answer：Alright, so I have this problem where an NGO wants to optimize their delivery routes to minimize both costs and carbon emissions. They operate in a region with 5 cities: A, B, C, D, and E. The problem is divided into two sub-problems. The first one is to find the optimal route starting from city A, visiting all other cities exactly once, and returning to A, such that the total cost is minimized. The second part is to calculate the total carbon emissions for that optimal route.Let me start by understanding the data provided. There are three matrices given: Distance, Cost, and Carbon. Each is a 5x5 matrix representing the connections between the cities. The distance matrix shows the distances in kilometers between each pair of cities. The cost factor matrix gives the monetary cost per kilometer for each road, and the carbon emission factor matrix gives the grams of CO2 emitted per kilometer for each road.For the first sub-problem, I need to find the route that minimizes the total cost. Since the delivery must start at A, visit all other cities exactly once, and return to A, this is essentially the Traveling Salesman Problem (TSP) with the objective of minimizing the total cost. The TSP is a classic optimization problem where the goal is to find the shortest possible route that visits each city once and returns to the origin city.Given that there are 5 cities, the number of possible routes is (5-1)! = 24. That's manageable because it's a small number, so maybe I can list all possible permutations and calculate the total cost for each to find the minimum. However, that might take a while, but since it's only 24 routes, it's feasible.Alternatively, I could use an algorithm like the nearest neighbor or dynamic programming to find the optimal route. But since the number of cities is small, brute force might be acceptable here.Let me outline the steps I need to take:1. Generate all possible permutations of the cities B, C, D, E (since the route starts and ends at A).2. For each permutation, calculate the total cost by summing the cost factors multiplied by the distances for each segment of the route.3. Identify the permutation with the minimum total cost.Wait, actually, the cost is given as a factor per kilometer, so the total cost for each segment is the distance multiplied by the cost factor. So, for each road between two cities, I need to multiply the distance by the corresponding cost factor to get the cost for that segment. Then, sum all these costs for the entire route.Similarly, for the carbon emissions, it's the distance multiplied by the carbon emission factor for each segment.But for now, focusing on the first sub-problem, let's structure this.First, list all possible routes. Since the route starts at A and ends at A, the permutations are of the cities B, C, D, E. There are 4! = 24 permutations.I can list them as follows:1. A -> B -> C -> D -> E -> A2. A -> B -> C -> E -> D -> A3. A -> B -> D -> C -> E -> A4. A -> B -> D -> E -> C -> A5. A -> B -> E -> C -> D -> A6. A -> B -> E -> D -> C -> A7. A -> C -> B -> D -> E -> A8. A -> C -> B -> E -> D -> A9. A -> C -> D -> B -> E -> A10. A -> C -> D -> E -> B -> A11. A -> C -> E -> B -> D -> A12. A -> C -> E -> D -> B -> A13. A -> D -> B -> C -> E -> A14. A -> D -> B -> E -> C -> A15. A -> D -> C -> B -> E -> A16. A -> D -> C -> E -> B -> A17. A -> D -> E -> B -> C -> A18. A -> D -> E -> C -> B -> A19. A -> E -> B -> C -> D -> A20. A -> E -> B -> D -> C -> A21. A -> E -> C -> B -> D -> A22. A -> E -> C -> D -> B -> A23. A -> E -> D -> B -> C -> A24. A -> E -> D -> C -> B -> ANow, for each of these 24 routes, I need to calculate the total cost. Let's pick one route as an example to understand how to compute it.Take route 1: A -> B -> C -> D -> E -> ACompute the cost for each segment:A to B: distance is 10 km, cost factor is 2. So cost = 10 * 2 = 20.B to C: distance is 35 km, cost factor is 5. Cost = 35 * 5 = 175.C to D: distance is 30 km, cost factor is 4. Cost = 30 * 4 = 120.D to E: distance is 25 km, cost factor is 4. Cost = 25 * 4 = 100.E to A: distance is 10 km, cost factor is 2. Cost = 10 * 2 = 20.Total cost for this route: 20 + 175 + 120 + 100 + 20 = 435.Similarly, I need to compute this for all 24 routes. This will take some time, but let's see if there's a smarter way or if I can find patterns.Alternatively, since the cost is distance multiplied by cost factor, maybe I can precompute the cost for each direct road and then use those precomputed values to calculate the total cost for each route.Let me create a Cost matrix where each entry (i,j) is distance[i][j] * cost[i][j]. That might make it easier.Given the distance matrix:Row A: 0, 10, 15, 20, 10Row B: 10, 0, 35, 25, 15Row C: 15, 35, 0, 30, 20Row D: 20, 25, 30, 0, 25Row E: 10, 15, 20, 25, 0And the cost factor matrix:Row A: 0, 2, 3, 4, 2Row B: 2, 0, 5, 3, 3Row C: 3, 5, 0, 4, 3Row D: 4, 3, 4, 0, 4Row E: 2, 3, 3, 4, 0So, the precomputed cost matrix (Cost_total) would be:For each i and j, Cost_total[i][j] = Distance[i][j] * Cost[i][j]Let's compute this:Row A:A to A: 0A to B: 10 * 2 = 20A to C: 15 * 3 = 45A to D: 20 * 4 = 80A to E: 10 * 2 = 20Row B:B to A: 10 * 2 = 20B to B: 0B to C: 35 * 5 = 175B to D: 25 * 3 = 75B to E: 15 * 3 = 45Row C:C to A: 15 * 3 = 45C to B: 35 * 5 = 175C to C: 0C to D: 30 * 4 = 120C to E: 20 * 3 = 60Row D:D to A: 20 * 4 = 80D to B: 25 * 3 = 75D to C: 30 * 4 = 120D to D: 0D to E: 25 * 4 = 100Row E:E to A: 10 * 2 = 20E to B: 15 * 3 = 45E to C: 20 * 3 = 60E to D: 25 * 4 = 100E to E: 0So, the Cost_total matrix is:Row A: 0, 20, 45, 80, 20Row B: 20, 0, 175, 75, 45Row C: 45, 175, 0, 120, 60Row D: 80, 75, 120, 0, 100Row E: 20, 45, 60, 100, 0Now, with this precomputed matrix, calculating the total cost for each route is easier.For example, route 1: A->B->C->D->E->ATotal cost = Cost_total[A][B] + Cost_total[B][C] + Cost_total[C][D] + Cost_total[D][E] + Cost_total[E][A]Which is 20 + 175 + 120 + 100 + 20 = 435, as calculated earlier.Similarly, let's compute for another route, say route 2: A->B->C->E->D->ATotal cost = Cost_total[A][B] + Cost_total[B][C] + Cost_total[C][E] + Cost_total[E][D] + Cost_total[D][A]Which is 20 + 175 + 60 + 100 + 80 = 435.Wait, same total cost as route 1.Hmm, interesting. Let me check another one.Route 3: A->B->D->C->E->ATotal cost = Cost_total[A][B] + Cost_total[B][D] + Cost_total[D][C] + Cost_total[C][E] + Cost_total[E][A]Which is 20 + 75 + 120 + 60 + 20 = 295.Wait, that's significantly lower. So 295 is less than 435.Wait, is that correct? Let me double-check.A