with parameter ( lambda ), so ( E[N] = lambda ). Thus,[E[X] = 0.6 lambda]Okay, that seems straightforward.Now, the variance. Using the law of total variance:[text{Var}(X) = E[text{Var}(X | N)] + text{Var}(E[X | N])]First, compute ( text{Var}(X | N = n) ). Since ( X | N = n ) is binomial, its variance is ( n times 0.6 times (1 - 0.6) = n times 0.6 times 0.4 = 0.24 n ).So,[E[text{Var}(X | N)] = E[0.24 N] = 0.24 E[N] = 0.24 lambda]Next, compute ( text{Var}(E[X | N]) ). Since ( E[X | N] = 0.6 N ), the variance is:[text{Var}(0.6 N) = (0.6)^2 text{Var}(N)]( N ) is Poisson, so ( text{Var}(N) = lambda ). Therefore,[text{Var}(E[X | N]) = 0.36 lambda]Putting it all together:[text{Var}(X) = 0.24 lambda + 0.36 lambda = 0.6 lambda]Wait, that's interesting. The variance of ( X ) is 0.6 times ( lambda ). So, both the expectation and variance are 0.6 ( lambda ). That seems a bit too clean, but let me verify.Alternatively, is there another way to think about this? Since ( X ) is the number of successes in a Poisson number of trials with success probability 0.6, then ( X ) itself is a Poisson random variable with parameter ( 0.6 lambda ). Because if you have a Poisson number of trials, each with success probability ( p ), then the number of successes is Poisson with parameter ( p lambda ).If that's the case, then both the mean and variance of ( X ) would be ( 0.6 lambda ), which matches what I got earlier.So, that seems to confirm my result. Therefore, the expected value is ( 0.6 lambda ) and the variance is also ( 0.6 lambda ).Let me just recap:1. For the first part, since ( N ) is Poisson, the probability is the Poisson PMF.2. For the second part, using the total expectation and variance, I found that both the mean and variance of ( X ) are ( 0.6 lambda ). Alternatively, recognizing that ( X ) is Poisson with parameter ( 0.6 lambda ) also gives the same result.I think that's solid. I don't see any mistakes in my reasoning.**Final Answer**1. The probability that exactly ( k ) individuals are referred in a month is boxed{dfrac{e^{-lambda} lambda^k}{k!}}.2. The expected value of ( X ) is boxed{0.6lambda} and the variance of ( X ) is boxed{0.6lambda}.

answer：Alright, so I have this problem about a public relations manager trying to improve customer privacy protocols. There are two parts: one about minimizing encryption time and another about minimizing the probability of a data breach. Let me tackle each part step by step.Starting with the first part: Encryption Time Analysis. The company uses two algorithms, A and B. Algorithm A encrypts data at a rate of ( f_A(N) = 2 cdot log(N) ) and Algorithm B at ( f_B(N) = 3 cdot N^{1/3} ). We need to find the value of ( N ) that minimizes the total encryption time.Hmm, so total encryption time would be the sum of the times taken by both algorithms, right? So, the total time function ( T(N) ) is:[ T(N) = 2 cdot log(N) + 3 cdot N^{1/3} ]To find the minimum, I remember from calculus that I need to take the derivative of ( T(N) ) with respect to ( N ), set it equal to zero, and solve for ( N ). That should give me the critical points, which I can then test to see if they're minima.So, let's compute the derivative ( T'(N) ):First, the derivative of ( 2 cdot log(N) ) with respect to ( N ). I think the derivative of ( log(N) ) is ( 1/N ), so:[ frac{d}{dN} [2 cdot log(N)] = 2 cdot frac{1}{N} = frac{2}{N} ]Next, the derivative of ( 3 cdot N^{1/3} ). Using the power rule, the derivative of ( N^{k} ) is ( k cdot N^{k - 1} ). So here, ( k = 1/3 ):[ frac{d}{dN} [3 cdot N^{1/3}] = 3 cdot frac{1}{3} cdot N^{-2/3} = N^{-2/3} ]So putting it all together, the derivative of ( T(N) ) is:[ T'(N) = frac{2}{N} + N^{-2/3} ]Wait, actually, hold on. Is that correct? Let me double-check. The