🧠:<think>Okay, so I have this problem about analyzing trade volumes and market welfare between three regions: A, B, and C. It's divided into two parts. Let me try to tackle them step by step.Starting with part 1: I need to find the time ( t^* ) when the total trade volume among the three regions reaches 1000 units for the first time. The trade volumes between each pair of regions are modeled by exponential functions. Specifically, the trade volume between region i and j is given by ( V_{ij}(t) = k_{ij} cdot e^{r_{ij}t} ). Given values:- ( k_{AB} = 100 ), ( r_{AB} = 0.05 )- ( k_{BC} = 150 ), ( r_{BC} = 0.03 )- ( k_{CA} = 120 ), ( r_{CA} = 0.04 )So, the total trade volume ( V(t) ) is the sum of all these individual trade volumes. That would be:( V(t) = V_{AB}(t) + V_{BC}(t) + V_{CA}(t) )Plugging in the given functions:( V(t) = 100e^{0.05t} + 150e^{0.03t} + 120e^{0.04t} )We need to find the smallest ( t^* ) such that ( V(t^*) = 1000 ).Hmm, this equation involves the sum of exponentials, which might not have an analytical solution. So, I think I'll need to solve this numerically. Maybe using methods like Newton-Raphson or just trial and error with some estimation.First, let me see what the trade volume looks like at different times. Maybe I can get a rough estimate.At t=0:( V(0) = 100 + 150 + 120 = 370 ). That's way below 1000.At t=10:( V(10) = 100e^{0.5} + 150e^{0.3} + 120e^{0.4} )Calculating each term:- 100e^{0.5} ≈ 100 * 1.6487 ≈ 164.87- 150e^{0.3} ≈ 150 * 1.3499 ≈ 202.485- 120e^{0.4} ≈ 120 * 1.4918 ≈ 179.016Adding them up: 164.87 + 202.485 + 179.016 ≈ 546.371. Still below 1000.At t=20:( V(20) = 100e^{1} + 150e^{0.6} + 120e^{0.8} )Calculating:- 100e ≈ 100 * 2.7183 ≈ 271.83- 150e^{0.6} ≈ 150 * 1.8221 ≈ 273.315- 120e^{0.8} ≈ 120 * 2.2255 ≈ 267.06Total: 271.83 + 273.315 + 267.06 ≈ 812.205. Closer, but still below 1000.At t=25:( V(25) = 100e^{1.25} + 150e^{0.75} + 120e^{1} )Calculating:- 100e^{1.25} ≈ 100 * 3.4903 ≈ 349.03- 150e^{0.75} ≈ 150 * 2.117 ≈ 317.55- 120e^{1} ≈ 120 * 2.7183 ≈ 326.196Total: 349.03 + 317.55 + 326.196 ≈ 992.776. Almost there, just under 1000.At t=26:( V(26) = 100e^{1.3} + 150e^{0.78} + 120e^{1.04} )Calculating each term:- 100e^{1.3} ≈ 100 * 3.6693 ≈ 366.93- 150e^{0.78} ≈ 150 * 2.1823 ≈ 327.345- 120e^{1.04} ≈ 120 * 2.828 ≈ 339.36Total: 366.93 + 327.345 + 339.36 ≈ 1033.635. That's above 1000.So, between t=25 and t=26, the total trade volume crosses 1000. To find the exact ( t^* ), I can use linear approximation or a more precise method.Let me denote ( V(t) = 100e^{0.05t} + 150e^{0.03t} + 120e^{0.04t} ). We know that at t=25, V=992.776 and at t=26, V≈1033.635.The difference between t=25 and t=26 is 1 year, and the increase in V is about 1033.635 - 992.776 ≈ 40.859.We need to find how much time after t=25 is needed to reach 1000. The required increase from t=25 is 1000 - 992.776 ≈ 7.224.Assuming the rate of increase is roughly linear over this small interval, the time needed would be approximately (7.224 / 40.859) * 1 year ≈ 0.1768 years.So, ( t^* ≈ 25 + 0.1768 ≈ 25.1768 ) years.To get a better approximation, maybe I can use the derivative at t=25 to perform a linear approximation.Compute ( V'(t) = 100*0.05 e^{0.05t} + 150*0.03 e^{0.03t} + 120*0.04 e^{0.04t} )At t=25:- 100*0.05 e^{1.25} ≈ 5 * 3.4903 ≈ 17.4515- 150*0.03 e^{0.75} ≈ 4.5 * 2.117 ≈ 9.5265- 120*0.04 e^{1} ≈ 4.8 * 2.7183 ≈ 12.9998Total derivative: 17.4515 + 9.5265 + 12.9998 ≈ 39.9778 ≈ 40.So, the rate of change at t=25 is approximately 40 units per year.We need to cover the remaining 7.224 units at this rate, so time needed is 7.224 / 40 ≈ 0.1806 years.Therefore, ( t^* ≈ 25 + 0.1806 ≈ 25.1806 ) years.To check, let's compute V(25.1806):Compute each term:- 100e^{0.05*25.1806} = 100e^{1.25903} ≈ 100 * 3.525 ≈ 352.5- 150e^{0.03*25.1806} = 150e^{0.7554} ≈ 150 * 2.128 ≈ 319.2- 120e^{0.04*25.1806} = 120e^{1.0072} ≈ 120 * 2.736 ≈ 328.32Total: 352.5 + 319.2 + 328.32 ≈ 1000.02. That's very close to 1000. So, t^* ≈25.18 years.So, approximately 25.18 years after the trade agreement was implemented, the total trade volume reaches 1000 units.Moving on to part 2: Analyzing the impact on global market welfare. The welfare function is given by ( W(t) = sum_{i,j} alpha_{ij} cdot V_{ij}(t) ). The weights are:- ( alpha_{AB} = 2 )- ( alpha_{BC} = 1.5 )- ( alpha_{CA} = 1.8 )So, the welfare function is:( W(t) = 2V_{AB}(t) + 1.5V_{BC}(t) + 1.8V_{CA}(t) )We need to calculate the rate of change of W(t) at t = t^* (which we found to be approximately 25.18 years). The rate of change is the derivative ( W'(t) ).First, let's express W(t):( W(t) = 2*100e^{0.05t} + 1.5*150e^{0.03t} + 1.8*120e^{0.04t} )Simplify:( W(t) = 200e^{0.05t} + 225e^{0.03t} + 216e^{0.04t} )Then, the derivative ( W'(t) ) is:( W'(t) = 200*0.05 e^{0.05t} + 225*0.03 e^{0.03t} + 216*0.04 e^{0.04t} )Simplify:( W'(t) = 10e^{0.05t} + 6.75e^{0.03t} + 8.64e^{0.04t} )Now, we need to evaluate this at t = t^* ≈25.18.Compute each term:First, compute the exponents:- 0.05*25.18 ≈1.259- 0.03*25.18 ≈0.7554- 0.04*25.18 ≈1.0072Compute each exponential:- e^{1.259} ≈3.525- e^{0.7554} ≈2.128- e^{1.0072} ≈2.736Now, compute each term:- 10 * 3.525 ≈35.25- 6.75 * 2.128 ≈14.376- 8.64 * 2.736 ≈23.63Adding them up: 35.25 + 14.376 + 23.63 ≈73.256So, the rate of change of global market welfare at t = t^* is approximately 73.26 units per year.Interpretation: The positive rate of change indicates that global market welfare is increasing at the time when the total trade volume reaches 1000 units. This suggests that further regional integration could continue to enhance economic welfare. However, the rate might be slowing down or speeding up depending on the second derivative, but since we only have the first derivative, we can say that at that specific time, welfare is still growing. This might inform policymakers that maintaining or enhancing trade agreements could be beneficial for continued welfare improvement.But wait, let me double-check the calculations for W'(t). Maybe I made an arithmetic error.First, 200*0.05 is 10, correct.225*0.03 is 6.75, correct.216*0.04 is 8.64, correct.Then, exponents:At t=25.18,e^{0.05*25.18}=e^{1.259}=approx 3.525e^{0.03*25.18}=e^{0.7554}=approx 2.128e^{0.04*25.18}=e^{1.0072}=approx 2.736Multiplying:10*3.525=35.256.75*2.128= Let's compute 6*2.128=12.768 and 0.75*2.128=1.596, so total 12.768+1.596=14.3648.64*2.736: Let's compute 8*2.736=21.888 and 0.64*2.736≈1.753, so total≈21.888+1.753≈23.641Adding: 35.25 +14.364=49.614 +23.641≈73.255. So, yes, approximately 73.26.So, the rate is about 73.26 units per year.In terms of policy, this positive rate suggests that the welfare is still improving, so maintaining or enhancing trade agreements could be beneficial. However, if the rate is decreasing over time, it might indicate diminishing returns, but since we only have the first derivative at a specific point, we can't conclude about the trend. But at least at t^*, the welfare is increasing, which is a good sign for ongoing integration.I think that's about it. Let me just recap:1. Calculated total trade volume as the sum of exponentials.2. Found t^* numerically between 25 and 26, approximated to 25.18 years.3. Expressed welfare function, took its derivative, and evaluated it at t^* to find the rate of change, which was positive, indicating increasing welfare.I don't see any mistakes in the calculations, so I think this is solid.Final Answer1. The time ( t^* ) at which the total trade volume reaches 1000 units is approximately boxed{25.18} years.2. The rate of change of global market welfare at ( t = t^* ) is approximately boxed{73.26} units per year.</think>

🧠:<think>Alright, so I have this problem about scheduling movies in a theater with 5 screens to maximize attendance and then also considering energy costs. Let me try to break it down step by step.Starting with part 1: I need to assign each movie M_j to a screen S_i such that the total audience attendance is maximized. The attendance for each screen-movie combination is given by A_ij = 50 + 30 cos(π(i - j)/3). First, I should probably compute A_ij for all possible combinations of i and j. Since there are 5 screens (i=1 to 5) and 6 movies (j=1 to 6), that's 30 different combinations. Let me make a table for that.Let me recall the cosine function: cos(πθ) oscillates between -1 and 1. So, depending on the value of (i - j)/3, the cosine term will vary. Let me compute A_ij for each i and j.Let's list out the values:For each i from 1 to 5 and j from 1 to 6:Compute (i - j)/3, then multiply by π, take cosine, multiply by 30, add 50.Let me compute this for each pair.Starting with i=1:j=1: (1-1)/3 = 0; cos(0) = 1; A_11=50+30*1=80j=2: (1-2)/3 = -1/3; cos(-π/3)=cos(π/3)=0.5; A_12=50+30*0.5=65j=3: (1-3)/3 = -2/3; cos(-2π/3)=cos(2π/3)=-0.5; A_13=50+30*(-0.5)=35j=4: (1-4)/3 = -1; cos(-π)=cos(π)=-1; A_14=50+30*(-1)=20j=5: (1-5)/3 = -4/3; cos(-4π/3)=cos(4π/3)=-0.5; A_15=50+30*(-0.5)=35j=6: (1-6)/3 = -5/3; cos(-5π/3)=cos(5π/3)=0.5; A_16=50+30*0.5=65So for i=1, the attendances are: 80, 65, 35, 20, 35, 65Similarly, for i=2:j=1: (2-1)/3=1/3; cos(π/3)=0.5; A_21=50+15=65j=2: (2-2)/3=0; cos(0)=1; A_22=80j=3: (2-3)/3=-1/3; cos(-π/3)=0.5; A_23=65j=4: (2-4)/3=-2/3; cos(-2π/3)=-0.5; A_24=50-15=35j=5: (2-5)/3=-1; cos(-π)=-1; A_25=50-30=20j=6: (2-6)/3=-4/3; cos(-4π/3)=-0.5; A_26=50-15=35So for i=2: 65, 80, 65, 35, 20, 35i=3:j=1: (3-1)/3=2/3; cos(2π/3)=-0.5; A_31=50-15=35j=2: (3-2)/3=1/3; cos(π/3)=0.5; A_32=65j=3: (3-3)/3=0; cos(0)=1; A_33=80j=4: (3-4)/3=-1/3; cos(-π/3)=0.5; A_34=65j=5: (3-5)/3=-2/3; cos(-2π/3)=-0.5; A_35=35j=6: (3-6)/3=-1; cos(-π)=-1; A_36=20So i=3: 35, 65, 80, 65, 35, 20i=4:j=1: (4-1)/3=1; cos(π)=-1; A_41=50-30=20j=2: (4-2)/3=2/3; cos(2π/3)=-0.5; A_42=35j=3: (4-3)/3=1/3; cos(π/3)=0.5; A_43=65j=4: (4-4)/3=0; cos(0)=1; A_44=80j=5: (4-5)/3=-1/3; cos(-π/3)=0.5; A_45=65j=6: (4-6)/3=-2/3; cos(-2π/3)=-0.5; A_46=35So i=4: 20, 35, 65, 80, 65, 35i=5:j=1: (5-1)/3=4/3; cos(4π/3)=-0.5; A_51=35j=2: (5-2)/3=1; cos(π)=-1; A_52=20j=3: (5-3)/3=2/3; cos(2π/3)=-0.5; A_53=35j=4: (5-4)/3=1/3; cos(π/3)=0.5; A_54=65j=5: (5-5)/3=0; cos(0)=1; A_55=80j=6: (5-6)/3=-1/3; cos(-π/3)=0.5; A_56=65So i=5: 35, 20, 35, 65, 80, 65Alright, so compiling all these attendances into a matrix:Rows are screens (i=1 to 5), columns are movies (j=1 to 6):i=1: [80, 65, 35, 20, 35, 65]i=2: [65, 80, 65, 35, 20, 35]i=3: [35, 65, 80, 65, 35, 20]i=4: [20, 35, 65, 80, 65, 35]i=5: [35, 20, 35, 65, 80, 65]Now, the problem is to assign each movie to a screen such that each screen shows exactly one movie, and each movie is shown at most once (since there are 5 screens and 6 movies, one movie will not be shown). Wait, actually, the problem says "each screen can show any movie", but it doesn't specify whether a movie can be shown on multiple screens or not. Hmm, the wording is a bit ambiguous.Wait, the problem says "schedule the movies M_j to screens S_i". So, each screen is assigned one movie, but movies can be assigned to multiple screens? Or is it that each movie can be assigned to only one screen? Hmm.Wait, the first part says "the technician needs to schedule the movies M_j to screens S_i that maximizes the total audience attendance across all screens." So, it's possible that multiple screens can show the same movie, but each screen must show one movie. So, movies can be repeated across screens.But, wait, the problem says "each screen S_i can show any movie M_j". So, it's allowed for multiple screens to show the same movie. So, in that case, we can assign any movie to any screen, with possible repetition.But wait, in that case, the maximum total attendance would be achieved by assigning the movie with the highest attendance to all screens. But looking at the attendance function, for each screen, the attendance depends on the movie assigned. So, for each screen, we can choose the movie that gives the highest attendance for that screen.Wait, but is that the case? Let's see.Wait, if we can assign any movie to any screen, regardless of other screens, then for each screen, we can choose the movie that gives the maximum attendance for that screen, and the total attendance would be the sum of these maxima.But let's check if that's possible.Looking at the attendance matrix:For each screen, find the maximum attendance:i=1: max(80,65,35,20,35,65)=80 (j=1)i=2: max(65,80,65,35,20,35)=80 (j=2)i=3: max(35,65,80,65,35,20)=80 (j=3)i=4: max(20,35,65,80,65,35)=80 (j=4)i=5: max(35,20,35,65,80,65)=80 (j=5)Wait, so for each screen, the maximum attendance is 80, achieved when j=i. So, if we assign movie j=i to screen i, each screen gets 80 attendance, so total attendance is 5*80=400.But wait, is that possible? Because each movie can be assigned to multiple screens. So, for example, movie j=1 can be assigned to screen 1, movie j=2 to screen 2, etc., up to j=5 to screen 5, and movie j=6 isn't used. So, that would give a total attendance of 400.Alternatively, could we get a higher total attendance by assigning some movies to multiple screens? For example, if a movie has high attendance on multiple screens, maybe assigning it to multiple screens would give a higher total.Wait, let's check for each movie, what is the maximum attendance it can get on any screen.Looking at the attendance matrix:For each movie j, find the maximum A_ij across screens i.j=1: max(80,65,35,20,35)=80 (i=1)j=2: max(65,80,65,35,20)=80 (i=2)j=3: max(35,65,80,65,35)=80 (i=3)j=4: max(20,35,65,80,65)=80 (i=4)j=5: max(35,20,35,65,80)=80 (i=5)j=6: max(65,35,20,35,65)=65 (i=1 or i=5)So, each movie j=1 to 5 has a maximum attendance of 80 on their respective screen i=j, and movie j=6 has a maximum of 65.Therefore, the maximum total attendance is achieved by assigning each movie j=1 to 5 to their respective screens i=1 to 5, giving 80 each, totaling 400. Movie j=6 is not assigned, as it would only give 65 on any screen, which is less than 80.Therefore, the optimal schedule is:Screen 1: Movie 1Screen 2: Movie 2Screen 3: Movie 3Screen 4: Movie 4Screen 5: Movie 5Total attendance: 400.So that's part 1.Now, moving on to part 2: We need to minimize the total operating cost while achieving at least 90% of the maximum audience attendance from part 1.First, 90% of 400 is 360. So, the total attendance must be at least 360.The operating cost is based on energy consumption, which is given by E_ij = 20 + 5|i - j| kWh per day. Each kWh costs 0.12, so the cost per screen is 0.12*(20 + 5|i - j|).Our goal is to assign movies to screens such that the total attendance is at least 360, and the total cost is minimized.This is an optimization problem with two objectives: maximize attendance (but at least 360) and minimize cost. But since we're to achieve at least 90% of the max attendance, we can frame it as a constrained optimization where total attendance >= 360, and minimize total cost.So, we need to find an assignment of movies to screens (each screen gets one movie, movies can be repeated) such that the sum of A_ij >= 360, and the sum of E_ij is minimized.Alternatively, since we can represent this as an integer linear programming problem, where variables x_ij are binary (1 if screen i shows movie j, 0 otherwise), with the constraints:Sum over j of x_ij = 1 for each screen i (each screen shows exactly one movie)Sum over i of A_ij * x_ij >= 360 (total attendance >= 360)And the objective is to minimize Sum over i,j of E_ij * x_ij.But since this is a bit complex, maybe we can approach it by considering that we need to find assignments where the total attendance is at least 360, and among those, find the one with the lowest energy cost.Given that the maximum attendance is 400, 360 is 90% of that, so we need to find assignments where the total attendance is 360 or more, but with minimal cost.Alternatively, perhaps we can find a way to slightly deviate from the optimal assignment in part 1 to reduce energy costs, but not drop the total attendance below 360.In part 1, the optimal assignment was each screen i showing movie j=i, giving 80 per screen, total 400. The energy consumption for each screen in this case is E_ii = 20 + 5|i - i| = 20 + 0 = 20 kWh. So, total energy is 5*20=100 kWh, costing 100*0.12=12.But if we can find another assignment where the total attendance is still 360 or more, but with lower energy consumption, that would be better.Wait, but in the optimal assignment, the energy cost is already minimized because |i - j|=0 for each screen, so E_ij=20, which is the minimum possible for each screen. Therefore, any deviation from this assignment would increase |i - j|, thus increasing E_ij, which would increase the total cost.But wait, perhaps we can have some screens showing movies that are not their optimal j=i, but in such a way that the total attendance is still 360, but the total energy cost is lower? Wait, no, because any deviation would require at least one screen to show a different movie, which would either decrease attendance or increase energy cost.Wait, but if we can have some screens show movies that are close to their optimal, maybe with |i - j|=1, which would increase E_ij by 5, but perhaps the attendance doesn't drop too much.Wait, let's think. For example, if screen 1 shows movie 2 instead of 1, the attendance would be 65 instead of 80, a drop of 15. But the energy cost would be E_12=20 +5*1=25, which is 5 more than 20. So, the cost increases by 0.60 (since 5 kWh * 0.12). But the attendance drops by 15, which is a significant drop. So, to compensate, we might need to increase attendance elsewhere.Alternatively, maybe we can have some screens show movies that are not their optimal, but others show movies that give higher than optimal attendance? Wait, but the maximum attendance per screen is 80, so you can't get higher than that.Wait, perhaps if we have some screens show movies that are not their optimal, but others show movies that are their optimal, such that the total attendance is still 360.Wait, let's calculate how much we can afford to drop in attendance.Total required attendance: 360Original total: 400So, we can drop 40 in total.Each screen can contribute to this drop. For example, if we have one screen drop by 15 (from 80 to 65), another drop by 15, and another drop by 10, that would total 40.But we need to see if such a combination is possible without overlapping movies or something.Wait, but movies can be assigned to multiple screens, so we can have multiple screens showing the same movie.Wait, but in the optimal assignment, each screen shows a different movie (j=1 to 5). If we allow multiple screens to show the same movie, we can potentially have some screens show a movie that gives a higher attendance, but others show a movie that gives lower attendance, but overall, the total is still 360.Wait, but the maximum attendance per screen is 80, so if we have some screens showing 80, and others showing less, but the total is 360.Wait, 5 screens, 360 total. So, average per screen is 72.So, perhaps, if we have some screens at 80, and others at lower, but the total is 360.Let me think: Suppose we have four screens at 80, and one screen at 40. Then total is 4*80 +40=360. But is 40 possible? Looking at the attendance matrix, the minimum attendance is 20, so 40 is possible if we have two screens at 20 each, but that would require two screens to drop by 60 each, which is not possible because the maximum drop per screen is 60 (from 80 to 20). But we can't have two screens at 20 because that would require two screens to show movies that give 20, but each screen can only show one movie.Wait, maybe it's better to think in terms of how much we can reduce the total attendance by moving movies around, while keeping the total above 360.Alternatively, perhaps we can have some screens show movies that are not their optimal, but others show movies that are their optimal, such that the total is still 360.Wait, let's consider the original assignment: each screen i shows movie j=i, total 400.If we change one screen to show a different movie, say screen 1 shows movie 2 instead of 1. Then, attendance for screen 1 drops from 80 to 65, a loss of 15. To compensate, we need to gain 15 elsewhere. But since the maximum attendance per screen is 80, we can't gain more than 80 on any screen. So, perhaps we can have another screen show a movie that gives higher attendance elsewhere.Wait, but in the original assignment, all screens are already at their maximum. So, if we move a movie from one screen to another, we might have to decrease one and possibly increase another, but not necessarily.Wait, for example, if screen 1 shows movie 2 (attendance 65) and screen 2 shows movie 1 (attendance 65). Then, both screens 1 and 2 would have 65 instead of 80 each, so total attendance drops by 30. That's not good.Alternatively, if screen 1 shows movie 2 (65) and screen 2 shows movie 3 (65). Then, screen 1 drops by 15, screen 2 drops by 15, total drop 30. Not good.Alternatively, if screen 1 shows movie 2 (65), and screen 3 shows movie 1 (35). Then, screen 1 drops by 15, screen 3 drops by 45 (from 80 to 35). Total drop 60, which is too much.Alternatively, if screen 1 shows movie 6 (65), which is the same as movie 2's attendance on screen 1. So, screen 1 would still have 65, but movie 6 is different. Then, screen 6 isn't a screen, so maybe that's not relevant.Wait, maybe I'm overcomplicating. Let's think differently.We need to find an assignment where the total attendance is at least 360, and the total energy cost is minimized.Since the original assignment gives 400 attendance with 100 kWh, costing 12.If we can find an assignment where the total attendance is 360, but the energy cost is less than 12, that would be better. But wait, the original energy cost is already the minimum possible because each screen is showing the movie that gives the minimum |i - j|=0, so E_ij=20. Any deviation would require |i - j|>=1, so E_ij>=25, which would increase the cost.Wait, but maybe we can have some screens show movies that are not their optimal, but others show movies that are their optimal, such that the total attendance is still 360, but the total energy cost is less than 12. But that seems impossible because any deviation from the optimal assignment would require at least one screen to have a higher E_ij, thus increasing the total cost.Wait, unless we can have some screens show movies that are not their optimal, but the increase in E_ij is offset by a decrease in E_ij elsewhere. But since in the optimal assignment, all E_ij are already minimized, any change would require at least one screen to have a higher E_ij, thus increasing the total cost.Therefore, the minimal total cost is achieved when all screens show their optimal movies, giving total attendance 400 and cost 12.But wait, the problem says "while still achieving at least 90% of the maximum possible audience attendance found in sub-problem 1." So, 90% of 400 is 360. So, we can have a total attendance of 360 or more, but we need to minimize the cost.But if the minimal cost is achieved at 400 attendance, which is above 360, then that would be the solution. However, perhaps there is a cheaper way to achieve 360 attendance.Wait, let's think. Maybe by not showing all 5 optimal movies, but instead showing some movies that are not optimal, but in such a way that the total attendance is 360, but the total energy cost is lower than 12.Wait, but in the optimal assignment, each screen is showing a movie that gives 80 attendance, and the energy cost is 20 per screen. If we can have some screens show movies that give less than 80, but others show movies that give more than 80, but overall, the total is 360, but the total energy cost is less than 100 kWh.Wait, but the maximum attendance per screen is 80, so you can't get more than 80. So, if you have some screens at 80, and others at lower, the total can be 360.But the energy cost for screens showing 80 is 20, which is the minimum. If you have some screens showing 80, and others showing less, but the total energy cost is still 100 kWh or less, that would be better. But since the screens showing less would have higher energy costs, it's unlikely.Wait, let's try to find an assignment where the total attendance is 360, but the total energy cost is less than 100 kWh.Wait, for example, suppose we have four screens showing their optimal movies (80 each), and one screen showing a movie that gives 40 attendance. Then, total attendance is 4*80 +40=360.But what's the energy cost for the screen showing 40? Let's see.Looking at the attendance matrix, which combinations give 40? Wait, in the matrix, the attendances are 80,65,35,20, etc. So, 40 isn't directly achieved. The closest is 35 or 20.Wait, perhaps I made a mistake. Let me check.Wait, the attendance function is A_ij=50 +30 cos(π(i-j)/3). So, cos can be -1, -0.5, 0, 0.5, 1.So, A_ij can be:50 +30*1=8050 +30*0.5=6550 +30*0=5050 +30*(-0.5)=3550 +30*(-1)=20So, possible attendances are 80,65,50,35,20.So, 40 isn't possible. So, to get a total of 360, we need to have a combination of these attendances.Let me see: 360 divided by 5 screens is 72 per screen on average.So, perhaps, four screens at 80 (total 320) and one screen at 40, but since 40 isn't possible, maybe 35 and 45? But 45 isn't possible either.Alternatively, three screens at 80 (240), and two screens at 60 each (120), total 360. But 60 isn't a possible attendance.Wait, possible attendances are 80,65,50,35,20.So, let's see combinations:We need sum A_ij >=360.Let me try to find combinations of 5 attendances from the set {20,35,50,65,80} that sum to at least 360.The maximum sum is 5*80=400.We need to find a combination where the sum is at least 360, but with minimal total energy cost.But since the energy cost is minimized when |i-j| is minimized, which is 0, giving E_ij=20. So, the minimal total energy cost is 100 kWh, which is 12.But if we can find a combination where the total attendance is 360, but the total energy cost is less than 100 kWh, that would be better. But I don't think that's possible because any deviation from the optimal assignment would require at least one screen to have a higher E_ij, thus increasing the total cost.Wait, unless we can have some screens show movies that are not their optimal, but the increase in E_ij is offset by a decrease in E_ij elsewhere. But since in the optimal assignment, all E_ij are already minimized, any change would require at least one screen to have a higher E_ij, thus increasing the total cost.Therefore, the minimal total cost is achieved when all screens show their optimal movies, giving total attendance 400 and cost 12.But wait, the problem says "while still achieving at least 90% of the maximum possible audience attendance". So, 90% of 400 is 360. So, we can have a total attendance of 360 or more, but we need to minimize the cost.But since the minimal cost is achieved at 400 attendance, which is above 360, that would be the solution. However, perhaps there is a cheaper way to achieve 360 attendance.Wait, let me think differently. Maybe we can have some screens show movies that are not their optimal, but the total attendance is still 360, but the total energy cost is less than 12.Wait, for example, suppose we have four screens showing their optimal movies (80 each), and one screen showing a movie that gives 40 attendance. But as we saw, 40 isn't possible. The closest is 35 or 20.If we have four screens at 80 and one at 35, total attendance is 4*80 +35=355, which is less than 360. Not enough.If we have four screens at 80 and one at 65, total is 4*80 +65=385, which is above 360. But the energy cost for the fifth screen showing 65 would be higher.Wait, let's see. If screen 1 shows movie 2 (65 attendance), then E_12=20 +5*1=25 kWh. So, total energy cost would be 4*20 +25=105 kWh, costing 12.60, which is more than the original 12.Alternatively, if screen 1 shows movie 6 (65 attendance), E_16=20 +5*5=20+25=45 kWh. That's even worse.Wait, so any screen showing a non-optimal movie would have higher E_ij, thus increasing the total cost.Therefore, the minimal total cost is achieved when all screens show their optimal movies, giving total attendance 400 and cost 12.But wait, the problem says "while still achieving at least 90% of the maximum possible audience attendance". So, 360 is the minimum required. But if we can achieve 360 with a lower cost, that would be better. But as we saw, any deviation from the optimal assignment would require at least one screen to have higher E_ij, thus increasing the total cost.Therefore, the minimal total cost is 12, achieved by the optimal assignment.Wait, but maybe there's a way to have some screens show movies that are not their optimal, but the total attendance is still 360, and the total energy cost is less than 12. But I don't think that's possible because any deviation would require at least one screen to have higher E_ij, thus increasing the total cost.Therefore, the minimal total cost is 12, achieved by the optimal assignment.But wait, let me double-check. Suppose we have some screens show movies that are not their optimal, but the total attendance is still 360, and the total energy cost is less than 12.Wait, for example, suppose we have three screens showing their optimal movies (80 each), and two screens showing movies that give 60 each. But 60 isn't a possible attendance. The closest is 65.So, three screens at 80 (240) and two screens at 65 (130), total 370. The energy cost for the two screens showing 65 would be higher.For example, if screen 1 shows movie 2 (65), E=25, and screen 2 shows movie 1 (65), E=25. So, total energy cost is 3*20 +2*25=60+50=110 kWh, costing 13.20, which is more than 12.Alternatively, if we have two screens showing 80, and three screens showing 60 each, but again, 60 isn't possible.Wait, perhaps if we have some screens show movies that are not their optimal, but the total attendance is still 360, but the total energy cost is less than 12. But I don't think that's possible because any deviation from the optimal assignment would require at least one screen to have higher E_ij, thus increasing the total cost.Therefore, the minimal total cost is 12, achieved by the optimal assignment.But wait, the problem says "minimizing the total operating cost while still achieving at least 90% of the maximum possible audience attendance". So, if the minimal cost is achieved at 400 attendance, which is above 360, then that's acceptable.Therefore, the solution is the same as part 1: assign each screen i to movie j=i, giving total attendance 400 and total cost 12.But wait, let me think again. Maybe there's a way to have some screens show movies that are not their optimal, but the total attendance is still 360, and the total energy cost is less than 12.Wait, for example, suppose we have four screens showing their optimal movies (80 each), and one screen showing a movie that gives 40 attendance. But as we saw, 40 isn't possible. The closest is 35 or 20.If we have four screens at 80 and one at 35, total attendance is 355, which is less than 360. Not enough.If we have four screens at 80 and one at 65, total is 385, which is above 360. But the energy cost for the fifth screen showing 65 would be higher.Wait, let's calculate the energy cost for that.Suppose screen 1 shows movie 2 (65 attendance), E=25 kWh.Then, total energy cost is 4*20 +25=105 kWh, costing 12.60, which is more than 12.Alternatively, if screen 1 shows movie 6 (65 attendance), E=45 kWh, which is even worse.Therefore, any screen showing a non-optimal movie would increase the total cost.Therefore, the minimal total cost is 12, achieved by the optimal assignment.Therefore, the answer to part 2 is the same as part 1: assign each screen i to movie j=i, giving total attendance 400 and total cost 12.But wait, the problem says "minimizing the total operating cost while still achieving at least 90% of the maximum possible audience attendance". So, if we can achieve 360 with a lower cost, that would be better. But as we saw, it's not possible because any deviation from the optimal assignment would increase the total cost.Therefore, the minimal total cost is 12, achieved by the optimal assignment.So, summarizing:1. Assign each screen i to movie j=i, total attendance 400.2. The same assignment gives total cost 12, which is minimal while achieving 400 attendance, which is above 360.Therefore, the solution is the same for both parts.But wait, the problem says "Formulate and solve an optimization problem that minimizes the total operating cost while still achieving at least 90% of the maximum possible audience attendance found in sub-problem 1."So, perhaps, the minimal cost is achieved by the same assignment, but maybe there's a way to have a lower cost by allowing some screens to show non-optimal movies, but still meeting the attendance requirement.Wait, but as we saw, any non-optimal assignment would require at least one screen to have higher E_ij, thus increasing the total cost.Therefore, the minimal total cost is 12, achieved by the optimal assignment.So, the final answer is:1. Assign each screen i to movie j=i, total attendance 400.2. The same assignment gives total cost 12, which is minimal while achieving 400 attendance, which is above 360.Therefore, the solution is the same for both parts.But let me double-check the energy cost calculation.In part 1, each screen shows movie j=i, so E_ij=20 for each screen. 5 screens, so 5*20=100 kWh. At 0.12 per kWh, total cost is 100*0.12=12.Yes, that's correct.Therefore, the final answer is:1. Assign each screen i to movie j=i, total attendance 400.2. The same assignment gives total cost 12, which is minimal while achieving 400 attendance, which is above 360.Therefore, the solution is the same for both parts.</think>

🧠:<think>Alright, so I'm trying to help this portfolio manager decide where to allocate funds between Company A and Company B. Let me start by understanding each part of the problem.First, for Company A, they mine copper with a fixed production cost of 150,000 per month. The price of copper is modeled by a stochastic differential equation (SDE): ( dP_t = mu P_t dt + sigma P_t dW_t ). The parameters given are ( mu = 0.02 ), ( sigma = 0.1 ), and the initial price ( P_0 = 10,000 ) per ton. I need to find the expected price after one year and the variance of the price.Hmm, okay. I remember that this kind of SDE is a geometric Brownian motion, which is commonly used to model stock prices and other assets. The solution to this SDE is given by:( P_t = P_0 expleft( left( mu - frac{sigma^2}{2} right) t + sigma W_t right) )But since we're dealing with expectations and variances, I don't need to worry about the exact distribution, just the moments.For the expected value, ( E[P_t] ), since the drift term is ( mu ), the expected growth rate is ( mu ). However, because of the volatility, the expectation isn't just ( P_0 e^{mu t} ). Wait, actually, I think it is. Let me recall: for geometric Brownian motion, the expected value is ( E[P_t] = P_0 e^{mu t} ). Is that correct?Yes, because the drift term is ( mu ), and even though the process is multiplicative, the expectation still grows exponentially at rate ( mu ). So, for one year, which is 12 months, but wait, actually, in the SDE, the time is in years, right? Because the drift and volatility are annualized rates. So, if the time is one year, ( t = 1 ).So, ( E[P_1] = 10,000 e^{0.02 * 1} ). Let me compute that.First, compute ( 0.02 * 1 = 0.02 ). Then, ( e^{0.02} ) is approximately 1.02020134. So, multiplying by 10,000 gives approximately 10,202.01 dollars per ton.Okay, that's the expected price. Now, for the variance. The variance of ( P_t ) in geometric Brownian motion is given by ( Var(P_t) = P_0^2 e^{2mu t} left( e^{sigma^2 t} - 1 right) ).Let me verify that formula. Since ( P_t ) is log-normally distributed, the variance can be calculated using the properties of lognormal variables. If ( X ) is lognormal with parameters ( mu ) and ( sigma ), then ( Var(X) = e^{2mu + sigma^2} (e^{sigma^2} - 1) ). But in our case, the process is over time ( t ), so the parameters are scaled by ( t ). So, the variance should be ( P_0^2 e^{2mu t} (e^{sigma^2 t} - 1) ).Yes, that seems right. So, plugging in the numbers:( P_0 = 10,000 ), ( mu = 0.02 ), ( sigma = 0.1 ), ( t = 1 ).First, compute ( 2mu t = 2 * 0.02 * 1 = 0.04 ). Then, ( e^{0.04} ) is approximately 1.040810774.Next, compute ( sigma^2 t = 0.1^2 * 1 = 0.01 ). Then, ( e^{0.01} ) is approximately 1.010050167.So, ( e^{sigma^2 t} - 1 = 1.010050167 - 1 = 0.010050167 ).Now, multiply all together:( Var(P_1) = (10,000)^2 * 1.040810774 * 0.010050167 ).First, ( (10,000)^2 = 100,000,000 ).Then, 1.040810774 * 0.010050167 ≈ 0.0104603.So, 100,000,000 * 0.0104603 ≈ 1,046,030.Therefore, the variance is approximately 1,046,030. To find the standard deviation, we could take the square root, but since the question only asks for variance, we can leave it at that.Wait, let me double-check the variance formula. I think sometimes it's expressed differently. Let me recall that for a lognormal variable ( Y = ln(X) ), the variance of ( Y ) is ( sigma^2 t ). But the variance of ( X ) is more complicated.Yes, the variance of ( X ) is ( E[X^2] - (E[X])^2 ). For geometric Brownian motion, ( E[X^2] = P_0^2 e^{2mu t + sigma^2 t} ). Therefore, ( Var(X) = E[X^2] - (E[X])^2 = P_0^2 e^{2mu t + sigma^2 t} - (P_0 e^{mu t})^2 = P_0^2 e^{2mu t} (e^{sigma^2 t} - 1) ). So, yes, my formula was correct.So, the variance is approximately 1,046,030.Moving on to Company B. Their production cost is variable and depends on the prices of gold and silver. The cost function is ( C_B(G, S) = 50,000 + 0.2G + 0.1S ). The prices of gold and silver follow joint SDEs:( dG_t = mu_G G_t dt + sigma_G G_t dW_t^G )( dS_t = mu_S S_t dt + sigma_S S_t dW_t^S )With ( mu_G = 0.015 ), ( sigma_G = 0.08 ), ( mu_S = 0.01 ), ( sigma_S = 0.12 ), and the Wiener processes ( W_t^G ) and ( W_t^S ) are correlated with ( rho = 0.5 ). The initial prices are ( G_0 = 1,800 ) and ( S_0 = 25 ) per ounce.I need to find the expected production cost after one year. So, the expected value of ( C_B(G_1, S_1) ).Since expectation is linear, ( E[C_B(G_1, S_1)] = 50,000 + 0.2 E[G_1] + 0.1 E[S_1] ).So, I just need to compute ( E[G_1] ) and ( E[S_1] ).Again, both gold and silver prices follow geometric Brownian motions, so similar to Company A, their expected prices after one year are:( E[G_1] = G_0 e^{mu_G t} )( E[S_1] = S_0 e^{mu_S t} )With ( t = 1 ) year.So, let's compute these.First, ( E[G_1] = 1,800 e^{0.015 * 1} ).Compute ( 0.015 * 1 = 0.015 ). ( e^{0.015} ) is approximately 1.015113.So, ( 1,800 * 1.015113 ≈ 1,800 * 1.015113 ).Calculating that: 1,800 * 1 = 1,800; 1,800 * 0.015113 ≈ 27.2034. So total is approximately 1,827.2034 dollars per ounce.Similarly, ( E[S_1] = 25 e^{0.01 * 1} ).Compute ( 0.01 * 1 = 0.01 ). ( e^{0.01} ≈ 1.010050167 ).So, 25 * 1.010050167 ≈ 25.251254 dollars per ounce.Now, plug these into the cost function:( E[C_B] = 50,000 + 0.2 * 1,827.2034 + 0.1 * 25.251254 ).Compute each term:0.2 * 1,827.2034 ≈ 365.440680.1 * 25.251254 ≈ 2.5251254So, adding these to 50,000:50,000 + 365.44068 + 2.5251254 ≈ 50,367.9658So, approximately 50,367.97.Wait, but hold on. Is that all? The question mentions that the Wiener processes are correlated with ( rho = 0.5 ). Does that affect the expected value?Hmm, in the expectation, since expectation is linear, the correlation between G and S doesn't affect the expected value of the cost function because the expectation of the sum is the sum of expectations, regardless of correlation. So, even though G and S are correlated, their individual expectations are just based on their own drifts. So, I think my calculation is correct.But just to be thorough, let me think about it. The cost function is linear in G and S, so expectation is linear, so yes, the correlation doesn't matter for the expectation. It would matter if we were calculating variance or covariance, but since we're only asked for the expected production cost, which is an expectation, we don't need to consider the correlation.Therefore, the expected production cost is approximately 50,367.97.Wait, but let me double-check the calculations for E[G1] and E[S1].For E[G1]:1,800 * e^{0.015} ≈ 1,800 * 1.015113 ≈ 1,827.2034. That seems correct.For E[S1]:25 * e^{0.01} ≈ 25 * 1.010050167 ≈ 25.251254. That also seems correct.Then, 0.2 * 1,827.2034 ≈ 365.440680.1 * 25.251254 ≈ 2.5251254Adding to 50,000: 50,000 + 365.44068 + 2.5251254 ≈ 50,367.9658, which is approximately 50,367.97.Yes, that seems right.So, summarizing:1. For Company A, the expected price of copper after one year is approximately 10,202.01, and the variance is approximately 1,046,030.2. For Company B, the expected production cost after one year is approximately 50,367.97.I think that covers both parts of the problem. I don't see any mistakes in my reasoning, but let me just recap to ensure I didn't miss anything.For Company A, using the properties of geometric Brownian motion, the expected price is straightforward with the drift term, and the variance formula accounts for both drift and volatility. For Company B, since the cost function is linear in G and S, and expectation is linear, the correlation between G and S doesn't affect the expected value. So, calculating each expected price separately and plugging them into the cost function gives the expected production cost.Yes, I think that's solid.Final Answer1. The expected price of copper after one year is boxed{10202.01} dollars per ton, and the variance of the price is boxed{1046030}.2. The expected production cost for Company B after one year is boxed{50367.97} dollars.</think>

🧠:<think>Okay, so I have this problem about an entrepreneur running a cremation services business. It's split into two parts, and I need to figure out both. Let's start with the first part.1. Profit Equation and Break-Even PointAlright, the business has fixed costs and variable costs. Fixed costs are 15,000 per month, which includes things like rent, utilities, and equipment maintenance. Variable costs are 200 per cremation service, which covers labor and materials. The entrepreneur charges 1,000 per service. I need to write an equation for the monthly profit, P(x), as a function of x, where x is the number of cremation services performed in a month. Then, I have to find the minimum number of services needed to break even.First, profit is generally calculated as total revenue minus total costs. So, I should figure out both the revenue and the total costs.Revenue is straightforward: it's the number of services multiplied by the price per service. So, revenue R(x) = 1000x.Total costs have two parts: fixed and variable. Fixed costs are constant at 15,000. Variable costs depend on the number of services, so that's 200x. Therefore, total costs C(x) = 15,000 + 200x.So, profit P(x) = Revenue - Costs = 1000x - (15,000 + 200x). Let me write that out:P(x) = 1000x - 15,000 - 200xSimplify that:1000x - 200x is 800x, so P(x) = 800x - 15,000.Okay, that seems right. So the profit function is linear, which makes sense because both revenue and costs are linear in x.Now, to find the break-even point, which is when profit is zero. So, set P(x) = 0 and solve for x.0 = 800x - 15,000Let me solve for x:800x = 15,000x = 15,000 / 800Let me compute that. 15,000 divided by 800. Hmm, 800 goes into 15,000 how many times?Well, 800 x 18 = 14,400, because 800 x 10 = 8,000, 800 x 20 = 16,000, so 18 x 800 is 14,400. Then, 15,000 - 14,400 = 600. So, 600 / 800 = 0.75.So, x = 18.75.But you can't perform a fraction of a cremation service, right? So, the business needs to perform at least 19 services to break even. Because 18 services would give a negative profit, and 19 would be the first point where profit is positive.Wait, let me verify that:If x = 18:Profit = 800*18 - 15,000 = 14,400 - 15,000 = -600. So, loss of 600.x = 19:Profit = 800*19 - 15,000 = 15,200 - 15,000 = +200. So, profit of 200.Yes, so 19 is the minimum number needed to break even.So, summarizing:Profit function: P(x) = 800x - 15,000Break-even point: x = 19.2. Probability of Performing at Least 70 Services After MarketingNow, the second part is about probability. The entrepreneur is considering a new marketing strategy that's expected to increase the number of cremation services by 20%. Currently, the monthly demand follows a normal distribution with a mean of 50 and a standard deviation of 10.So, first, I need to figure out what the new mean will be after a 20% increase. Then, calculate the probability that the number of services performed is at least 70.Let me break it down.First, current mean is 50. A 20% increase would mean the new mean is 50 * 1.20.50 * 1.20 = 60.So, the new mean is 60. The standard deviation remains the same unless stated otherwise. The problem doesn't mention changing the standard deviation, so I assume it's still 10.Therefore, after the marketing strategy, the monthly demand follows a normal distribution with mean μ = 60 and standard deviation σ = 10.We need to find P(X ≥ 70), where X is the number of services.In a normal distribution, to find probabilities, we usually convert the value to a z-score and then use the standard normal distribution table or a calculator.The z-score is calculated as:z = (X - μ) / σSo, plugging in the numbers:z = (70 - 60) / 10 = 10 / 10 = 1.So, z = 1.Now, we need to find the probability that Z is greater than or equal to 1.Looking at standard normal distribution tables, the area to the left of z=1 is approximately 0.8413. Therefore, the area to the right (which is what we need) is 1 - 0.8413 = 0.1587.So, approximately 15.87% chance.But let me double-check the z-table value. For z=1.00, the cumulative probability is 0.8413, so yes, the upper tail is 0.1587.Alternatively, using a calculator or more precise method, it's about 0.15865525, which is roughly 15.87%.So, the probability is approximately 15.87%.Wait, but let me think again. The original distribution is for the current demand, which is N(50,10). After a 20% increase, the new distribution is N(60,10). So, 70 is 10 units above the new mean, which is 1 standard deviation away. So, yes, the z-score is 1, and the probability is about 15.87%.Alternatively, if I use technology, I can compute it more precisely, but for the purposes of this problem, 15.87% is sufficient.So, summarizing:After the marketing strategy, the mean increases to 60, standard deviation remains 10. The probability of performing at least 70 services is approximately 15.87%.Wait, just to make sure, is 70 a reasonable number? 70 is 10 more than the new mean of 60, which is 1 standard deviation. In a normal distribution, about 68% of the data is within 1 standard deviation, so about 16% is above 1 standard deviation. So, that aligns with our previous calculation.Yes, that seems consistent.So, to recap:1. Profit function: P(x) = 800x - 15,000. Break-even at x=19.2. After a 20% increase in demand, the new mean is 60, standard deviation 10. Probability of at least 70 services is approximately 15.87%.I think that's it. I don't see any mistakes in my reasoning. Let me just check the profit function again.Fixed costs: 15,000. Variable cost per service: 200. Price per service: 1000.So, profit per service is 1000 - 200 = 800. So, total profit is 800x - 15,000. Yep, that's correct.Break-even when 800x = 15,000. So, x = 15,000 / 800 = 18.75, rounded up to 19. Correct.For the probability, 20% increase on 50 is 60, standard deviation 10. 70 is 10 away, z=1, probability ~15.87%. Correct.I think I'm confident with these answers.Final Answer1. The monthly profit function is ( P(x) = 800x - 15,000 ) and the break-even point is boxed{19} services.2. The probability of performing at least 70 services is approximately boxed{0.1587} (or 15.87%).</think>