to B: 20B to D: 75D to C: 120C to E: 60E to A: 20Total: 20+75=95, 95+120=215, 215+60=275, 275+20=295. Yes, that's correct.So, route 3 has a total cost of 295, which is lower than the previous two.Let me try route 4: A->B->D->E->C->ATotal cost = 20 (A-B) + 75 (B-D) + 100 (D-E) + 60 (E-C) + 45 (C-A)Wait, E to C is 60, and C to A is 45.So total: 20+75=95, +100=195, +60=255, +45=300.So 300, which is higher than 295.Route 5: A->B->E->C->D->ATotal cost = 20 (A-B) + 45 (B-E) + 60 (E-C) + 120 (C-D) + 80 (D-A)20+45=65, +60=125, +120=245, +80=325.Higher than 295.Route 6: A->B->E->D->C->ATotal cost = 20 (A-B) + 45 (B-E) + 100 (E-D) + 120 (D-C) + 45 (C-A)20+45=65, +100=165, +120=285, +45=330.Still higher.Route 7: A->C->B->D->E->ATotal cost = 45 (A-C) + 175 (C-B) + 75 (B-D) + 100 (D-E) + 20 (E-A)45+175=220, +75=295, +100=395, +20=415.Higher.Route 8: A->C->B->E->D->ATotal cost = 45 (A-C) + 175 (C-B) + 45 (B-E) + 100 (E-D) + 80 (D-A)45+175=220, +45=265, +100=365, +80=445.Higher.Route 9: A->C->D->B->E->ATotal cost = 45 (A-C) + 120 (C-D) + 75 (D-B) + 45 (B-E) + 20 (E-A)45+120=165, +75=240, +45=285, +20=305.Higher than 295.Route 10: A->C->D->E->B->ATotal cost = 45 (A-C) + 120 (C-D) + 100 (D-E) + 45 (E-B) + 20 (B-A)45+120=165, +100=265, +45=310, +20=330.Higher.Route 11: A->C->E->B->D->ATotal cost = 45 (A-C) + 60 (C-E) + 45 (E-B) + 75 (B-D) + 80 (D-A)45+60=105, +45=150, +75=225, +80=305.Higher.Route 12: A->C->E->D->B->ATotal cost = 45 (A-C) + 60 (C-E) + 100 (E-D) + 75 (D-B) + 20 (B-A)45+60=105, +100=205, +75=280, +20=300.Higher.Route 13: A->D->B->C->E->ATotal cost = 80 (A-D) + 75 (D-B) + 175 (B-C) + 60 (C-E) + 20 (E-A)80+75=155, +175=330, +60=390, +20=410.Higher.Route 14: A->D->B->E->C->ATotal cost = 80 (A-D) + 75 (D-B) + 45 (B-E) + 60 (E-C) + 45 (C-A)80+75=155, +45=200, +60=260, +45=305.Higher.Route 15: A->D->C->B->E->ATotal cost = 80 (A-D) + 120 (D-C) + 175 (C-B) + 45 (B-E) + 20 (E-A)80+120=200, +175=375, +45=420, +20=440.Higher.Route 16: A->D->C->E->B->ATotal cost = 80 (A-D) + 120 (D-C) + 60 (C-E) + 45 (E-B) + 20 (B-A)80+120=200, +60=260, +45=305, +20=325.Higher.Route 17: A->D->E->B->C->ATotal cost = 80 (A-D) + 100 (D-E) + 45 (E-B) + 175 (B-C) + 45 (C-A)80+100=180, +45=225, +175=400, +45=445.Higher.Route 18: A->D->E->C->B->ATotal cost = 80 (A-D) + 100 (D-E) + 60 (E-C) + 175 (C-B) + 20 (B-A)80+100=180, +60=240, +175=415, +20=435.Higher.Route 19: A->E->B->C->D->ATotal cost = 20 (A-E) + 45 (E-B) + 175 (B-C) + 120 (C-D) + 80 (D-A)20+45=65, +175=240, +120=360, +80=440.Higher.Route 20: A->E->B->D->C->ATotal cost = 20 (A-E) + 45 (E-B) + 75 (B-D) + 120 (D-C) + 45 (C-A)20+45=65, +75=140, +120=260, +45=305.Higher.Route 21: A->E->C->B->D->ATotal cost = 20 (A-E) + 60 (E-C) + 175 (C-B) + 75 (B-D) + 80 (D-A)20+60=80, +175=255, +75=330, +80=410.Higher.Route 22: A->E->C->D->B->ATotal cost = 20 (A-E) + 60 (E-C) + 120 (C-D) + 75 (D-B) + 20 (B-A)20+60=80, +120=200, +75=275, +20=295.Wait, this is 295, same as route 3.So, route 22: A->E->C->D->B->A has a total cost of 295.Similarly, let's check route 23: A->E->D->B->C->ATotal cost = 20 (A-E) + 100 (E-D) + 75 (D-B) + 175 (B-C) + 45 (C-A)20+100=120, +75=195, +175=370, +45=415.Higher.Route 24: A->E->D->C->B->ATotal cost = 20 (A-E) + 100 (E-D) + 120 (D-C) + 175 (C-B) + 20 (B-A)20+100=120, +120=240, +175=415, +20=435.Higher.So, from all 24 routes, the minimum total cost is 295, achieved by two routes:1. Route 3: A->B->D->C->E->A2. Route 22: A->E->C->D->B->AWait, let me confirm the total cost for route 22:A->E: 20E->C: 60C->D: 120D->B: 75B->A: 20Total: 20+60=80, +120=200, +75=275, +20=295. Yes, correct.Similarly, route 3:A->B:20B->D:75D->C:120C->E:60E->A:20Total:20+75=95, +120=215, +60=275, +20=295.So both routes have the same total cost of 295.Now, since both routes have the same cost, we can consider either one as the optimal route. However, the problem states "the optimal route", implying there might be only one, but in reality, there can be multiple optimal routes with the same minimal cost.But perhaps I made a mistake in calculating the total cost for route 3. Let me double-check.Route 3: A->B->D->C->E->AA to B:20B to D:75D to C:120C to E:60E to A:20Total:20+75=95, +120=215, +60=275, +20=295. Correct.Similarly, route 22: A->E->C->D->B->AA to E:20E to C:60C to D:120D to B:75B to A:20Total:20+60=80, +120=200, +75=275, +20=295. Correct.So both routes are equally optimal in terms of cost.But the problem says "the optimal route", so maybe I need to choose one. Perhaps the one that also has lower carbon emissions? But that's part of the second sub-problem.Alternatively, maybe I made a mistake in considering the direction of the roads. Wait, the distance matrix is symmetric, but the cost and carbon matrices are also symmetric? Let me check.Looking at the cost matrix:Row A: 0,2,3,4,2Row B:2,0,5,3,3Row C:3,5,0,4,3Row D:4,3,4,0,4Row E:2,3,3,4,0Yes, it's symmetric. So Cost[i][j] = Cost[j][i]. Similarly, the distance matrix is symmetric.Therefore, the cost from A to B is the same as B to A, etc. So, the total cost for a route and its reverse will be the same.Wait, but in our case, route 3 is A->B->D->C->E->A, and route 22 is A->E->C->D->B->A. These are not reverses of each other. Route 3 goes A-B-D-C-E-A, while route 22 goes A-E-C-D-B-A.So, they are different routes, but both have the same total cost.Therefore, both are optimal.But the problem asks for "the optimal route", so perhaps I should present both, but since the problem might expect a single route, maybe I should choose one.Alternatively, perhaps I made a mistake in calculating the total cost for some routes, thinking that 295 is the minimum, but let me check if there's any route with a lower total cost.Looking back at the total costs:The lowest I found was 295, achieved by routes 3 and 22.Is there any route with a lower total cost?Looking at route 3 and 22, both have 295.Let me check another route, say route 19: A->E->B->C->D->ATotal cost was 440, which is higher.Route 15: 440.Route 24: 435.So, 295 is indeed the minimum.Therefore, the optimal route(s) have a total cost of 295.Now, for the second sub-problem, I need to calculate the total carbon emissions for the optimal route found in sub-problem 1.Since there are two optimal routes, I need to calculate the carbon emissions for both and see if they are the same or different.But let's proceed step by step.First, let's compute the carbon emissions for route 3: A->B->D->C->E->A.To do this, I need to calculate the carbon emissions for each segment, which is distance multiplied by carbon emission factor.Given the distance matrix and the carbon emission factor matrix, I can precompute a Carbon_total matrix similar to the Cost_total matrix.Let me compute that.Carbon_total[i][j] = Distance[i][j] * Carbon[i][j]Given the distance matrix:Row A: 0,10,15,20,10Row B:10,0,35,25,15Row C:15,35,0,30,20Row D:20,25,30,0,25Row E:10,15,20,25,0Carbon emission factor matrix:Row A:0,50,70,90,50Row B:50,0,100,75,60Row C:70,100,0,85,65Row D:90,75,85,0,80Row E:50,60,65,80,0So, Carbon_total matrix:Row A:A to A:0A to B:10*50=500A to C:15*70=1050A to D:20*90=1800A to E:10*50=500Row B:B to A:10*50=500B to B:0B to C:35*100=3500B to D:25*75=1875B to E:15*60=900Row C:C to A:15*70=1050C to B:35*100=3500C to C:0C to D:30*85=2550C to E:20*65=1300Row D:D to A:20*90=1800D to B:25*75=1875D to C:30*85=2550D to D:0D to E:25*80=2000Row E:E to A:10*50=500E to B:15*60=900E to C:20*65=1300E to D:25*80=2000E to E:0So, the Carbon_total matrix is:Row A: 0, 500, 1050, 1800, 500Row B: 500, 0, 3500, 1875, 900Row C: 1050, 3500, 0, 2550, 1300Row D: 1800, 1875, 2550, 0, 2000Row E: 500, 900, 1300, 2000, 0Now, let's compute the total carbon emissions for route 3: A->B->D->C->E->ATotal carbon = Carbon_total[A][B] + Carbon_total[B][D] + Carbon_total[D][C] + Carbon_total[C][E] + Carbon_total[E][A]Which is 500 (A-B) + 1875 (B-D) + 2550 (D-C) + 1300 (C-E) + 500 (E-A)Calculating:500 + 1875 = 23752375 + 2550 = 49254925 + 1300 = 62256225 + 500 = 6725 grams of CO2.Now, let's compute for route 22: A->E->C->D->B->ATotal carbon = Carbon_total[A][E] + Carbon_total[E][C] + Carbon_total[C][D] + Carbon_total[D][B] + Carbon_total[B][A]Which is 500 (A-E) + 1300 (E-C) + 2550 (C-D) + 1875 (D-B) + 500 (B-A)Calculating:500 + 1300 = 18001800 + 2550 = 43504350 + 1875 = 62256225 + 500 = 6725 grams of CO2.So, both routes have the same total carbon emissions of 6725 grams.Therefore, regardless of which optimal route we take, the total carbon emissions are the same.But let me double-check the calculations for route 3:A->B:500B->D:1875D->C:2550C->E:1300E->A:500Total:500+1875=2375, +2550=4925, +1300=6225, +500=6725. Correct.Similarly for route 22:A->E:500E->C:1300C->D:2550D->B:1875B->A:500Total:500+1300=1800, +2550=4350, +1875=6225, +500=6725. Correct.So, both routes have the same total carbon emissions.Therefore, the optimal route(s) have a total cost of 295 and total carbon emissions of 6725 grams.But wait, the problem says "the optimal route", so perhaps I should present one of them, but since both are equally optimal, I can mention both.However, the problem might expect a single route, so I can choose one, say route 3: A->B->D->C->E->A.Alternatively, since both have the same cost and carbon emissions, it doesn't matter which one I choose.So, to summarize:Sub-problem 1: The optimal route(s) are A->B->D->C->E->A and A->E->C->D->B->A, both with a total cost of 295.Sub-problem 2: The total carbon emissions for these routes are 6725 grams.But let me check if there's any route with a lower carbon emission. Wait, since both optimal routes have the same carbon emissions, and they are the minimal cost routes, perhaps there isn't a route with lower carbon emissions that also has a lower cost. But the problem only asks for the carbon emissions of the optimal route found in sub-problem 1, which is the minimal cost route.Therefore, the answer is:1. The optimal route has a total cost of 295.2. The total carbon emissions for this route are 6725 grams.But wait, the problem might expect the route itself, not just the cost and emissions. Let me check the problem statement again."Sub-problems:1. Determine the optimal route for a delivery starting from city A and visiting all other cities exactly once before returning to city A, such that the total cost is minimized. Use the given distance and cost factor matrices to calculate the total cost of the optimal route.2. For the optimal route found in sub-problem 1, calculate the total carbon emissions. Use the given distance and carbon emission factor matrices to determine the total carbon emissions for the route."So, for sub-problem 1, I need to determine the optimal route (i.e., the specific sequence of cities) and calculate its total cost.Similarly, for sub-problem 2, using that route, calculate the total carbon emissions.Since there are two optimal routes, I can present both, but perhaps the problem expects one. Alternatively, since both have the same cost and emissions, it's acceptable to present either.But to be thorough, I should mention both routes and their total cost and emissions.However, in the context of an exam or problem-solving, sometimes only one route is expected, even if multiple exist.Alternatively, perhaps I made a mistake in considering the direction of the roads. Wait, in the distance matrix, the distance from A to B is 10, and from B to A is also 10, so it's symmetric. Similarly, the cost and carbon matrices are symmetric, so the total cost and emissions for a route and its reverse are the same.But in our case, the two optimal routes are not reverses of each other. Route 3 is A-B-D-C-E-A, and route 22 is A-E-C-D-B-A. They are different sequences but result in the same total cost and emissions.Therefore, both are valid optimal routes.But perhaps the problem expects the route with the lexicographically smallest sequence or something, but that's not specified.Alternatively, perhaps I should present both routes as optimal.But in the answer, I need to provide the route and the total cost for sub-problem 1, and then the total carbon emissions for that route.Since both routes are equally optimal, I can choose one, say route 3: A-B-D-C-E-A.So, final answers:1. The optimal route is A -> B -> D -> C -> E -> A with a total cost of 295.2. The total carbon emissions for this route are 6725 grams.Alternatively, if I choose route 22, it would be the same total cost and emissions.But to be precise, I should mention both routes if possible, but since the problem asks for "the optimal route", perhaps it's acceptable to present one.Alternatively, perhaps I should present both routes as equally optimal.But in the context of the problem, I think presenting one is sufficient, as both have the same cost and emissions.Therefore, I will proceed with route 3: A-B-D-C-E-A.So, to recap:Sub-problem 1:Optimal route: A -> B -> D -> C -> E -> ATotal cost: 295Sub-problem 2:Total carbon emissions: 6725 gramsTherefore, the answers are:1. The optimal route is A-B-D-C-E-A with a total cost of 295.2. The total carbon emissions are 6725 grams.I think that's it.

answer：Okay, so I have this problem about a hemp farmer with a 100-acre plot. The first part is to find the total yield of hemp using definite integration. The yield per acre is given by the function Y(x) = 50 + 10 sin(πx/50), where x is the number of acres. Hmm, so I need to integrate this function from 0 to 100 to get the total yield.Let me write that down. The total yield would be the integral from 0 to 100 of Y(x) dx, which is ∫₀¹⁰⁰ [50 + 10 sin(πx/50)] dx.Alright, let's break this integral into two parts: the integral of 50 dx and the integral of 10 sin(πx/50) dx. That should make it easier to compute.First, integrating 50 with respect to x from 0 to 100. That's straightforward. The integral of a constant is just the constant times x. So, ∫50 dx from 0 to 100 is 50x evaluated from 0 to 100, which is 50*100 - 50*0 = 5000. So that part is 5000 pounds.Now, the second part is ∫₀¹⁰⁰ 10 sin(πx/50) dx. Let me factor out the 10 first, so it becomes 10 ∫₀¹⁰⁰ sin(πx/50) dx.To integrate sin(ax), the integral is (-1/a) cos(ax) + C. So, applying that here, the integral of sin(πx/50) is (-50/π) cos(πx/50). Therefore, multiplying by 10, we get 10*(-50/π) cos(πx/50) evaluated from 0 to 100.Let me compute that. So, it's 10*(-50/π)[cos(π*100/50) - cos(π*0/50)]. Simplifying inside the cosine functions: π*100/50 is π*2, which is 2π, and π*0/50 is 0.So, cos(2π) is 1, and cos(0) is also 1. Therefore, the expression becomes 10*(-50/π)[1 - 1] = 10*(-50/π)(0) = 0.Wait, so the integral of the sine function over this interval is zero? That makes sense because the sine function is symmetric over its period, and integrating over a whole number of periods would cancel out the positive and negative areas.So, putting it all together, the total yield is 5000 + 0 = 5000 pounds. Hmm, that seems straightforward. But let me double-check my steps.First, the integral of 50 from 0 to 100 is indeed 5000. Then, for the sine part, the integral over 0 to 100. The period of sin(πx/50) is 2π/(π/50) = 100. So, over 0 to 100, it's exactly one full period. Therefore, the integral over one full period of a sine function is zero, which confirms that the second part is zero.So, the total yield is 5000 pounds. That seems correct.Moving on to the second part. The farmer needs to ensure consistent CBD concentration across the entire plot. The concentration C(x) is given by 0.05x + 0.02 cos(πx/50). We need to calculate the average CBD concentration over the 100-acre plot.The average value of a function over an interval [a, b] is given by (1/(b-a)) ∫ₐᵇ C(x) dx. In this case, a is 0 and b is 100, so the average concentration is (1/100) ∫₀¹⁰⁰ [0.05x + 0.02 cos(πx/50)] dx.Let me compute this integral step by step. First, let's split the integral into two parts: ∫₀¹⁰⁰ 0.05x dx and ∫₀¹⁰⁰ 0.02 cos(πx/50) dx.Starting with the first integral: ∫₀¹⁰⁰ 0.05x dx. The integral of x is (1/2)x², so multiplying by 0.05 gives 0.025x². Evaluating from 0 to 100, it's 0.025*(100)² - 0.025*(0)² = 0.025*10000 = 250.Now, the second integral: ∫₀¹⁰⁰ 0.02 cos(πx/50) dx. Let's factor out the 0.02, so it becomes 0.02 ∫₀¹⁰⁰ cos(πx/50) dx.The integral of cos(ax) is (1/a) sin(ax). So, integrating cos(πx/50) gives (50/π) sin(πx/50). Therefore, multiplying by 0.02, we get 0.02*(50/π) sin(πx/50) evaluated from 0 to 100.Calculating that: 0.02*(50/π)[sin(π*100/50) - sin(π*0/50)]. Simplifying inside the sine functions: π*100/50 is 2π, and π*0/50 is 0. So, sin(2π) is 0 and sin(0) is 0. Therefore, the entire expression becomes 0.02*(50/π)*(0 - 0) = 0.So, the integral of the cosine part is zero. That makes sense because, similar to the sine function, the integral over one full period is zero.Therefore, the total integral ∫₀¹⁰⁰ C(x) dx is 250 + 0 = 250.Now, the average concentration is (1/100)*250 = 2.5.Wait, 250 divided by 100 is 2.5. So, the average CBD concentration is 2.5. But let me check the units. The concentration function C(x) is given in what units? The problem says it's modeled by C(x) = 0.05x + 0.02 cos(πx/50). It doesn't specify units, but since the yield was in pounds, perhaps concentration is in some percentage or fraction? It just says "CBD concentration," so maybe it's a decimal or a percentage.But regardless, the average is 2.5. So, if it's a percentage, that would be 2.5%, or if it's a decimal, 0.025, but since the function is 0.05x + 0.02 cos(...), which for x=100 would be 0.05*100 + 0.02 cos(2π) = 5 + 0.02*1 = 5.02. So, the concentration varies from 0.05*0 + 0.02*1 = 0.02 to 5.02. So, the average is 2.5, which is in the same units as the function.Therefore, the average CBD concentration is 2.5.Wait, but let me think again. The average is 250 divided by 100, which is 2.5. So, yes, that's correct.But hold on, let me verify the integral calculations again.First integral: ∫₀¹⁰⁰ 0.05x dx. The antiderivative is 0.025x². At 100, it's 0.025*(10000) = 250. At 0, it's 0. So, 250 - 0 = 250. Correct.Second integral: ∫₀¹⁰⁰ 0.02 cos(πx/50) dx. Antiderivative is 0.02*(50/π) sin(πx/50). Evaluating from 0 to 100: sin(2π) - sin(0) = 0 - 0 = 0. So, 0.02*(50/π)*0 = 0. Correct.Therefore, the average is (250 + 0)/100 = 2.5. So, yes, that's correct.Wait, but let me think about the concentration function again. It's 0.05x + 0.02 cos(πx/50). So, as x increases from 0 to 100, the concentration increases linearly from 0.02 to 5.02, but with a cosine oscillation. The average of the cosine part over the interval is zero, so the average concentration is just the average of the linear part, which is 0.05x averaged over 0 to 100.Wait, actually, the average of 0.05x over 0 to 100 is 0.05*(average of x). The average of x over 0 to 100 is (0 + 100)/2 = 50. So, 0.05*50 = 2.5. That's another way to see it. So, that confirms the result.Therefore, the average CBD concentration is 2.5.So, summarizing:1. Total yield is 5000 pounds.2. Average CBD concentration is 2.5.I think that's it. Let me just make sure I didn't make any calculation mistakes.For the first integral:∫₀¹⁰⁰ [50 + 10 sin(πx/50)] dx= ∫₀¹⁰⁰ 50 dx + ∫₀¹⁰⁰ 10 sin(πx/50) dx= 50x|₀¹⁰⁰ + 10*(-50/π cos(πx/50))|₀¹⁰⁰= 5000 + 10*(-50/π)(cos(2π) - cos(0))= 5000 + 10*(-50/π)(1 - 1)= 5000 + 0= 5000. Correct.For the second integral:(1/100) ∫₀¹⁰⁰ [0.05x + 0.02 cos(πx/50)] dx= (1/100)[ ∫₀¹⁰⁰ 0.05x dx + ∫₀¹⁰⁰ 0.02 cos(πx/50) dx ]= (1/100)[0.025x²|₀¹⁰⁰ + 0.02*(50/π sin(πx/50))|₀¹⁰⁰ ]= (1/100)[250 + 0.02*(50/π)(sin(2π) - sin(0)) ]= (1/100)[250 + 0]= 2.5. Correct.Yes, everything checks out.**Final Answer**1. The total yield of hemp is boxed{5000} pounds.2. The average CBD concentration is boxed{2.5}.

answer：Okay, so I have this problem about a striker who has taken 200 shots and scored 60 goals. I need to figure out the probability ( p ) that he scores on any given shot, and then use that to find the expected number of goals if he takes 250 shots next season. Then, I also need to find the probability that he scores at least 5 goals in the next 10 shots. Hmm, let me think step by step.First, part 1. They mention that the probability of scoring follows a binomial distribution. I remember that in a binomial distribution, the probability of success is constant for each trial, which in this case is each shot. So, the probability ( p ) is the same for each shot.Given that he took 200 shots and scored 60 times, I can calculate ( p ) by dividing the number of goals by the total number of shots. So, ( p = frac{60}{200} ). Let me compute that. 60 divided by 200 is 0.3. So, ( p = 0.3 ). That means he has a 30% chance of scoring on any given shot.Now, the expected number of goals if he takes 250 shots next season. The expected value in a binomial distribution is given by ( E[X] = n times p ), where ( n ) is the number of trials. So, plugging in the numbers, ( E[X] = 250 times 0.3 ). Let me calculate that. 250 times 0.3 is 75. So, the expected number of goals is 75. That seems straightforward.Okay, moving on to part 2. I need to find the probability that the striker will score at least 5 goals in the next 10 shots. So, this is a binomial probability problem where we need the probability of scoring 5 or more goals in 10 shots.The binomial probability formula is ( P(X = k) = C(n, k) times p^k times (1 - p)^{n - k} ), where ( C(n, k) ) is the combination of n things taken k at a time.But since we need the probability of scoring at least 5 goals, that means we need to calculate the probabilities for ( k = 5, 6, 7, 8, 9, 10 ) and sum them up. Alternatively, sometimes it's easier to calculate the complement and subtract from 1, but in this case, since 5 is more than half of 10, maybe it's just as easy to compute each probability and add them.Wait, actually, let me think. The complement of scoring at least 5 goals is scoring fewer than 5 goals, which is 0, 1, 2, 3, or 4 goals. So, if I calculate the probability of scoring 0 to 4 goals and subtract that from 1, I'll get the probability of scoring at least 5 goals. That might be fewer calculations because 5 terms instead of 6. Let me go with that.So, ( P(X geq 5) = 1 - P(X leq 4) ). Therefore, I need to compute ( P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) ) and subtract that sum from 1.Let me write down the formula for each term:For each ( k ) from 0 to 4:( P(X = k) = C(10, k) times (0.3)^k times (0.7)^{10 - k} )I can compute each of these probabilities and then add them up.Let me compute each term step by step.First, ( P(X = 0) ):( C(10, 0) = 1 )( (0.3)^0 = 1 )( (0.7)^{10} approx 0.0282475249 )So, ( P(X = 0) approx 1 times 1 times 0.0282475249 = 0.0282475249 )Next, ( P(X = 1) ):( C(10, 1) = 10 )( (0.3)^1 = 0.3 )( (0.7)^9 approx 0.040353607 )So, ( P(X = 1) approx 10 times 0.3 times 0.040353607 = 10 times 0.0121060821 = 0.121060821 )Wait, hold on, let me verify that calculation. 0.3 times 0.040353607 is approximately 0.0121060821, and then multiplied by 10 gives 0.121060821. That seems correct.Moving on, ( P(X = 2) ):( C(10, 2) = 45 )( (0.3)^2 = 0.09 )( (0.7)^8 approx 0.05764801 )So, ( P(X = 2) approx 45 times 0.09 times 0.05764801 )First, 45 times 0.09 is 4.05, then 4.05 times 0.05764801 is approximately 0.2334729405Wait, let me compute that again. 45 * 0.09 = 4.05. Then, 4.05 * 0.05764801. Let me compute 4 * 0.05764801 = 0.23059204, and 0.05 * 0.05764801 = 0.0028824005. So, adding those together, 0.23059204 + 0.0028824005 ≈ 0.2334744405. So, approximately 0.2334744405.Next, ( P(X = 3) ):( C(10, 3) = 120 )( (0.3)^3 = 0.027 )( (0.7)^7 approx 0.0823543 )So, ( P(X = 3) approx 120 times 0.027 times 0.0823543 )First, 120 * 0.027 = 3.24Then, 3.24 * 0.0823543 ≈ 0.266788332Wait, let me compute 3.24 * 0.0823543. 3 * 0.0823543 = 0.2470629, and 0.24 * 0.0823543 ≈ 0.019765032. Adding them together, 0.2470629 + 0.019765032 ≈ 0.266827932. So, approximately 0.266827932.Moving on, ( P(X = 4) ):( C(10, 4) = 210 )( (0.3)^4 = 0.0081 )( (0.7)^6 approx 0.117649 )So, ( P(X = 4) approx 210 times 0.0081 times 0.117649 )First, 210 * 0.0081 = 1.701Then, 1.701 * 0.117649 ≈ 0.200000Wait, let me compute that more accurately. 1.701 * 0.117649. Let's break it down:1 * 0.117649 = 0.1176490.7 * 0.117649 = 0.08235430.001 * 0.117649 = 0.000117649Adding those together: 0.117649 + 0.0823543 = 0.2000033 + 0.000117649 ≈ 0.200120949So, approximately 0.200120949.Now, let me sum up all these probabilities:( P(X = 0) ≈ 0.0282475249 )( P(X = 1) ≈ 0.121060821 )( P(X = 2) ≈ 0.2334744405 )( P(X = 3) ≈ 0.266827932 )( P(X = 4) ≈ 0.200120949 )Adding them together:First, 0.0282475249 + 0.121060821 = 0.1493083459Then, 0.1493083459 + 0.2334744405 = 0.3827827864Next, 0.3827827864 + 0.266827932 = 0.6496107184Then, 0.6496107184 + 0.200120949 = 0.8497316674So, the total probability of scoring 0 to 4 goals is approximately 0.8497316674.Therefore, the probability of scoring at least 5 goals is ( 1 - 0.8497316674 ≈ 0.1502683326 ).So, approximately 15.03%.Wait, let me double-check my calculations because 15% seems a bit low for scoring at least 5 goals in 10 shots with a 30% chance per shot. Maybe I made a mistake in computing the individual probabilities.Let me verify each step.First, ( P(X = 0) ): 1 * 1 * (0.7)^10. 0.7^10 is approximately 0.0282475249. Correct.( P(X = 1) ): 10 * 0.3 * (0.7)^9. 0.7^9 is approximately 0.040353607. 10 * 0.3 is 3, so 3 * 0.040353607 ≈ 0.121060821. Correct.( P(X = 2) ): 45 * 0.09 * (0.7)^8. 0.7^8 is approximately 0.05764801. 45 * 0.09 is 4.05. 4.05 * 0.05764801 ≈ 0.2334744405. Correct.( P(X = 3) ): 120 * 0.027 * (0.7)^7. 0.7^7 is approximately 0.0823543. 120 * 0.027 is 3.24. 3.24 * 0.0823543 ≈ 0.266827932. Correct.( P(X = 4) ): 210 * 0.0081 * (0.7)^6. 0.7^6 is approximately 0.117649. 210 * 0.0081 is 1.701. 1.701 * 0.117649 ≈ 0.200120949. Correct.Adding them up: 0.0282475249 + 0.121060821 = 0.14930834590.1493083459 + 0.2334744405 = 0.38278278640.3827827864 + 0.266827932 = 0.64961071840.6496107184 + 0.200120949 = 0.8497316674So, 1 - 0.8497316674 ≈ 0.1502683326, which is approximately 15.03%.Hmm, so that seems correct. So, the probability of scoring at least 5 goals in the next 10 shots is approximately 15.03%.Alternatively, maybe I can use the binomial cumulative distribution function (CDF) to compute this more accurately. But since I don't have a calculator here, I think my manual calculations are correct.Alternatively, I can use the normal approximation, but since n=10 is small, the normal approximation might not be very accurate. So, sticking with the exact binomial calculation is better.Wait, another way to check is to compute the expected number of goals in 10 shots, which is 10 * 0.3 = 3. So, the mean is 3. So, scoring at least 5 is above the mean, but not extremely high. So, 15% seems plausible.Alternatively, let me compute the probability of scoring exactly 5 goals and see if that aligns.( P(X = 5) = C(10,5) * (0.3)^5 * (0.7)^5 )C(10,5) is 252.(0.3)^5 ≈ 0.00243(0.7)^5 ≈ 0.16807So, 252 * 0.00243 * 0.16807 ≈ 252 * 0.0004084101 ≈ 0.103032123So, approximately 10.3% chance of scoring exactly 5 goals.Similarly, P(X=6) = C(10,6)*(0.3)^6*(0.7)^4C(10,6)=210(0.3)^6≈0.000729(0.7)^4≈0.2401So, 210 * 0.000729 * 0.2401 ≈ 210 * 0.0001750229 ≈ 0.036754809Similarly, P(X=7)=C(10,7)*(0.3)^7*(0.7)^3C(10,7)=120(0.3)^7≈0.0002187(0.7)^3≈0.343So, 120 * 0.0002187 * 0.343 ≈ 120 * 0.0000750561 ≈ 0.009006732P(X=8)=C(10,8)*(0.3)^8*(0.7)^2C(10,8)=45(0.3)^8≈0.00006561(0.7)^2=0.49So, 45 * 0.00006561 * 0.49 ≈ 45 * 0.0000321489 ≈ 0.0014467005P(X=9)=C(10,9)*(0.3)^9*(0.7)^1C(10,9)=10(0.3)^9≈0.000019683(0.7)^1=0.7So, 10 * 0.000019683 * 0.7 ≈ 10 * 0.0000137781 ≈ 0.000137781P(X=10)=C(10,10)*(0.3)^10*(0.7)^0C(10,10)=1(0.3)^10≈0.0000059049(0.7)^0=1So, 1 * 0.0000059049 * 1 ≈ 0.0000059049Now, adding up P(X=5) to P(X=10):0.103032123 (X=5)+ 0.036754809 (X=6) = 0.139786932+ 0.009006732 (X=7) = 0.148793664+ 0.0014467005 (X=8) = 0.1502403645+ 0.000137781 (X=9) = 0.1503781455+ 0.0000059049 (X=10) ≈ 0.1503840504So, approximately 0.1503840504, which is about 15.04%. So, that's consistent with my earlier calculation of 15.03%. So, that seems correct.Therefore, the probability of scoring at least 5 goals in the next 10 shots is approximately 15.04%.So, summarizing:1. The probability ( p ) is 0.3, and the expected number of goals in 250 shots is 75.2. The probability of scoring at least 5 goals in 10 shots is approximately 15.04%.I think that's it. I don't see any mistakes in my calculations now.

answer：Alright, so I have this problem about a parent whose child is attending a workshop for autism spectrum disorder. The workshop has both individual counseling and group activities, and the effectiveness is measured by improvement in social interaction skills. There are two parts to the problem. Let me tackle them one by one.Starting with part 1: They've given me a function for improvement, I(t) = 100(1 - e^{-kt}), where t is the total hours spent in the workshop, and k is a positive constant depending on the child's responsiveness. The parent observed that after 10 hours, the improvement is 50. I need to find the value of k.Okay, so let's plug in the values we know into the equation. When t = 10, I(t) = 50. So,50 = 100(1 - e^{-k*10})Let me write that down:50 = 100(1 - e^{-10k})Hmm, I can simplify this equation. First, divide both sides by 100 to make it easier:50 / 100 = 1 - e^{-10k}Which simplifies to:0.5 = 1 - e^{-10k}Now, let's solve for e^{-10k}. Subtract 0.5 from both sides:0.5 = e^{-10k}Wait, actually, let me rearrange it:1 - 0.5 = e^{-10k}So, 0.5 = e^{-10k}To solve for k, I need to take the natural logarithm of both sides. Remember that ln(e^x) = x.So, ln(0.5) = ln(e^{-10k})Which simplifies to:ln(0.5) = -10kI know that ln(0.5) is equal to -ln(2), so:-ln(2) = -10kMultiply both sides by -1:ln(2) = 10kTherefore, k = ln(2) / 10Let me compute that. ln(2) is approximately 0.6931, so:k ≈ 0.6931 / 10 ≈ 0.06931So, k is approximately 0.0693. But since the question doesn't specify rounding, I can leave it in terms of ln(2). So, k = (ln 2)/10.Alright, that seems straightforward. Let me just double-check my steps:1. Plugged in t=10 and I=50 into the equation.2. Simplified to 0.5 = e^{-10k}.3. Took natural log of both sides, which gave me ln(0.5) = -10k.4. Recognized ln(0.5) is -ln(2), so k = ln(2)/10.Yes, that looks correct.Moving on to part 2: The social worker recommends an optimal allocation of time between individual counseling (C) and group activities (G) to maximize the improvement. The total time is constrained by C + G = 12 hours. The effectiveness functions are f(C) = 10√C and g(G) = 15 ln(G + 1). We need to maximize I(C, G) = f(C) + g(G).So, the problem is to maximize I = 10√C + 15 ln(G + 1) subject to C + G = 12.Since C + G = 12, we can express G as 12 - C. Then, substitute G into the function I.So, I(C) = 10√C + 15 ln((12 - C) + 1) = 10√C + 15 ln(13 - C)Now, we have I as a function of C alone. To find the maximum, we can take the derivative of I with respect to C, set it equal to zero, and solve for C.Let me compute the derivative I’(C):dI/dC = derivative of 10√C + derivative of 15 ln(13 - C)Derivative of 10√C is 10*(1/(2√C)) = 5 / √CDerivative of 15 ln(13 - C) is 15*( -1 / (13 - C) ) = -15 / (13 - C)So, putting it together:I’(C) = 5 / √C - 15 / (13 - C)Set this equal to zero for maximization:5 / √C - 15 / (13 - C) = 0Let me write that equation:5 / √C = 15 / (13 - C)Divide both sides by 5:1 / √C = 3 / (13 - C)Cross-multiplying:(13 - C) = 3√CLet me write that:13 - C = 3√CHmm, this is a nonlinear equation. Let me try to solve for C.Let me denote √C as x, so C = x². Then, substituting:13 - x² = 3xBring all terms to one side:x² + 3x - 13 = 0Wait, that's quadratic in x. Let me write it as:x² + 3x - 13 = 0Using the quadratic formula:x = [-b ± sqrt(b² - 4ac)] / (2a)Here, a = 1, b = 3, c = -13.So,x = [-3 ± sqrt(9 + 52)] / 2 = [-3 ± sqrt(61)] / 2Since x = √C must be positive, we discard the negative root:x = [ -3 + sqrt(61) ] / 2Compute sqrt(61): approximately 7.81So,x ≈ ( -3 + 7.81 ) / 2 ≈ (4.81)/2 ≈ 2.405Therefore, x ≈ 2.405, so C = x² ≈ (2.405)^2 ≈ 5.78So, C ≈ 5.78 hours, and G = 12 - C ≈ 12 - 5.78 ≈ 6.22 hours.But let me verify if this is indeed a maximum. We can check the second derivative or analyze the behavior.Alternatively, since it's the only critical point in the domain, and the function I(C) is defined for C in [0,12), with I(C) approaching 10√12 + 15 ln(13) as C approaches 12, and I(C) approaching 15 ln(13) as C approaches 0, it's likely that this critical point is a maximum.Alternatively, we can compute the second derivative.Compute I''(C):First derivative: I’(C) = 5 / √C - 15 / (13 - C)Second derivative:Derivative of 5 / √C is 5 * (-1/2) * C^(-3/2) = -5 / (2 C^(3/2))Derivative of -15 / (13 - C) is -15 * (1 / (13 - C)^2 )So,I''(C) = -5 / (2 C^(3/2)) - 15 / (13 - C)^2Since both terms are negative for C in (0,13), the second derivative is negative, which means the function is concave down, so the critical point is indeed a maximum.Therefore, the optimal allocation is approximately C ≈ 5.78 hours and G ≈ 6.22 hours.But let me express the exact value. Earlier, we had x = [ -3 + sqrt(61) ] / 2, so C = x².Compute x exactly:x = [ -3 + sqrt(61) ] / 2So,C = ( [ -3 + sqrt(61) ] / 2 )²Let me compute that:C = ( (-3 + sqrt(61))² ) / 4Expanding the numerator:(-3 + sqrt(61))² = 9 - 6 sqrt(61) + 61 = 70 - 6 sqrt(61)Therefore,C = (70 - 6 sqrt(61)) / 4 = (35 - 3 sqrt(61)) / 2So, exact value is C = (35 - 3 sqrt(61))/2Similarly, G = 12 - C = 12 - (35 - 3 sqrt(61))/2 = (24 - 35 + 3 sqrt(61))/2 = (-11 + 3 sqrt(61))/2So, G = ( -11 + 3 sqrt(61) ) / 2But let me compute approximate numerical values to check.sqrt(61) ≈ 7.81So,C ≈ (35 - 3*7.81)/2 ≈ (35 - 23.43)/2 ≈ (11.57)/2 ≈ 5.785G ≈ ( -11 + 3*7.81 ) / 2 ≈ ( -11 + 23.43 ) / 2 ≈ (12.43)/2 ≈ 6.215Which matches our earlier approximate calculation.Therefore, the optimal allocation is approximately 5.78 hours of counseling and 6.22 hours of group activities.But let me check if these are the exact values or if I can represent them more neatly.Alternatively, since the problem might expect an exact form, perhaps we can leave it in terms of sqrt(61), but maybe they want decimal approximations.But since the problem says "find the optimal values," it might be acceptable to present both exact and approximate values.But in the context of the problem, since time is usually given in decimal hours, perhaps the approximate decimal is more useful.Alternatively, maybe we can rationalize or present it as fractions.But 5.78 is approximately 5 hours and 47 minutes, and 6.22 is approximately 6 hours and 13 minutes. But unless the problem specifies, decimal hours are probably fine.Wait, let me see if I did everything correctly.We started with I = 10√C + 15 ln(13 - C)Took derivative: 5 / √C - 15 / (13 - C) = 0Solved for C: 5 / √C = 15 / (13 - C)Which led to 13 - C = 3√CThen, substitution x = √C, quadratic equation, solved for x, got x ≈ 2.405, so C ≈ 5.78, G ≈ 6.22.Yes, that seems correct.Alternatively, let me test the endpoints.If C = 0, then I = 0 + 15 ln(13) ≈ 15 * 2.5649 ≈ 38.47If C = 12, then I = 10√12 + 15 ln(1) ≈ 10*3.464 + 0 ≈ 34.64At C ≈5.78, let's compute I:10√5.78 ≈ 10*2.405 ≈24.0515 ln(13 - 5.78) =15 ln(7.22) ≈15*1.976 ≈29.64Total I≈24.05 +29.64≈53.69Which is higher than both endpoints, so that makes sense.Hence, the optimal allocation is approximately 5.78 hours of counseling and 6.22 hours of group activities.So, summarizing:1. The value of k is ln(2)/10.2. The optimal values are approximately C ≈5.78 hours and G≈6.22 hours.I think that's it.**Final Answer**1. The value of ( k ) is boxed{dfrac{ln 2}{10}}.2. The optimal values are ( C = boxed{dfrac{35 - 3sqrt{61}}{2}} ) hours and ( G = boxed{dfrac{-11 + 3sqrt{61}}{2}} ) hours.

answer：Okay, so I have two problems to solve here about Penang F.C.'s performance over 20 seasons. Let me take them one at a time.**Problem 1: Total Number of Wins**They've given me a function ( f(t) = 5t - frac{t^2}{4} + sinleft(frac{pi t}{10}right) ) which represents the number of wins per season for each year ( t ) from 1 to 20. I need to find the total number of wins over these 20 seasons.Hmm, so for each season ( t ), I calculate ( f(t) ) and then sum all those up from ( t = 1 ) to ( t = 20 ). That makes sense. So, the total wins ( W ) would be:[W = sum_{t=1}^{20} f(t) = sum_{t=1}^{20} left(5t - frac{t^2}{4} + sinleft(frac{pi t}{10}right)right)]I can split this sum into three separate sums:[W = 5sum_{t=1}^{20} t - frac{1}{4}sum_{t=1}^{20} t^2 + sum_{t=1}^{20} sinleft(frac{pi t}{10}right)]Alright, let me compute each part one by one.**First Sum: ( 5sum_{t=1}^{20} t )**The sum of the first ( n ) natural numbers is given by ( frac{n(n+1)}{2} ). So, for ( n = 20 ):[sum_{t=1}^{20} t = frac{20 times 21}{2} = 210]Multiply by 5:[5 times 210 = 1050]**Second Sum: ( -frac{1}{4}sum_{t=1}^{20} t^2 )**The sum of the squares of the first ( n ) natural numbers is ( frac{n(n+1)(2n+1)}{6} ). Plugging in ( n = 20 ):[sum_{t=1}^{20} t^2 = frac{20 times 21 times 41}{6}]Let me compute that step by step:First, 20 divided by 6 is ( frac{10}{3} ), but maybe it's better to compute numerator first:20 x 21 = 420420 x 41 = Let's compute 420 x 40 = 16,800 and 420 x 1 = 420, so total is 17,220.Then divide by 6:17,220 ÷ 6 = 2,870.So, the sum is 2,870.Multiply by ( -frac{1}{4} ):[-frac{1}{4} times 2,870 = -717.5]Wait, that's a decimal. Hmm, but the number of wins should be an integer, right? But maybe the function allows for fractional wins, but in reality, you can't have half a win. But since it's a mathematical model, maybe it's okay. I'll keep it as is for now.**Third Sum: ( sum_{t=1}^{20} sinleft(frac{pi t}{10}right) )**This seems a bit trickier. I need to compute the sum of sine terms for ( t = 1 ) to ( 20 ) where each term is ( sinleft(frac{pi t}{10}right) ).Let me see if I can find a pattern or a formula for this sum.I recall that the sum of sine functions with arguments in arithmetic progression can be expressed using a formula. The general formula is:[sum_{k=0}^{n-1} sin(a + kd) = frac{sinleft(frac{nd}{2}right) cdot sinleft(a + frac{(n-1)d}{2}right)}{sinleft(frac{d}{2}right)}]In our case, the argument inside the sine is ( frac{pi t}{10} ), so ( a = frac{pi}{10} ) when ( t = 1 ), and the common difference ( d = frac{pi}{10} ) as each term increases by ( frac{pi}{10} ).But wait, in the formula, the sum starts at ( k = 0 ), but our sum starts at ( t = 1 ). So, maybe I can adjust it accordingly.Let me reindex the sum. Let ( k = t - 1 ), so when ( t = 1 ), ( k = 0 ), and when ( t = 20 ), ( k = 19 ). So, the sum becomes:[sum_{k=0}^{19} sinleft(frac{pi (k + 1)}{10}right) = sum_{k=0}^{19} sinleft(frac{pi}{10} + frac{pi k}{10}right)]So, now it's in the form of the formula with ( a = frac{pi}{10} ), ( d = frac{pi}{10} ), and ( n = 20 ).Applying the formula:[sum_{k=0}^{19} sinleft(frac{pi}{10} + frac{pi k}{10}right) = frac{sinleft(frac{20 times frac{pi}{10}}{2}right) cdot sinleft(frac{pi}{10} + frac{(19) times frac{pi}{10}}{2}right)}{sinleft(frac{frac{pi}{10}}{2}right)}]Simplify each part:First, compute ( frac{20 times frac{pi}{10}}{2} ):20 x (π/10) = 2π, so divided by 2 is π.Next, compute the second sine term:( frac{pi}{10} + frac{19 times frac{pi}{10}}{2} )Let me compute 19 x (π/10) = (19π)/10, then divide by 2: (19π)/20.So, adding to π/10:π/10 + 19π/20 = (2π/20 + 19π/20) = 21π/20.So, the numerator becomes:sin(π) * sin(21π/20)But sin(π) is 0, so the entire numerator is 0.Therefore, the sum is 0.Wait, that can't be right. If the numerator is zero, the sum is zero? Let me verify.Yes, because sin(π) is zero, so regardless of the other terms, the entire expression is zero. So, the sum of the sine terms from t=1 to t=20 is zero.Hmm, interesting. So, that third sum is zero.So, putting it all together:Total wins ( W = 1050 - 717.5 + 0 = 1050 - 717.5 = 332.5 )But wait, 332.5 wins over 20 seasons? That seems like an average of about 16.625 wins per season. Is that reasonable?Looking back at the function ( f(t) = 5t - frac{t^2}{4} + sinleft(frac{pi t}{10}right) ). Let me check for t=1:f(1) = 5(1) - (1)^2/4 + sin(π/10) ≈ 5 - 0.25 + 0.3090 ≈ 5.059t=20:f(20) = 5(20) - (20)^2 /4 + sin(2π) = 100 - 100 + 0 = 0Wait, so in the 20th season, they have zero wins? That seems odd, but maybe it's a trough in their performance.But the total is 332.5. Since the number of wins should be an integer, but since we're summing a function that can give fractional values, the total can be a decimal. So, 332.5 is acceptable in this context.But let me double-check the sum of the sine terms. Maybe I made a mistake there.I used the formula for the sum of sines in arithmetic progression, which gave me zero because sin(π) is zero. Let me compute the sum manually for a few terms to see if it makes sense.Compute sin(π t /10) for t=1 to 20:t=1: sin(π/10) ≈ 0.3090t=2: sin(π/5) ≈ 0.5878t=3: sin(3π/10) ≈ 0.8090t=4: sin(2π/5) ≈ 0.9511t=5: sin(π/2) = 1t=6: sin(3π/5) ≈ 0.9511t=7: sin(7π/10) ≈ 0.8090t=8: sin(4π/5) ≈ 0.5878t=9: sin(9π/10) ≈ 0.3090t=10: sin(π) = 0t=11: sin(11π/10) ≈ -0.3090t=12: sin(12π/10) = sin(6π/5) ≈ -0.5878t=13: sin(13π/10) ≈ -0.8090t=14: sin(14π/10) = sin(7π/5) ≈ -0.9511t=15: sin(15π/10) = sin(3π/2) = -1t=16: sin(16π/10) = sin(8π/5) ≈ -0.9511t=17: sin(17π/10) ≈ -0.8090t=18: sin(18π/10) = sin(9π/5) ≈ -0.5878t=19: sin(19π/10) ≈ -0.3090t=20: sin(20π/10) = sin(2π) = 0Now, let's add these up:Positive terms (t=1 to t=10):0.3090 + 0.5878 + 0.8090 + 0.9511 + 1 + 0.9511 + 0.8090 + 0.5878 + 0.3090 + 0Let me compute step by step:Start with 0.3090+0.5878 = 0.8968+0.8090 = 1.7058+0.9511 = 2.6569+1 = 3.6569+0.9511 = 4.6080+0.8090 = 5.4170+0.5878 = 6.0048+0.3090 = 6.3138+0 = 6.3138Negative terms (t=11 to t=20):-0.3090 -0.5878 -0.8090 -0.9511 -1 -0.9511 -0.8090 -0.5878 -0.3090 +0Compute step by step:Start with -0.3090-0.5878 = -0.8968-0.8090 = -1.7058-0.9511 = -2.6569-1 = -3.6569-0.9511 = -4.6080-0.8090 = -5.4170-0.5878 = -6.0048-0.3090 = -6.3138+0 = -6.3138So, total sum is 6.3138 (positive) + (-6.3138) (negative) = 0.Ah, so that's why the sum is zero. The positive and negative terms cancel each other out. So, the sum of the sine terms is indeed zero. That makes sense.Therefore, the total number of wins is 1050 - 717.5 = 332.5.But since we can't have half a win, maybe the function is designed such that the total is a whole number. Wait, 332.5 is half-integer. Maybe I made a mistake in the second sum.Wait, let's check the second sum again.Sum of t^2 from t=1 to 20 is 2,870. Then, multiplied by -1/4 is -717.5. That's correct.First sum is 1050, correct.Third sum is 0, correct.So, 1050 - 717.5 = 332.5. Hmm. Maybe the function allows for fractional wins, so the total can be a decimal. Alternatively, perhaps the function is defined such that the total is an integer, but in this case, it's 332.5. Maybe I need to round it? The problem doesn't specify, so perhaps it's acceptable as 332.5.But let me check if I did the sum of t^2 correctly.Sum of squares formula: ( frac{n(n+1)(2n+1)}{6} )For n=20:20 x 21 x 41 / 620/6 = 10/3 ≈ 3.333, but let's compute 20 x 21 = 420, 420 x 41 = 17,220, 17,220 /6 = 2,870. Correct.So, no mistake there.So, total wins is 332.5.But the problem says "the total number of wins over the 20 seasons." Since it's a mathematical model, fractional wins are acceptable in the calculation, even though in reality, you can't have half a win. So, I think 332.5 is the answer.But let me see if the function f(t) is supposed to give integer values. Let me compute f(t) for a few t's.For t=1: 5(1) - (1)^2/4 + sin(π/10) ≈ 5 - 0.25 + 0.3090 ≈ 5.059t=2: 10 - 1 + sin(π/5) ≈ 9 + 0.5878 ≈ 9.5878t=3: 15 - 2.25 + sin(3π/10) ≈ 12.75 + 0.8090 ≈ 13.559t=4: 20 - 4 + sin(2π/5) ