🧠:<think>Okay, so I have this problem where Alex and Mr. Smith are discussing population growth in Connecticut using a mathematical model. It involves differential equations and Fourier series. Hmm, let me try to break this down step by step.First, the logistic growth equation is given by dP/dt = rP(1 - P/K). That's a standard model where the population grows exponentially but is limited by the carrying capacity K. Now, Alex added an extra term, h sin(ωt), which represents the impact of cultural policies. So the equation becomes:dP/dt = rP(1 - P/K) + h sin(ωt)I need to analyze the stability of this population model. Stability in differential equations usually refers to whether the solutions approach a fixed point or not. So, without the h sin(ωt) term, the logistic equation has a stable equilibrium at P = K. But with the added term, it's a non-autonomous equation because of the time-dependent sine function.I remember that for linear differential equations, we can use methods like finding homogeneous and particular solutions, but this is nonlinear because of the P(1 - P/K) term. So maybe I need to look into perturbation methods or consider the system near the equilibrium.Let me consider the equilibrium points first. If I set dP/dt = 0, then:0 = rP(1 - P/K) + h sin(ωt)But since sin(ωt) is oscillating, the equilibrium points are time-dependent. That complicates things because the system isn't autonomous. So instead of fixed points, the system might have oscillatory behavior or other dynamics.Maybe I can linearize the equation around the carrying capacity K. Let me set P(t) = K + x(t), where x(t) is a small perturbation. Then substitute into the equation:d(K + x)/dt = r(K + x)(1 - (K + x)/K) + h sin(ωt)Simplify:dK/dt + dx/dt = r(K + x)(1 - 1 - x/K) + h sin(ωt)Since dK/dt = 0 (because K is a constant), we have:dx/dt = r(K + x)(-x/K) + h sin(ωt)Expanding the right-hand side:dx/dt = -r(K + x)(x/K) + h sin(ωt)Since x is small, we can approximate (K + x) ≈ K, so:dx/dt ≈ -r(K)(x/K) + h sin(ωt) = -r x + h sin(ωt)So the linearized equation is:dx/dt = -r x + h sin(ωt)This is a linear nonhomogeneous differential equation. The homogeneous solution is x_h = C e^{-rt}, which decays to zero as t increases because r is positive. The particular solution can be found using the method of undetermined coefficients.Assume a particular solution of the form x_p = A cos(ωt) + B sin(ωt). Then, dx_p/dt = -A ω sin(ωt) + B ω cos(ωt).Substitute into the equation:- A ω sin(ωt) + B ω cos(ωt) = -r (A cos(ωt) + B sin(ωt)) + h sin(ωt)Group the terms:[ -A ω sin(ωt) + B ω cos(ωt) ] = [ -r A cos(ωt) - r B sin(ωt) + h sin(ωt) ]Equate coefficients:For cos(ωt):B ω = -r AFor sin(ωt):- A ω = -r B + hSo we have the system:1. B ω = -r A2. - A ω = -r B + hFrom equation 1, B = (-r A)/ωSubstitute into equation 2:- A ω = -r (-r A / ω) + h- A ω = (r² A)/ω + hMultiply both sides by ω:- A ω² = r² A + h ωBring terms with A to one side:- A ω² - r² A = h ωFactor A:A (-ω² - r²) = h ωSo,A = - h ω / (ω² + r²)Then, from equation 1,B = (-r A)/ω = (-r / ω)( - h ω / (ω² + r²)) = (r h) / (ω² + r²)Therefore, the particular solution is:x_p = A cos(ωt) + B sin(ωt) = [ - h ω / (ω² + r²) ] cos(ωt) + [ r h / (ω² + r²) ] sin(ωt)So the general solution is:x(t) = C e^{-rt} + [ - h ω / (ω² + r²) ] cos(ωt) + [ r h / (ω² + r²) ] sin(ωt)As t approaches infinity, the homogeneous solution C e^{-rt} tends to zero, so the population approaches:P(t) ≈ K + [ - h ω / (ω² + r²) ] cos(ωt) + [ r h / (ω² + r²) ] sin(ωt)This is a steady-state oscillation around the carrying capacity K. So the population doesn't stabilize to a fixed point but oscillates around K with amplitude depending on h, ω, and r.To find the amplitude of these oscillations, we can write the particular solution as a single sinusoid. The amplitude is sqrt( ( - h ω / (ω² + r²) )² + ( r h / (ω² + r²) )² )Which simplifies to:sqrt( h² ω² + r² h² ) / (ω² + r²) ) = h sqrt(ω² + r²) / (ω² + r²) ) = h / sqrt(ω² + r²)So the amplitude is h / sqrt(ω² + r²). Therefore, the oscillations are damped if h is small or if ω is large (high frequency policies) or r is large (strong growth rate). Conversely, if h is large or ω is small (low frequency policies) or r is small, the oscillations can be larger.So, for the population to stabilize, the oscillations need to be minimal. That would happen when the amplitude is small, i.e., when h is small, ω is large, or r is large.But wait, in the general solution, the homogeneous part dies out, so regardless of the parameters, the population will approach these oscillations. So maybe the system doesn't truly stabilize but instead reaches a steady oscillation. So in terms of stability, it's neutrally stable around K, oscillating with a certain amplitude.But if we consider the original equation, the logistic term tends to bring the population back to K, but the sinusoidal term perturbs it. So depending on the relative strength of the perturbation, the population could either oscillate around K or diverge.Wait, but in the linearized model, the homogeneous solution decays, so the population approaches the oscillatory solution. So unless the perturbation is so strong that the linearization breaks down, the population remains bounded near K.So, in terms of conditions:- If the amplitude h / sqrt(ω² + r²) is small, the population remains close to K, oscillating slightly.- If h is very large compared to r and ω, the oscillations could be large, potentially causing the population to vary widely.But since the logistic term is nonlinear, if the population deviates too far from K, the logistic term might dominate again, pulling it back. So maybe even with large h, the population remains bounded.But in the linearized model, we assumed x is small, so if h is too large, the linear approximation might not hold. So perhaps the full nonlinear equation needs to be considered for large h.But for small h, the linear analysis is sufficient. So, in summary, the population will oscillate around K with an amplitude dependent on h, ω, and r. The system doesn't stabilize to a fixed point but reaches a steady oscillation. So the conditions for oscillation are always present when h ≠ 0, but the amplitude depends on the parameters.For the second part, Mr. Smith wants to express the cultural policy impact as a Fourier series. The given function is f(t) = sum_{n=1}^∞ (h_n / n cos(n ω t) + k_n / n sin(n ω t)). They want to find the Fourier coefficients h_n and k_n over one period T = 2π / ω.Wait, but the function f(t) is already given as a Fourier series. So maybe they want to express a general periodic function as a Fourier series and find the coefficients h_n and k_n. But the problem says "determine the Fourier coefficients h_n and k_n that best approximate the impact of the cultural policy over one period T = 2π / ω."So, if f(t) is a periodic function with period T, then its Fourier series is given by:f(t) = a_0 / 2 + sum_{n=1}^∞ [a_n cos(n ω t) + b_n sin(n ω t)]where ω = 2π / T.The coefficients are:a_n = (1/T) ∫_{0}^{T} f(t) cos(n ω t) dtb_n = (1/T) ∫_{0}^{T} f(t) sin(n ω t) dtBut in the problem, the series is written as sum_{n=1}^∞ (h_n / n cos(n ω t) + k_n / n sin(n ω t)). So comparing, we have:a_n = h_n / nb_n = k_n / nTherefore, h_n = n a_nk_n = n b_nSo, to find h_n and k_n, we need to compute the Fourier coefficients a_n and b_n first, then multiply by n.Given that, the process is:1. Determine the period T = 2π / ω.2. Compute a_n = (1/T) ∫_{0}^{T} f(t) cos(n ω t) dt3. Compute b_n = (1/T) ∫_{0}^{T} f(t) sin(n ω t) dt4. Then, h_n = n a_n and k_n = n b_nBut since f(t) is given as a complex periodic function, unless we have an explicit form, we can't compute the integrals. So perhaps the question is more about expressing the impact as a Fourier series and identifying the coefficients, rather than computing them numerically.Alternatively, if f(t) is given as a specific function, we could compute the integrals, but since it's general, the answer is that h_n and k_n are n times the Fourier cosine and sine coefficients respectively.So, in conclusion, to best approximate f(t) over one period, compute the standard Fourier coefficients a_n and b_n, then scale them by n to get h_n and k_n.But I'm not sure if I'm interpreting the question correctly. It says "the cultural policy follows a more complex periodic function f(t)" and they want to express this as a Fourier series. So, yes, the coefficients are found by integrating f(t) multiplied by cos(n ω t) and sin(n ω t) over one period, then scaling by n.So, summarizing:For the first part, the population oscillates around K with amplitude h / sqrt(ω² + r²). The system doesn't stabilize to a fixed point but maintains oscillations. The conditions for oscillation are always present when h ≠ 0, with the amplitude depending on h, ω, and r.For the second part, the Fourier coefficients h_n and k_n are obtained by computing the standard Fourier series coefficients a_n and b_n, then scaling them by n. So h_n = n a_n and k_n = n b_n, where a_n and b_n are the integrals of f(t) multiplied by cos(n ω t) and sin(n ω t) over one period.I think that's the gist of it. I might have missed some nuances, especially in the stability analysis, but this is my understanding based on the linearization approach.Final Answer1. The population oscillates around the carrying capacity ( K ) with amplitude ( frac{h}{sqrt{omega^2 + r^2}} ). The system does not stabilize to a fixed point but exhibits sustained oscillations. The conditions for oscillation are always present when ( h neq 0 ), with the amplitude depending on the parameters ( h ), ( omega ), and ( r ).2. The Fourier coefficients are given by ( h_n = n a_n ) and ( k_n = n b_n ), where ( a_n ) and ( b_n ) are the standard Fourier coefficients computed as: [ a_n = frac{1}{T} int_{0}^{T} f(t) cos(n omega t) , dt ] [ b_n = frac{1}{T} int_{0}^{T} f(t) sin(n omega t) , dt ] with ( T = frac{2pi}{omega} ).The final answers are:1. The population exhibits oscillatory behavior around ( K ) with amplitude ( boxed{frac{h}{sqrt{omega^2 + r^2}}} ).2. The Fourier coefficients are ( h_n = n a_n ) and ( k_n = n b_n ), where ( a_n ) and ( b_n ) are computed as above.boxed{frac{h}{sqrt{omega^2 + r^2}}}</think>

🧠:<think>Okay, so I have this problem about a material supplier who provides reclaimed and recycled materials for sustainable furniture production. They need to deliver 10 tons of materials, and the materials are in the ratio of 3 parts reclaimed wood, 2 parts recycled metal, and 1 part recycled plastic. First, I need to figure out how much of each material they're delivering. Since the total is 10 tons and the ratio is 3:2:1, that adds up to 6 parts. So each part is 10 tons divided by 6, which is approximately 1.6667 tons. So, reclaimed wood is 3 parts, which would be 3 * 1.6667 ≈ 5 tons. Recycled metal is 2 parts, so 2 * 1.6667 ≈ 3.3333 tons. Recycled plastic is 1 part, so about 1.6667 tons.Next, I need to calculate the transportation cost for each material. The transportation cost is given per ton-mile. For wood, it's 50 per ton-mile, metal is 70, and plastic is 40. The distances are 200 miles for wood, 150 miles for metal, and 100 miles for plastic.So, for each material, I can calculate the cost by multiplying the cost per ton-mile by the distance and then by the tons delivered.Starting with wood: 5 tons * 200 miles * 50 per ton-mile. Let me compute that. 5 * 200 = 1000 ton-miles. 1000 * 50 = 50,000.For metal: 3.3333 tons * 150 miles * 70 per ton-mile. Let's see, 3.3333 * 150 ≈ 500 ton-miles. 500 * 70 = 35,000.For plastic: 1.6667 tons * 100 miles * 40 per ton-mile. 1.6667 * 100 ≈ 166.67 ton-miles. 166.67 * 40 ≈ 6,666.80.Adding all these up: 50,000 + 35,000 + 6,666.80 ≈ 91,666.80. So, the total transportation cost is approximately 91,666.80.Moving on to the second part. The contract requires at least 60% of the materials to be sourced within a 100-mile radius. Currently, the distances are 200, 150, and 100 miles for wood, metal, and plastic respectively. So, only the plastic is within 100 miles. First, let's check what percentage of the materials are sourced within 100 miles. The plastic is 1.6667 tons, so the percentage is (1.6667 / 10) * 100 ≈ 16.6667%. That's way below the 60% requirement. So, they need to adjust the sourcing distances.The goal is to adjust the distances so that at least 60% of the materials are within 100 miles without increasing the total transportation cost. So, we need to figure out how much more material needs to be sourced within 100 miles and how to do that without increasing the cost.First, 60% of 10 tons is 6 tons. Currently, only 1.6667 tons are within 100 miles. So, they need an additional 6 - 1.6667 ≈ 4.3333 tons to be sourced within 100 miles.They can adjust the sourcing locations for each material. So, they can try to source more of the materials from within 100 miles. But, they have to make sure that the transportation cost doesn't increase.Let me think about how to approach this. The transportation cost is dependent on both the distance and the amount. So, if they can source more of the cheaper materials (in terms of transportation cost per ton-mile) from closer locations, they can potentially meet the 60% requirement without increasing the total cost.Looking at the transportation costs: wood is 50/ton-mile, metal is 70, and plastic is 40. So, plastic is the cheapest to transport, followed by wood, then metal. So, to minimize cost, they should prioritize sourcing more plastic within 100 miles, then wood, then metal.But currently, all the plastic is already within 100 miles. So, they can't get more plastic within 100 miles unless they increase the amount of plastic. But the ratio is fixed at 3:2:1, so they can't change the proportions. They have to maintain 5 tons wood, 3.3333 tons metal, and 1.6667 tons plastic.Wait, actually, the problem says they can adjust the sourcing locations for each material type. So, they can source some of each material from within 100 miles, but the total amount of each material is fixed.So, for each material, they can source some portion from within 100 miles and the rest from farther away. But the transportation cost is per ton-mile, so if they source a portion from closer, the cost for that portion would be lower, but the rest would still be sourced from farther, so the cost for that portion would be higher.But the total transportation cost should not increase. So, the savings from sourcing some material closer should offset the increased cost from sourcing the remaining farther away.Wait, but in the original setup, all materials are sourced from their respective distances. If they change the sourcing, they might have to source some from closer and some from farther, but the overall cost should remain the same.Alternatively, maybe they can just move some of the farther materials to within 100 miles, but that might increase the cost unless they can compensate by reducing the distance for more expensive materials.This is getting a bit complicated. Let me try to structure it.Let me denote:For each material, let x be the amount sourced within 100 miles, and (total amount - x) sourced from the original distance.But wait, the original distances are 200, 150, and 100 miles. So, for wood, original distance is 200, metal is 150, plastic is 100.If they adjust the sourcing, they can have some amount sourced within 100 miles, and the rest from the original location. But the cost for the within 100 miles would be distance <=100, but what distance exactly? Maybe we can assume that the within 100 miles is 100 miles, but actually, it's any distance up to 100. To minimize cost, they would source as much as possible from the closest possible, which is 0 miles, but of course, they can't do that. Alternatively, to just meet the 100-mile requirement, they can source from 100 miles.But perhaps the cost is based on the distance, so if they source from 100 miles, the cost would be 100 miles * cost per ton-mile.But in the original cost, wood was sourced from 200 miles, metal from 150, and plastic from 100.So, if they source some amount of wood from 100 miles instead of 200, the cost for that portion would be 100 * 50 = 5,000 per ton, whereas before it was 200 * 50 = 10,000 per ton. Wait, no, that's not correct. Wait, the cost is per ton-mile, so for each ton, it's 50 per mile. So, for 1 ton, 200 miles would be 200 * 50 = 10,000. If sourced from 100 miles, it's 100 * 50 = 5,000. So, the cost per ton would be halved.Similarly for metal: original cost per ton is 150 * 70 = 10,500. If sourced from 100 miles, it's 100 * 70 = 7,000 per ton.For plastic, it's already sourced from 100 miles, so no change.So, if they can source some of the wood and metal from within 100 miles, they can reduce the transportation cost for those portions, but they have to make sure that the total transportation cost doesn't increase. Wait, but the problem says they need to comply without increasing the total transportation cost. So, they need to adjust the distances such that the total cost remains the same as in part 1, but at least 60% of the materials are sourced within 100 miles.So, let's denote:Let’s define variables:For wood: let x be the tons sourced within 100 miles, so (5 - x) tons sourced from 200 miles.For metal: let y be the tons sourced within 100 miles, so (3.3333 - y) tons sourced from 150 miles.Plastic is already all within 100 miles, so no change.The total transportation cost should be equal to the original cost, which was 91,666.80.So, the cost for wood would be x * 100 * 50 + (5 - x) * 200 * 50.Similarly, for metal: y * 100 * 70 + (3.3333 - y) * 150 * 70.Plastic remains the same: 1.6667 * 100 * 40.So, let's write the equation:Total cost = [x * 100 * 50 + (5 - x) * 200 * 50] + [y * 100 * 70 + (3.3333 - y) * 150 * 70] + [1.6667 * 100 * 40] = 91,666.80Simplify each term:Wood cost: 5000x + 50000 - 10000x = -5000x + 50000Metal cost: 7000y + 36666.5 - 10500y = -3500y + 36666.5Plastic cost: 6666.80So, total cost equation:(-5000x + 50000) + (-3500y + 36666.5) + 6666.80 = 91,666.80Combine constants: 50000 + 36666.5 + 6666.80 ≈ 93,333.30So, equation becomes:-5000x -3500y + 93,333.30 = 91,666.80Subtract 93,333.30 from both sides:-5000x -3500y = 91,666.80 - 93,333.30 ≈ -1,666.50Multiply both sides by -1:5000x + 3500y = 1,666.50We can simplify this equation by dividing both sides by 50:100x + 70y = 33.33So, 100x + 70y = 33.33We also have the constraint that the total amount sourced within 100 miles should be at least 6 tons (60% of 10 tons). Currently, plastic is 1.6667 tons within 100 miles, so we need x + y + 1.6667 >= 6So, x + y >= 6 - 1.6667 ≈ 4.3333 tonsSo, we have two equations:1. 100x + 70y = 33.332. x + y >= 4.3333We need to find x and y such that these are satisfied, with x <=5 and y <=3.3333.Let me solve equation 1 for y:70y = 33.33 - 100xy = (33.33 - 100x)/70We can plug this into equation 2:x + (33.33 - 100x)/70 >= 4.3333Multiply both sides by 70 to eliminate denominator:70x + 33.33 - 100x >= 303.331Combine like terms:-30x + 33.33 >= 303.331-30x >= 303.331 - 33.33 ≈ 270x <= 270 / (-30) = -9But x cannot be negative, so this suggests that there is no solution where x + y >=4.3333 and 100x +70y=33.33. Because solving leads to x <= -9, which is impossible.This means that it's impossible to meet the 60% requirement without increasing the total transportation cost. Therefore, the supplier cannot comply with the clause without increasing the total transportation cost.Wait, but the problem says "if not, calculate the minimum adjustment needed in the sourcing distances to comply with the clause without increasing the total transportation cost calculated in sub-problem 1."But according to this, it's impossible. So, maybe I made a mistake in my approach.Alternatively, perhaps instead of keeping the same total cost, they can adjust the distances such that the total cost remains the same, but more material is sourced within 100 miles.Wait, but the total cost is fixed at 91,666.80. So, they need to find x and y such that the total cost remains the same, but x + y +1.6667 >=6.But as per the previous calculation, it's impossible because the equation leads to a negative x.Alternatively, maybe I need to consider that they can source some materials from closer than 100 miles, but that would reduce the cost, allowing them to source more within 100 miles without increasing the total cost.Wait, but the problem says "without increasing the total transportation cost calculated in sub-problem 1." So, they can't spend more than 91,666.80.But if they source some materials from closer, the cost for those would be less, so they could potentially source more within 100 miles and use the saved cost to source other materials from farther away, but that might complicate things.Alternatively, maybe they can only adjust the distances for some materials, not all.Wait, perhaps I need to approach this differently. Let's think about the total cost.Original cost:Wood: 5 tons * 200 miles * 50 = 50,000Metal: 3.3333 tons * 150 miles * 70 ≈ 35,000Plastic: 1.6667 tons * 100 miles * 40 ≈ 6,666.80Total: 91,666.80If they want to source more within 100 miles, they can reduce the distance for some materials, but they have to make sure that the total cost doesn't exceed 91,666.80.Let me consider moving some wood and metal to within 100 miles.Suppose they move z tons of wood from 200 miles to 100 miles. The cost for that z tons would decrease by (200 - 100)*50*z = 5000z.Similarly, moving w tons of metal from 150 miles to 100 miles would decrease the cost by (150 - 100)*70*w = 3500w.But to meet the 60% requirement, they need to have at least 6 tons within 100 miles. Currently, plastic is 1.6667 tons, so they need 6 -1.6667 ≈4.3333 tons more.So, z + w >=4.3333The total cost saved from moving z tons of wood and w tons of metal is 5000z + 3500w.But they need to keep the total cost the same, so the savings must be used to offset any increased costs elsewhere. However, in this case, moving materials closer only reduces the cost, so to keep the total cost the same, they would have to move some materials farther, which would increase the cost, offsetting the savings.But the problem says they can adjust the sourcing distances, so perhaps they can move some materials closer and others farther, but the total cost remains the same.Wait, but the total cost is fixed. So, the savings from moving some materials closer must be offset by increased costs from moving other materials farther.But in this case, since they are already sourcing from the farthest possible points (200, 150, 100), moving any closer would only save cost, but they can't move any farther because that would increase cost beyond the original total.Wait, perhaps they can only move materials closer, but to meet the 60% requirement, they need to move enough to make up the 4.3333 tons, but the savings from moving those materials closer would reduce the total cost below 91,666.80, which is not allowed. Therefore, they cannot meet the requirement without either increasing the total cost or moving some materials farther, which would increase the cost beyond the original.Therefore, it's impossible to meet the 60% requirement without increasing the total transportation cost.But the problem says "if not, calculate the minimum adjustment needed in the sourcing distances to comply with the clause without increasing the total transportation cost calculated in sub-problem 1."Wait, maybe I need to find the minimal adjustment in distances such that the total cost remains the same, but 60% of the materials are within 100 miles.Alternatively, perhaps they can adjust the distances for some materials to be within 100 miles, and for others, keep them at the same distance, but adjust the amounts sourced from each location.Wait, but the amounts are fixed by the ratio. They have to deliver 5 tons wood, 3.3333 tons metal, and 1.6667 tons plastic.So, they can't change the amounts, only the distances.Therefore, to meet the 60% requirement, they need to have at least 6 tons within 100 miles. Currently, plastic is 1.6667 tons. So, they need to have 4.3333 tons of wood and metal sourced within 100 miles.Let me denote:Let x be the tons of wood sourced within 100 miles, so (5 - x) tons sourced from 200 miles.Let y be the tons of metal sourced within 100 miles, so (3.3333 - y) tons sourced from 150 miles.We need x + y >=4.3333The total transportation cost should remain 91,666.80.So, the cost equation is:x*100*50 + (5 - x)*200*50 + y*100*70 + (3.3333 - y)*150*70 + 1.6667*100*40 = 91,666.80Simplify each term:Wood: 5000x + 50000 - 10000x = -5000x + 50000Metal: 7000y + 36666.5 - 10500y = -3500y + 36666.5Plastic: 6666.80Total cost: (-5000x + 50000) + (-3500y + 36666.5) + 6666.80 = 91,666.80Combine constants: 50000 + 36666.5 + 6666.80 ≈93,333.30So, -5000x -3500y +93,333.30 =91,666.80Thus, -5000x -3500y = -1,666.50Multiply both sides by -1:5000x + 3500y =1,666.50Divide both sides by 50:100x +70y =33.33So, 100x +70y =33.33We also have x + y >=4.3333We need to find x and y such that 100x +70y =33.33 and x + y >=4.3333Let me solve for y from the first equation:70y =33.33 -100xy=(33.33 -100x)/70Now, plug into the second inequality:x + (33.33 -100x)/70 >=4.3333Multiply both sides by70:70x +33.33 -100x >=303.331Combine like terms:-30x +33.33 >=303.331-30x >=270x<=-9But x cannot be negative, so this is impossible.Therefore, there is no solution where x + y >=4.3333 and 100x +70y=33.33. Hence, it's impossible to meet the 60% requirement without increasing the total transportation cost.Therefore, the supplier cannot comply with the clause without increasing the total transportation cost.But the problem says "if not, calculate the minimum adjustment needed in the sourcing distances to comply with the clause without increasing the total transportation cost calculated in sub-problem 1."Wait, maybe I need to find the minimal adjustment in distances such that the total cost remains the same, but 60% of the materials are within 100 miles.Alternatively, perhaps they can adjust the distances for some materials to be within 100 miles, and for others, keep them at the same distance, but adjust the amounts sourced from each location.Wait, but the amounts are fixed by the ratio. They have to deliver 5 tons wood, 3.3333 tons metal, and 1.6667 tons plastic.So, they can't change the amounts, only the distances.Therefore, to meet the 60% requirement, they need to have at least 6 tons within 100 miles. Currently, plastic is 1.6667 tons. So, they need to have 4.3333 tons of wood and metal sourced within 100 miles.But as per the previous calculation, it's impossible because the equation leads to a negative x.Therefore, the conclusion is that the current setup does not meet the requirement, and it's impossible to adjust the distances without increasing the total transportation cost.But the problem says "calculate the minimum adjustment needed in the sourcing distances to comply with the clause without increasing the total transportation cost calculated in sub-problem 1."Wait, maybe I need to consider that they can move some materials closer and others farther, but in such a way that the total cost remains the same.Let me think about this.Suppose they move some wood closer and some metal farther, or vice versa, so that the total cost remains the same.But the problem is that moving wood closer reduces the cost, so to keep the total cost the same, they would have to move some other material farther, increasing its cost.But the goal is to have more materials within 100 miles, so moving wood closer is good, but moving metal farther would be bad because it would increase the cost.Alternatively, maybe they can move some metal closer and some wood farther, but that would increase the cost for wood, which is cheaper per ton-mile.Wait, let's try to model this.Let’s denote:For wood: let x be the tons sourced within 100 miles, and (5 - x) tons sourced from a farther distance D1.For metal: let y be the tons sourced within 100 miles, and (3.3333 - y) tons sourced from a farther distance D2.Plastic remains at 100 miles.Total cost remains 91,666.80.We need x + y +1.6667 >=6, so x + y >=4.3333.The cost equation is:x*100*50 + (5 - x)*D1*50 + y*100*70 + (3.3333 - y)*D2*70 +1.6667*100*40 =91,666.80We have two variables D1 and D2, which are the new distances for the remaining wood and metal.But this seems too complex because we have two variables and one equation.Alternatively, perhaps we can assume that the remaining wood and metal are sourced from their original distances, but that would mean no adjustment, which doesn't help.Alternatively, maybe they can source some wood from within 100 miles and some from farther, but the farther distance would have to be such that the total cost remains the same.Wait, let me try to think differently.Suppose they move some wood closer to 100 miles, which reduces the cost, but to keep the total cost the same, they have to move some other material farther.But moving other materials farther would increase the cost, so the net effect is that the total cost remains the same.But we need to maximize the amount sourced within 100 miles.Let me consider moving some wood closer and some metal farther.Let’s say they move z tons of wood from 200 to 100 miles. The cost saved is z*(200 -100)*50 = z*5000.To keep the total cost the same, they have to increase the cost elsewhere by z*5000.Suppose they move w tons of metal from 150 to D miles, where D >150. The increased cost is w*(D -150)*70.So, we have:z*5000 = w*(D -150)*70We need to find z and w such that z + w >=4.3333 (since x + y >=4.3333, where x=z and y= w, but actually, y is the amount moved closer, but in this case, metal is moved farther, so y would decrease. Wait, this is getting confusing.Alternatively, perhaps it's better to consider that moving wood closer allows us to have more within 100 miles, but to keep the total cost the same, we have to move some metal farther, which would reduce the amount within 100 miles.Wait, but we need to have x + y >=4.3333, where x is wood within 100, y is metal within 100.If we move some wood closer, x increases, but if we move some metal farther, y decreases.So, the net effect on x + y could be positive or negative.We need to find z and w such that:z (wood moved closer) + (y - w) (metal moved closer minus those moved farther) >=4.3333But this is getting too convoluted.Perhaps a better approach is to set up an optimization problem where we maximize the amount within 100 miles subject to the total cost constraint.But since this is a thought process, let me try to find a way.Let’s assume that they move some wood closer and some metal closer, but to keep the total cost the same, they have to move some other materials farther.But since plastic is already at 100 miles, they can't move it farther without increasing the cost.Alternatively, maybe they can move some wood closer and some metal closer, and compensate by moving some wood farther, but that would complicate.Wait, perhaps the minimal adjustment is to move as much as possible of the cheaper materials (in terms of transportation cost) closer, which would allow more within 100 miles without increasing the total cost.Since plastic is already within 100 miles, the next cheapest is wood at 50/ton-mile, then metal at 70.So, moving wood closer gives more cost savings per ton moved.Therefore, to maximize the amount within 100 miles, we should move as much wood as possible closer.Let’s calculate how much wood we can move closer without increasing the total cost.Moving z tons of wood from 200 to 100 miles saves z*(200 -100)*50 =5000z.To keep the total cost the same, we can move some metal farther, increasing the cost by 5000z.Let’s say we move w tons of metal from 150 to D miles, where D >150.The increased cost is w*(D -150)*70.So, 5000z =70w(D -150)We need to find z and w such that z + y >=4.3333, where y is the amount of metal within 100 miles.But y is currently 0, since all metal is at 150 miles. If we move some metal closer, y increases, but if we move some farther, y decreases.Wait, this is getting too tangled.Alternatively, perhaps the minimal adjustment is to move some wood closer and some metal closer, but to keep the total cost the same, we have to move some other materials farther.But since plastic is already at 100 miles, we can't move it farther without increasing the cost.Alternatively, maybe the minimal adjustment is to move some wood closer and some metal closer, but the cost savings from moving wood closer can be used to move some metal closer as well.Wait, let me try to think in terms of cost per ton-mile.Wood: 50 per ton-mileMetal: 70 per ton-milePlastic: 40 per ton-mileSo, moving wood closer saves 50 per ton-mile, moving metal closer saves 70 per ton-mile.To maximize the amount within 100 miles, we should prioritize moving the materials with the highest cost per ton-mile first, because moving them closer gives more savings per ton.So, metal has higher cost per ton-mile than wood, so moving metal closer gives more savings per ton.Therefore, to maximize the amount within 100 miles, we should move as much metal as possible closer, then wood.But moving metal closer would require reducing the distance from 150 to 100, saving 50 miles per ton.So, moving w tons of metal closer saves w*50*70 =3500w.Similarly, moving z tons of wood closer saves z*100*50=5000z.We need to move enough to get x + y >=4.3333, where x is wood moved closer, y is metal moved closer.But the total savings from moving x and y should be used to offset any increased costs elsewhere. But since we can't move plastic farther, the only way is to move some materials closer and others farther, but the total cost remains the same.Wait, perhaps the minimal adjustment is to move some metal closer and some wood farther, but that would increase the cost for wood, which is cheaper, so the net effect might be a cost increase.Alternatively, maybe the minimal adjustment is to move some metal closer and some wood closer, but to keep the total cost the same, we have to move some other materials farther, but since plastic is already at 100, we can't move it farther.This is getting too complicated. Maybe the answer is that it's impossible to meet the requirement without increasing the total cost, so the minimal adjustment is to increase the total cost, but the problem says "without increasing the total transportation cost."Therefore, the conclusion is that the current setup does not meet the 60% requirement, and it's impossible to adjust the distances without increasing the total transportation cost. Therefore, the supplier cannot comply with the clause without increasing the total transportation cost.But the problem says "if not, calculate the minimum adjustment needed in the sourcing distances to comply with the clause without increasing the total transportation cost calculated in sub-problem 1."Wait, maybe I need to find the minimal adjustment in distances such that the total cost remains the same, but 60% of the materials are within 100 miles.Alternatively, perhaps they can adjust the distances for some materials to be within 100 miles, and for others, keep them at the same distance, but adjust the amounts sourced from each location.But the amounts are fixed by the ratio. They have to deliver 5 tons wood, 3.3333 tons metal, and 1.6667 tons plastic.So, they can't change the amounts, only the distances.Therefore, to meet the 60% requirement, they need to have at least 6 tons within 100 miles. Currently, plastic is 1.6667 tons. So, they need to have 4.3333 tons of wood and metal sourced within 100 miles.But as per the previous calculation, it's impossible because the equation leads to a negative x.Therefore, the answer is that the current setup does not meet the requirement, and it's impossible to adjust the distances without increasing the total transportation cost.</think>

🧠:<think>First, I need to calculate the total time the translator will spend on transcribing and translating all five interviews. I'll start by determining the total length of all the interviews combined.Next, I'll use the given rates to find out how long it takes to transcribe and translate each interview. For transcription, the translator can handle 150 words per minute, and the chefs speak at 130 words per minute. For translation, the rate is 80 words per minute.After calculating the time for each task, I'll sum them up to get the total time in minutes and then convert it to hours.Then, I'll calculate the total earnings by determining the number of words for each interview, multiplying them by the respective rates for transcription and translation, and summing up all the amounts.</think>

🧠:<think>Okay, so I have this problem about a beekeeper who wants to distribute his bees among his hives to maximize honey production. There are two parts: first, deriving a general formula, and then applying it with specific numbers. Let me try to work through this step by step.Starting with part 1: Derive the formula for the number of bees ( B_i ) that should be placed in each hive to maximize the total honey production. Then express the total honey production in terms of ( a_i ), ( b_i ), and ( N ).Hmm, each hive has a honey production function ( H_i(B_i) = a_i B_i - b_i B_i^2 ). So, this is a quadratic function in terms of ( B_i ). Since the coefficient of ( B_i^2 ) is negative (( -b_i )), the function is concave down, meaning it has a maximum point. That makes sense because too many bees in a hive might lead to diminishing returns or even reduced honey production.To find the maximum for each hive, I can take the derivative of ( H_i ) with respect to ( B_i ) and set it equal to zero. Let me do that.The derivative ( dH_i/dB_i = a_i - 2b_i B_i ). Setting this equal to zero gives:( a_i - 2b_i B_i = 0 )Solving for ( B_i ):( 2b_i B_i = a_i )( B_i = frac{a_i}{2b_i} )So, for each hive, the optimal number of bees is ( a_i/(2b_i) ). That seems straightforward.But wait, the problem mentions distributing a total of ( N ) bees among ( M ) hives. So, if I just set each ( B_i ) to ( a_i/(2b_i) ), the sum of all ( B_i ) might not equal ( N ). Therefore, I need to consider the distribution such that the total number of bees is ( N ) while maximizing the total honey production.This sounds like an optimization problem with a constraint. The total honey production ( H ) is the sum of ( H_i(B_i) ) for all hives, so:( H = sum_{i=1}^{M} (a_i B_i - b_i B_i^2) )Subject to the constraint:( sum_{i=1}^{M} B_i = N )To solve this, I can use the method of Lagrange multipliers. The idea is to maximize the total honey production while satisfying the constraint on the total number of bees.Let me set up the Lagrangian function:( mathcal{L} = sum_{i=1}^{M} (a_i B_i - b_i B_i^2) - lambda left( sum_{i=1}^{M} B_i - N right) )Here, ( lambda ) is the Lagrange multiplier. To find the maximum, I need to take the partial derivatives of ( mathcal{L} ) with respect to each ( B_i ) and ( lambda ), and set them equal to zero.Taking the partial derivative with respect to ( B_i ):( frac{partial mathcal{L}}{partial B_i} = a_i - 2b_i B_i - lambda = 0 )So, for each hive ( i ):( a_i - 2b_i B_i - lambda = 0 )Which can be rearranged to:( 2b_i B_i = a_i - lambda )( B_i = frac{a_i - lambda}{2b_i} )Hmm, that's interesting. So, each ( B_i ) is expressed in terms of ( a_i ), ( b_i ), and ( lambda ). Now, I need another equation to solve for ( lambda ). That comes from the constraint:( sum_{i=1}^{M} B_i = N )Substituting the expression for ( B_i ):( sum_{i=1}^{M} frac{a_i - lambda}{2b_i} = N )Let me write that as:( frac{1}{2} sum_{i=1}^{M} left( frac{a_i}{b_i} - frac{lambda}{b_i} right) = N )Factor out the constants:( frac{1}{2} left( sum_{i=1}^{M} frac{a_i}{b_i} - lambda sum_{i=1}^{M} frac{1}{b_i} right) = N )Multiply both sides by 2:( sum_{i=1}^{M} frac{a_i}{b_i} - lambda sum_{i=1}^{M} frac{1}{b_i} = 2N )Now, solve for ( lambda ):( lambda sum_{i=1}^{M} frac{1}{b_i} = sum_{i=1}^{M} frac{a_i}{b_i} - 2N )So,( lambda = frac{ sum_{i=1}^{M} frac{a_i}{b_i} - 2N }{ sum_{i=1}^{M} frac{1}{b_i} } )Once I have ( lambda ), I can plug it back into the expression for each ( B_i ):( B_i = frac{a_i - lambda}{2b_i} )So, that gives me the number of bees for each hive. Therefore, the formula for ( B_i ) is:( B_i = frac{a_i - lambda}{2b_i} ), where ( lambda = frac{ sum_{i=1}^{M} frac{a_i}{b_i} - 2N }{ sum_{i=1}^{M} frac{1}{b_i} } )Alternatively, this can be written as:( B_i = frac{a_i}{2b_i} - frac{ lambda }{2b_i } )But since ( lambda ) is a constant across all hives, this shows that each hive's optimal bee count is adjusted by the same term ( lambda/(2b_i) ).Now, moving on to expressing the total honey production in terms of ( a_i ), ( b_i ), and ( N ).First, let's compute the total honey production ( H ):( H = sum_{i=1}^{M} (a_i B_i - b_i B_i^2) )We can substitute ( B_i ) from above into this equation. But this might get a bit complicated. Alternatively, since we have the Lagrangian, we can use the fact that at the maximum, the derivative with respect to ( lambda ) gives the constraint, so the total honey can be expressed in terms of ( lambda ) and ( N ).Wait, another approach is to note that each hive's maximum is achieved when ( B_i = a_i/(2b_i) ), but since we have a constraint on the total number of bees, we have to adjust each hive's allocation so that the sum is ( N ). The total honey production will be the sum of each hive's production at their respective ( B_i ).Alternatively, perhaps we can express ( H ) in terms of ( lambda ). Let me think.From the Lagrangian, we have:( mathcal{L} = H - lambda ( sum B_i - N ) )At the maximum, the derivative with respect to ( lambda ) is zero, which gives the constraint. But perhaps we can express ( H ) in terms of ( lambda ).Wait, let's compute ( H ):( H = sum_{i=1}^{M} (a_i B_i - b_i B_i^2 ) )But from the first-order condition, ( a_i - 2b_i B_i = lambda ), so ( a_i B_i = 2b_i B_i^2 + lambda B_i )Therefore, substituting back into ( H ):( H = sum_{i=1}^{M} (2b_i B_i^2 + lambda B_i - b_i B_i^2 ) )Simplify:( H = sum_{i=1}^{M} (b_i B_i^2 + lambda B_i ) )But ( sum_{i=1}^{M} B_i = N ), so:( H = sum_{i=1}^{M} b_i B_i^2 + lambda N )Hmm, that's an expression for ( H ) in terms of ( lambda ) and ( N ). But I need to express ( H ) in terms of ( a_i ), ( b_i ), and ( N ).Alternatively, perhaps we can use the expression for ( lambda ) and substitute it into this equation.Wait, let's recall that:( lambda = frac{ sum_{i=1}^{M} frac{a_i}{b_i} - 2N }{ sum_{i=1}^{M} frac{1}{b_i} } )So, ( lambda ) is expressed in terms of ( a_i ), ( b_i ), and ( N ). Therefore, ( H ) can be written as:( H = sum_{i=1}^{M} b_i B_i^2 + lambda N )But ( B_i ) is also expressed in terms of ( a_i ), ( b_i ), and ( lambda ). So, perhaps substituting ( B_i ) into this equation would give ( H ) purely in terms of ( a_i ), ( b_i ), and ( N ).Let me try that.We have:( B_i = frac{a_i - lambda}{2b_i} )So, ( B_i^2 = left( frac{a_i - lambda}{2b_i} right)^2 = frac{(a_i - lambda)^2}{4b_i^2} )Therefore,( sum_{i=1}^{M} b_i B_i^2 = sum_{i=1}^{M} b_i cdot frac{(a_i - lambda)^2}{4b_i^2} = sum_{i=1}^{M} frac{(a_i - lambda)^2}{4b_i} )So, substituting back into ( H ):( H = sum_{i=1}^{M} frac{(a_i - lambda)^2}{4b_i} + lambda N )But ( lambda ) itself is a function of ( a_i ), ( b_i ), and ( N ). So, this expression is in terms of ( a_i ), ( b_i ), and ( N ), but it's a bit complex.Alternatively, maybe there's a more straightforward way to express ( H ). Let me think.From the Lagrangian, we have:( mathcal{L} = H - lambda ( sum B_i - N ) )At the maximum, the derivative with respect to ( lambda ) is zero, which gives the constraint.But perhaps we can express ( H ) as:( H = mathcal{L} + lambda ( sum B_i - N ) )But since ( sum B_i = N ), this simplifies to ( H = mathcal{L} ). Hmm, not sure if that helps.Wait, maybe another approach. Let's consider that each hive's production is maximized at ( B_i = a_i/(2b_i) ), but due to the constraint, we have to adjust each hive's allocation. The total maximum production without any constraints would be ( sum_{i=1}^{M} (a_i^2)/(4b_i) ). However, with the constraint, the total production is less than that.But I'm not sure if that's helpful. Alternatively, perhaps we can express ( H ) in terms of ( lambda ):From earlier, ( H = sum b_i B_i^2 + lambda N )But ( sum b_i B_i^2 ) can be expressed as:( sum b_i left( frac{a_i - lambda}{2b_i} right)^2 = sum frac{(a_i - lambda)^2}{4b_i} )So,( H = sum frac{(a_i - lambda)^2}{4b_i} + lambda N )But ( lambda ) is given by:( lambda = frac{ sum frac{a_i}{b_i} - 2N }{ sum frac{1}{b_i} } )Let me denote ( S = sum_{i=1}^{M} frac{1}{b_i} ) and ( T = sum_{i=1}^{M} frac{a_i}{b_i} ). Then,( lambda = frac{T - 2N}{S} )So, substituting back into ( H ):( H = sum frac{(a_i - lambda)^2}{4b_i} + lambda N )Let me compute ( (a_i - lambda)^2 ):( (a_i - lambda)^2 = a_i^2 - 2a_i lambda + lambda^2 )Therefore,( sum frac{(a_i - lambda)^2}{4b_i} = frac{1}{4} sum left( frac{a_i^2}{b_i} - 2a_i lambda / b_i + lambda^2 / b_i right ) )Which is:( frac{1}{4} left( sum frac{a_i^2}{b_i} - 2lambda sum frac{a_i}{b_i} + lambda^2 sum frac{1}{b_i} right ) )Substituting ( T ) and ( S ):( frac{1}{4} left( sum frac{a_i^2}{b_i} - 2lambda T + lambda^2 S right ) )Therefore, ( H ) becomes:( H = frac{1}{4} left( sum frac{a_i^2}{b_i} - 2lambda T + lambda^2 S right ) + lambda N )Now, substituting ( lambda = (T - 2N)/S ):First, compute each term:1. ( sum frac{a_i^2}{b_i} ) remains as is.2. ( -2lambda T = -2 cdot frac{T - 2N}{S} cdot T = -2T cdot frac{T - 2N}{S} )3. ( lambda^2 S = left( frac{T - 2N}{S} right )^2 cdot S = frac{(T - 2N)^2}{S} )4. ( lambda N = frac{T - 2N}{S} cdot N )Putting it all together:( H = frac{1}{4} left( sum frac{a_i^2}{b_i} - 2T cdot frac{T - 2N}{S} + frac{(T - 2N)^2}{S} right ) + frac{T - 2N}{S} cdot N )This is getting quite involved. Let me see if I can simplify this expression.First, let's compute the terms inside the brackets:- The first term is ( sum frac{a_i^2}{b_i} ).- The second term is ( -2T cdot frac{T - 2N}{S} ).- The third term is ( frac{(T - 2N)^2}{S} ).So, combining the second and third terms:( -2T cdot frac{T - 2N}{S} + frac{(T - 2N)^2}{S} = frac{ -2T(T - 2N) + (T - 2N)^2 }{S} )Factor out ( (T - 2N) ):( frac{(T - 2N)(-2T + T - 2N)}{S} = frac{(T - 2N)(-T - 2N)}{S} = frac{ - (T - 2N)(T + 2N) }{S} )Which simplifies to:( frac{ - (T^2 - (2N)^2) }{S} = frac{ - (T^2 - 4N^2) }{S} = frac{ -T^2 + 4N^2 }{S} )So, the expression inside the brackets becomes:( sum frac{a_i^2}{b_i} + frac{ -T^2 + 4N^2 }{S} )Therefore, ( H ) is:( H = frac{1}{4} left( sum frac{a_i^2}{b_i} - frac{T^2 - 4N^2}{S} right ) + frac{T - 2N}{S} cdot N )Simplify the last term:( frac{T - 2N}{S} cdot N = frac{TN - 2N^2}{S} )So, putting it all together:( H = frac{1}{4} sum frac{a_i^2}{b_i} - frac{T^2 - 4N^2}{4S} + frac{TN - 2N^2}{S} )Combine the last two terms:First, write them with a common denominator:( - frac{T^2 - 4N^2}{4S} + frac{4(TN - 2N^2)}{4S} = frac{ -T^2 + 4N^2 + 4TN - 8N^2 }{4S } = frac{ -T^2 + 4TN - 4N^2 }{4S } )Factor the numerator:( -T^2 + 4TN - 4N^2 = - (T^2 - 4TN + 4N^2 ) = - (T - 2N)^2 )So, the expression becomes:( H = frac{1}{4} sum frac{a_i^2}{b_i} - frac{(T - 2N)^2}{4S} )Therefore, the total honey production is:( H = frac{1}{4} left( sum_{i=1}^{M} frac{a_i^2}{b_i} - frac{( sum_{i=1}^{M} frac{a_i}{b_i} - 2N )^2 }{ sum_{i=1}^{M} frac{1}{b_i} } right ) )Wow, that's quite a formula. Let me verify if this makes sense.When ( N ) is such that the sum of ( a_i/(2b_i) ) equals ( N ), then ( T = 2N ), so ( T - 2N = 0 ), and the second term becomes zero. Then, ( H = frac{1}{4} sum a_i^2 / b_i ), which is the sum of each hive's maximum production. That makes sense because if we can allocate the bees optimally without constraint, we get the sum of individual maxima.But when ( N ) is different, the second term subtracts something, which adjusts the total production accordingly. So, this formula seems reasonable.Therefore, for part 1, the number of bees in each hive is:( B_i = frac{a_i - lambda}{2b_i} ), where ( lambda = frac{ sum_{i=1}^{M} frac{a_i}{b_i} - 2N }{ sum_{i=1}^{M} frac{1}{b_i} } )And the total honey production is:( H = frac{1}{4} left( sum_{i=1}^{M} frac{a_i^2}{b_i} - frac{( sum_{i=1}^{M} frac{a_i}{b_i} - 2N )^2 }{ sum_{i=1}^{M} frac{1}{b_i} } right ) )Okay, that's part 1 done. Now, moving on to part 2, where we have specific values:- ( N = 5000 )- ( M = 5 )- Hive 1: ( a_1 = 0.5 ), ( b_1 = 0.0001 )- Hive 2: ( a_2 = 0.45 ), ( b_2 = 0.00009 )- Hive 3: ( a_3 = 0.55 ), ( b_3 = 0.00008 )- Hive 4: ( a_4 = 0.52 ), ( b_4 = 0.00007 )- Hive 5: ( a_5 = 0.48 ), ( b_5 = 0.00011 )We need to calculate the number of bees ( B_i ) for each hive.First, let's compute ( S = sum_{i=1}^{5} frac{1}{b_i} ) and ( T = sum_{i=1}^{5} frac{a_i}{b_i} ).Let me compute each term step by step.Compute ( 1/b_i ) for each hive:- Hive 1: ( 1/0.0001 = 10000 )- Hive 2: ( 1/0.00009 approx 11111.1111 )- Hive 3: ( 1/0.00008 = 12500 )- Hive 4: ( 1/0.00007 approx 14285.7143 )- Hive 5: ( 1/0.00011 approx 9090.9091 )So, summing these up:( S = 10000 + 11111.1111 + 12500 + 14285.7143 + 9090.9091 )Let me compute this step by step:10000 + 11111.1111 = 21111.111121111.1111 + 12500 = 33611.111133611.1111 + 14285.7143 ≈ 47896.825447896.8254 + 9090.9091 ≈ 56987.7345So, ( S ≈ 56987.7345 )Now, compute ( a_i / b_i ) for each hive:- Hive 1: ( 0.5 / 0.0001 = 5000 )- Hive 2: ( 0.45 / 0.00009 = 5000 )- Hive 3: ( 0.55 / 0.00008 = 6875 )- Hive 4: ( 0.52 / 0.00007 ≈ 7428.5714 )- Hive 5: ( 0.48 / 0.00011 ≈ 4363.6364 )Now, sum these up:( T = 5000 + 5000 + 6875 + 7428.5714 + 4363.6364 )Compute step by step:5000 + 5000 = 1000010000 + 6875 = 1687516875 + 7428.5714 ≈ 24303.571424303.5714 + 4363.6364 ≈ 28667.2078So, ( T ≈ 28667.2078 )Now, compute ( lambda ):( lambda = frac{T - 2N}{S} = frac{28667.2078 - 2*5000}{56987.7345} )Compute numerator:28667.2078 - 10000 = 18667.2078So,( lambda = 18667.2078 / 56987.7345 ≈ 0.3277 )So, ( lambda ≈ 0.3277 )Now, compute ( B_i = (a_i - lambda)/(2b_i) ) for each hive.Let's compute each one:Hive 1:( B_1 = (0.5 - 0.3277)/(2*0.0001) = (0.1723)/0.0002 = 861.5 )Hive 2:( B_2 = (0.45 - 0.3277)/(2*0.00009) = (0.1223)/0.00018 ≈ 679.4444 )Hive 3:( B_3 = (0.55 - 0.3277)/(2*0.00008) = (0.2223)/0.00016 ≈ 1389.375 )Hive 4:( B_4 = (0.52 - 0.3277)/(2*0.00007) = (0.1923)/0.00014 ≈ 1373.5714 )Hive 5:( B_5 = (0.48 - 0.3277)/(2*0.00011) = (0.1523)/0.00022 ≈ 692.2727 )Now, let's sum these up to check if they total approximately 5000:861.5 + 679.4444 ≈ 1540.94441540.9444 + 1389.375 ≈ 2930.31942930.3194 + 1373.5714 ≈ 4303.89084303.8908 + 692.2727 ≈ 4996.1635Hmm, that's approximately 4996.16, which is slightly less than 5000. There might be some rounding errors because I approximated ( lambda ) and each ( B_i ). Let me check the exact values without rounding.Wait, perhaps I should carry more decimal places to get a more accurate sum.Let me recalculate ( lambda ) with more precision.Earlier, I had:( T ≈ 28667.2078 )( S ≈ 56987.7345 )So,( lambda = (28667.2078 - 10000)/56987.7345 = 18667.2078 / 56987.7345 )Let me compute this division more accurately.18667.2078 ÷ 56987.7345First, note that 56987.7345 × 0.3277 ≈ 18667.2078, as I had before. But let me compute it more precisely.Compute 56987.7345 × 0.3277:56987.7345 × 0.3 = 17096.3203556987.7345 × 0.02 = 1139.7546956987.7345 × 0.007 = 398.914141556987.7345 × 0.0007 ≈ 39.89141415Adding these up:17096.32035 + 1139.75469 ≈ 18236.0750418236.07504 + 398.9141415 ≈ 18634.9891818634.98918 + 39.89141415 ≈ 18674.8806But we have 18667.2078, which is slightly less. So, perhaps ( lambda ≈ 0.3277 - ) a small amount.Let me compute ( 56987.7345 × 0.3277 ≈ 18674.8806 ), but we need 18667.2078, which is 7.6728 less.So, 7.6728 / 56987.7345 ≈ 0.0001346Therefore, ( lambda ≈ 0.3277 - 0.0001346 ≈ 0.3275654 )So, more accurately, ( lambda ≈ 0.3275654 )Now, let's recalculate each ( B_i ) with this more precise ( lambda ).Hive 1:( B_1 = (0.5 - 0.3275654)/(2*0.0001) = (0.1724346)/0.0002 = 862.173 )Hive 2:( B_2 = (0.45 - 0.3275654)/(2*0.00009) = (0.1224346)/0.00018 ≈ 680.1922 )Hive 3:( B_3 = (0.55 - 0.3275654)/(2*0.00008) = (0.2224346)/0.00016 ≈ 1389.59125 )Hive 4:( B_4 = (0.52 - 0.3275654)/(2*0.00007) = (0.1924346)/0.00014 ≈ 1374.5329 )Hive 5:( B_5 = (0.48 - 0.3275654)/(2*0.00011) = (0.1524346)/0.00022 ≈ 692.8845 )Now, sum these up:862.173 + 680.1922 ≈ 1542.36521542.3652 + 1389.59125 ≈ 2931.956452931.95645 + 1374.5329 ≈ 4306.489354306.48935 + 692.8845 ≈ 4999.37385That's approximately 4999.37, which is very close to 5000. The slight discrepancy is due to rounding during calculations. So, we can consider this as approximately 5000.Therefore, the number of bees per hive are approximately:- Hive 1: 862- Hive 2: 680- Hive 3: 1389.59 ≈ 1390- Hive 4: 1374.53 ≈ 1375- Hive 5: 692.88 ≈ 693But let me check the exact values without rounding:Hive 1: 862.173 ≈ 862Hive 2: 680.1922 ≈ 680Hive 3: 1389.59125 ≈ 1390Hive 4: 1374.5329 ≈ 1375Hive 5: 692.8845 ≈ 693Adding these: 862 + 680 = 1542; 1542 + 1390 = 2932; 2932 + 1375 = 4307; 4307 + 693 = 5000.Perfect, that adds up exactly to 5000 when rounded to the nearest whole number.So, the optimal distribution is approximately:- Hive 1: 862 bees- Hive 2: 680 bees- Hive 3: 1390 bees- Hive 4: 1375 bees- Hive 5: 693 beesLet me double-check the calculations for each hive to ensure accuracy.Hive 1:( (0.5 - 0.3275654) = 0.1724346 )Divide by ( 2*0.0001 = 0.0002 ):0.1724346 / 0.0002 = 862.173Yes, correct.Hive 2:( (0.45 - 0.3275654) = 0.1224346 )Divide by ( 2*0.00009 = 0.00018 ):0.1224346 / 0.00018 ≈ 680.1922Correct.Hive 3:( (0.55 - 0.3275654) = 0.2224346 )Divide by ( 2*0.00008 = 0.00016 ):0.2224346 / 0.00016 ≈ 1389.59125Correct.Hive 4:( (0.52 - 0.3275654) = 0.1924346 )Divide by ( 2*0.00007 = 0.00014 ):0.1924346 / 0.00014 ≈ 1374.5329Correct.Hive 5:( (0.48 - 0.3275654) = 0.1524346 )Divide by ( 2*0.00011 = 0.00022 ):0.1524346 / 0.00022 ≈ 692.8845Correct.So, all calculations seem accurate. Therefore, the optimal distribution is approximately:- Hive 1: 862 bees- Hive 2: 680 bees- Hive 3: 1390 bees- Hive 4: 1375 bees- Hive 5: 693 beesLet me just verify that the sum is exactly 5000:862 + 680 = 15421542 + 1390 = 29322932 + 1375 = 43074307 + 693 = 5000Yes, perfect.Therefore, these are the optimal numbers of bees per hive to maximize honey production.</think>