derivative of ( 3 cdot N^{1/3} ) is indeed ( 3 cdot (1/3) cdot N^{-2/3} = N^{-2/3} ). So yes, that part is correct.So, to find the critical points, set ( T'(N) = 0 ):[ frac{2}{N} + N^{-2/3} = 0 ]Hmm, but both terms ( frac{2}{N} ) and ( N^{-2/3} ) are positive for ( N > 0 ). So their sum can't be zero. That seems odd. Did I make a mistake?Wait, maybe I messed up the signs. Let me check the derivatives again. The derivative of ( 2 cdot log(N) ) is positive because as ( N ) increases, the encryption time increases. Similarly, the derivative of ( 3 cdot N^{1/3} ) is also positive because as ( N ) increases, encryption time increases. So both terms are positive, meaning their sum is positive. Therefore, ( T'(N) ) is always positive, which would mean that ( T(N) ) is an increasing function for all ( N > 0 ).But that can't be right because if ( N ) is very small, say approaching zero, ( log(N) ) goes to negative infinity, but ( N^{1/3} ) approaches zero. Wait, but ( N ) represents the number of customers, so it must be a positive integer greater than zero. So, actually, when ( N ) is 1, ( log(1) = 0 ), and ( N^{1/3} = 1 ). As ( N ) increases, both ( log(N) ) and ( N^{1/3} ) increase.Wait, but if the derivative is always positive, that means the function is increasing, so the minimum would be at the smallest possible ( N ), which is 1. But that seems counterintuitive because if you have more data, encryption time should increase, but maybe for small ( N ), the encryption time is minimal.But let me think again. Maybe I misapplied the derivative. Let me write it again:Total time ( T(N) = 2 log N + 3 N^{1/3} )Derivative:( T'(N) = frac{2}{N} + frac{1}{3} cdot 3 N^{-2/3} )Wait, hold on, no. The derivative of ( 3 N^{1/3} ) is ( 3 cdot (1/3) N^{-2/3} = N^{-2/3} ). So that's correct.So, ( T'(N) = frac{2}{N} + N^{-2/3} ). Since both terms are positive, ( T'(N) ) is always positive for ( N > 0 ). Therefore, ( T(N) ) is an increasing function. So, the minimal encryption time occurs at the minimal ( N ), which is 1.But that seems odd because the problem is asking for a value of ( N ) such that the total encryption time is minimized. If it's minimized at ( N = 1 ), then that's the answer. But maybe I misunderstood the problem.Wait, perhaps the manager is trying to find the optimal ( N ) where the encryption time is minimized, but ( N ) is not a variable here. Wait, no, ( N ) is the number of customers, which is fixed. Wait, hold on, maybe I'm misinterpreting the problem.Wait, the problem says: "Determine the value of ( N ) such that the total encryption time using both algorithms is minimized." So, they want to choose ( N ) to minimize the total encryption time. But ( N ) is the number of customers, which is given as fixed. Wait, no, the company has a database containing sensitive information of ( N ) customers. So ( N ) is fixed, but the manager is working with the company to improve its customer privacy protocols, so maybe they can choose how to split the encryption between A and B? Or perhaps the problem is considering the total time as a function of ( N ), and we need to find the ( N ) that minimizes this total time.Wait, but ( N ) is the number of customers, which is a given number, not a variable. So perhaps I'm misunderstanding the problem. Maybe the manager can choose how much data to encrypt with each algorithm? Or perhaps the encryption rates are given per customer, so the total time is the sum over all customers, but that would make ( N ) a variable.Wait, let me read the problem again:"A public relations manager is working with a company to improve its customer privacy protocols. The company has a database containing sensitive information of ( N ) customers. To ensure data security, the manager suggests encrypting the data using a combination of two encryption algorithms: Algorithm A and Algorithm B.1. Encryption Time Analysis: - Algorithm A encrypts data at a rate of ( f_A(N) = k_A cdot log(N) ), where ( k_A ) is a constant. - Algorithm B encrypts data at a rate of ( f_B(N) = k_B cdot N^{1/3} ), where ( k_B ) is a constant. Determine the value of ( N ) such that the total encryption time using both algorithms is minimized. Assume ( k_A = 2 ) and ( k_B = 3 )."Wait, so the total encryption time is ( f_A(N) + f_B(N) = 2 log N + 3 N^{1/3} ). So, the manager can choose ( N ) to minimize this total time. But ( N ) is the number of customers, which is fixed. So, perhaps the manager can choose how to distribute the encryption between A and B? Or maybe the encryption rates are per unit data, and the total data is proportional to ( N ), so the total time is ( 2 log N + 3 N^{1/3} ). So, as ( N ) increases, the total encryption time increases.But if ( N ) is fixed, then the total encryption time is fixed as well. So, perhaps the problem is considering that the manager can choose how much data to encrypt with each algorithm, but the total data is ( N ). So, maybe the manager can split the data between A and B, such that the total encryption time is minimized.Wait, that makes more sense. So, perhaps instead of encrypting all data with both algorithms, the manager can split the data into two parts: one part encrypted with A and the other with B. So, if ( x ) is the amount of data encrypted with A, and ( N - x ) is encrypted with B, then the total encryption time would be ( 2 log x + 3 (N - x)^{1/3} ). Then, we can find the ( x ) that minimizes this time.But the problem says "the total encryption time using both algorithms", so maybe it's the sum of the times for each algorithm, regardless of how the data is split. Hmm, I'm confused.Wait, let me read the problem again: "the total encryption time using both algorithms is minimized." So, perhaps the manager is using both algorithms simultaneously, and the total time is the maximum of the two times, since both need to finish. Or maybe it's the sum of the times, but that's not clear.Alternatively, maybe the encryption is done in a pipeline, so the total time is the sum. But without more context, it's hard to say. But given the way the problem is phrased, it seems like the total encryption time is the sum of the times taken by each algorithm.But then, as I thought earlier, the derivative is always positive, so the minimal time is at the smallest ( N ), which is 1. But that seems trivial. Maybe the problem is considering that the manager can choose how much data to encrypt with each algorithm, so that the total time is minimized.Wait, let me think again. If the manager can split the data between A and B, then the total encryption time would be the maximum of the two times, since both need to complete before the encryption is done. So, if you split the data into two parts, one encrypted by A and the other by B, the total time would be the maximum of ( 2 log x ) and ( 3 (N - x)^{1/3} ). To minimize the total time, you would want to balance the two so that both times are equal. That way, neither algorithm is idle waiting for the other.So, perhaps the minimal total time occurs when ( 2 log x = 3 (N - x)^{1/3} ). Then, solving for ( x ) and ( N ). But the problem is asking for ( N ), so maybe I need to express this in terms of ( N ).Wait, but I'm getting confused. Let me try to clarify.If the manager can choose how to split the data between A and B, then the total encryption time is the maximum of the two individual times. To minimize this maximum, we set the two times equal:[ 2 log x = 3 (N - x)^{1/3} ]But we have two variables here: ( x ) and ( N ). But ( N ) is the total amount of data, so ( x ) is a fraction of ( N ). Let me denote ( x = a N ), where ( 0 < a < 1 ). Then, ( N - x = (1 - a) N ).Substituting into the equation:[ 2 log(a N) = 3 ((1 - a) N)^{1/3} ]Simplify:[ 2 (log a + log N) = 3 (1 - a)^{1/3} N^{1/3} ]This equation relates ( a ) and ( N ). But we need another equation to solve for both variables. However, the problem is asking for the value of ( N ) that minimizes the total encryption time. So, perhaps we can express ( a ) in terms of ( N ) and then find the ( N ) that minimizes the total time.Alternatively, maybe the manager cannot split the data and has to encrypt all data with both algorithms, so the total time is the sum ( 2 log N + 3 N^{1/3} ). In that case, as I initially thought, the derivative is always positive, so the minimal time is at ( N = 1 ). But that seems too straightforward, and the problem is probably expecting a more involved solution.Wait, maybe the manager can choose to use either A or B, not necessarily both. So, the total encryption time would be the minimum of ( 2 log N ) and ( 3 N^{1/3} ). But then, the problem says "using both algorithms," so it's not that.Alternatively, perhaps the encryption is done in a way that both algorithms are applied sequentially, so the total time is the sum. But in that case, again, the derivative is always positive, so minimal at ( N = 1 ).I'm getting stuck here. Maybe I need to consider that the manager can choose how much data to encrypt with each algorithm, but the total data is ( N ). So, let's say ( x ) is the amount encrypted with A, and ( N - x ) with B. Then, the total encryption time is ( 2 log x + 3 (N - x)^{1/3} ). To minimize this, we can take the derivative with respect to ( x ) and set it to zero.So, let's define ( T(x) = 2 log x + 3 (N - x)^{1/3} ). Then, the derivative ( T'(x) ) is:[ T'(x) = frac{2}{x} - 3 cdot frac{1}{3} (N - x)^{-2/3} cdot (-1) ][ T'(x) = frac{2}{x} + (N - x)^{-2/3} ]Set ( T'(x) = 0 ):[ frac{2}{x} + (N - x)^{-2/3} = 0 ]But both terms are positive, so their sum can't be zero. Therefore, the minimum occurs at the boundary. So, either ( x = 0 ) or ( x = N ). If ( x = 0 ), then ( T(x) = 3 N^{1/3} ). If ( x = N ), then ( T(x) = 2 log N ). So, the minimal total time is the minimum of these two. Therefore, the manager should choose to encrypt all data with the algorithm that gives the smaller time.So, we need to find ( N ) such that ( 2 log N = 3 N^{1/3} ). Because if ( 2 log N < 3 N^{1/3} ), then encrypting all with A is better, and vice versa.So, set ( 2 log N = 3 N^{1/3} ). Let me solve for ( N ).This is a transcendental equation, so it might not have an analytical solution. I'll need to solve it numerically.Let me define ( f(N) = 2 log N - 3 N^{1/3} ). We need to find ( N ) such that ( f(N) = 0 ).Let me test some values:- ( N = 1 ): ( f(1) = 0 - 3(1) = -3 )- ( N = 8 ): ( f(8) = 2 log 8 - 3(2) = 2 cdot 3 - 6 = 6 - 6 = 0 )Wait, that's interesting. So, ( N = 8 ) is a solution.Wait, let me check:( 2 log 8 = 2 cdot 3 = 6 )( 3 cdot 8^{1/3} = 3 cdot 2 = 6 )Yes, so ( N = 8 ) is a solution.Is there another solution? Let's check ( N = 1 ): ( f(1) = -3 ), ( N = 8 ): 0, ( N = 27 ): ( 2 log 27 - 3 cdot 3 = 2 cdot 3.2958 - 9 ≈ 6.5916 - 9 ≈ -2.4084 ). So, it goes from negative at ( N = 1 ), crosses zero at ( N = 8 ), and then becomes negative again at ( N = 27 ). So, the function ( f(N) ) is positive somewhere between ( N = 1 ) and ( N = 8 )?Wait, no. At ( N = 1 ), ( f(N) = -3 ). At ( N = 8 ), ( f(N) = 0 ). Let's check ( N = e ) (approximately 2.718):( f(e) = 2 cdot 1 - 3 cdot e^{1/3} ≈ 2 - 3 cdot 1.3956 ≈ 2 - 4.1868 ≈ -2.1868 )Still negative. How about ( N = 4 ):( f(4) = 2 log 4 - 3 cdot 4^{1/3} ≈ 2 cdot 1.3863 - 3 cdot 1.5874 ≈ 2.7726 - 4.7622 ≈ -1.9896 )Still negative. ( N = 5 ):( f(5) ≈ 2 cdot 1.6094 - 3 cdot 1.7099 ≈ 3.2188 - 5.1297 ≈ -1.9109 )Hmm, still negative. Wait, so the function goes from -3 at N=1, increases to 0 at N=8, but in between, it's still negative? Wait, no, because at N=8, it's zero, but before that, it's negative. So, the function is increasing from N=1 to N=8, crossing zero at N=8.Wait, let me check the derivative of ( f(N) ):( f(N) = 2 log N - 3 N^{1/3} )( f'(N) = frac{2}{N} - frac{1}{3} cdot 3 N^{-2/3} = frac{2}{N} - N^{-2/3} )At N=8, ( f'(8) = 2/8 - 8^{-2/3} = 0.25 - (4)^{-2/3} ≈ 0.25 - (0.25) = 0 ). So, the function has a critical point at N=8, which is a minimum or maximum?Wait, let's check the second derivative:( f''(N) = -2/N^2 + (2/3) N^{-5/3} )At N=8:( f''(8) = -2/64 + (2/3) cdot 8^{-5/3} ≈ -0.03125 + (2/3) cdot (1/32) ≈ -0.03125 + 0.0208 ≈ -0.01045 )Negative, so the function has a local maximum at N=8. But wait, that contradicts because at N=8, f(N)=0, and before that, it's negative, and after that, it becomes negative again. So, actually, N=8 is a local maximum? Wait, that can't be.Wait, let me plot the function mentally. At N=1, f(N)=-3. At N=8, f(N)=0. At N=27, f(N)≈-2.4. So, the function increases from N=1 to N=8, reaching zero, then decreases again. So, N=8 is a point where the function reaches zero after increasing. So, it's not a local maximum, but rather a point where the function crosses zero.Wait, but the derivative at N=8 is zero, which suggests it's a critical point. Since the function is increasing before N=8 and decreasing after N=8, it must be a maximum. But how can it be a maximum if the function is increasing before and decreasing after? That would make it a maximum. But at N=8, f(N)=0, which is higher than the values on either side (which are negative). So, yes, N=8 is a local maximum.But that seems contradictory because if the function is increasing up to N=8 and then decreasing, the maximum is at N=8. However, the function is negative everywhere except at N=8, where it's zero. So, the function has a maximum at N=8, but it's still negative except at N=8 where it's zero.Therefore, the only solution to ( f(N)=0 ) is N=8. So, at N=8, the times are equal. For N < 8, ( 2 log N < 3 N^{1/3} ), so encrypting all with A is better. For N > 8, ( 2 log N > 3 N^{1/3} ), so encrypting all with B is better.But the problem is asking for the value of N that minimizes the total encryption time using both algorithms. Wait, but if we have to use both algorithms, then the total time is the sum, which is always increasing. So, the minimal total time is at N=1.But that contradicts the earlier reasoning where splitting the data could lead to a lower total time. Wait, I'm getting confused again.Let me clarify the problem statement again: "Determine the value of ( N ) such that the total encryption time using both algorithms is minimized."So, the manager is using both algorithms, meaning the total time is the sum of the times for each algorithm. So, ( T(N) = 2 log N + 3 N^{1/3} ). As I initially thought, the derivative is always positive, so the minimal total time is at N=1.But that seems too trivial. Maybe the manager can choose to use either A or B, not necessarily both, but the problem says "using both algorithms," so it's not that.Alternatively, perhaps the manager can choose how much data to encrypt with each algorithm, but the total data is N. So, the total encryption time is the maximum of the two times, as both need to finish. So, to minimize the maximum time, we set the two times equal, which happens at N=8.Wait, that makes sense. So, if the manager splits the data such that both algorithms finish at the same time, the total encryption time is minimized. So, the minimal total time occurs when both times are equal, which is at N=8.Therefore, the value of N that minimizes the total encryption time is 8.Now, moving on to the second part: Data Breach Probability.The probability that Algorithm A is compromised is ( P_A = frac{1}{N+1} ), and for Algorithm B, ( P_B = frac{2}{N+2} ). The algorithms are independent, so the combined probability of at least one being compromised is ( P = P_A + P_B - P_A P_B ). We need to find N such that ( P leq 0.05 ).So, let's write the equation:[ frac{1}{N+1} + frac{2}{N+2} - frac{1}{N+1} cdot frac{2}{N+2} leq 0.05 ]Simplify this expression:First, let's compute ( P_A + P_B - P_A P_B ):[ P = frac{1}{N+1} + frac{2}{N+2} - frac{2}{(N+1)(N+2)} ]Combine the terms:Let me find a common denominator for all terms, which is ( (N+1)(N+2) ).So,[ P = frac{(N+2) + 2(N+1) - 2}{(N+1)(N+2)} ]Simplify the numerator:[ (N + 2) + 2(N + 1) - 2 = N + 2 + 2N + 2 - 2 = 3N + 2 ]So,[ P = frac{3N + 2}{(N+1)(N+2)} ]We need this to be less than or equal to 0.05:[ frac{3N + 2}{(N+1)(N+2)} leq 0.05 ]Multiply both sides by ( (N+1)(N+2) ) (which is positive for N > 0):[ 3N + 2 leq 0.05 (N+1)(N+2) ]Expand the right side:[ 0.05 (N^2 + 3N + 2) = 0.05N^2 + 0.15N + 0.1 ]So, the inequality becomes:[ 3N + 2 leq 0.05N^2 + 0.15N + 0.1 ]Bring all terms to one side:[ 0.05N^2 + 0.15N + 0.1 - 3N - 2 geq 0 ]Simplify:[ 0.05N^2 - 2.85N - 1.9 geq 0 ]Multiply both sides by 20 to eliminate decimals:[ N^2 - 57N - 38 geq 0 ]Now, solve the quadratic inequality ( N^2 - 57N - 38 geq 0 ).First, find the roots of the quadratic equation ( N^2 - 57N - 38 = 0 ).Using the quadratic formula:[ N = frac{57 pm sqrt{57^2 + 4 cdot 38}}{2} ][ N = frac{57 pm sqrt{3249 + 152}}{2} ][ N = frac{57 pm sqrt{3401}}{2} ]Calculate ( sqrt{3401} ):Since ( 58^2 = 3364 ) and ( 59^2 = 3481 ), so ( sqrt{3401} ) is between 58 and 59.Compute ( 58.3^2 = 58^2 + 2 cdot 58 cdot 0.3 + 0.3^2 = 3364 + 34.8 + 0.09 = 3398.89 )Still less than 3401. Try 58.35:( 58.35^2 = (58 + 0.35)^2 = 58^2 + 2 cdot 58 cdot 0.35 + 0.35^2 = 3364 + 40.6 + 0.1225 = 3404.7225 )That's more than 3401. So, the square root is between 58.3 and 58.35.Let me approximate:Let ( x = 58.3 ), ( x^2 = 3398.89 )We need ( x ) such that ( x^2 = 3401 ). The difference is 3401 - 3398.89 = 2.11.The derivative of ( x^2 ) is ( 2x ), so approximate the root:( x ≈ 58.3 + frac{2.11}{2 cdot 58.3} ≈ 58.3 + frac{2.11}{116.6} ≈ 58.3 + 0.018 ≈ 58.318 )So, ( sqrt{3401} ≈ 58.318 )Thus, the roots are:[ N = frac{57 pm 58.318}{2} ]Calculate both roots:1. ( N = frac{57 + 58.318}{2} = frac{115.318}{2} ≈ 57.659 )2. ( N = frac{57 - 58.318}{2} = frac{-1.318}{2} ≈ -0.659 )Since ( N ) represents the number of customers, it must be positive. So, the relevant root is approximately 57.659.The quadratic ( N^2 - 57N - 38 ) is a parabola opening upwards, so it is ≥ 0 when ( N leq -0.659 ) or ( N geq 57.659 ). Since ( N ) must be positive, we consider ( N geq 57.659 ).Therefore, ( N ) must be at least 58 to satisfy the inequality ( P leq 0.05 ).But let's verify this. Let me plug N=58 into the probability expression:[ P = frac{3 cdot 58 + 2}{(58 + 1)(58 + 2)} = frac{174 + 2}{59 cdot 60} = frac{176}{3540} ≈ 0.0497 ]Which is approximately 0.0497, which is less than 0.05.Now, check N=57:[ P = frac{3 cdot 57 + 2}{58 cdot 59} = frac{171 + 2}{3422} = frac{173}{3422} ≈ 0.0505 ]Which is approximately 0.0505, which is slightly above 0.05.Therefore, the minimal integer value of N that satisfies the condition is 58.So, summarizing:1. The value of N that minimizes the total encryption time is 8.2. The value of N that ensures the combined probability is no more than 0.05 is 58.**Final Answer**1. The value of ( N ) that minimizes the total encryption time is boxed{8}.2. The value of ( N ) that ensures the combined probability is no more than 0.05 is boxed{58}.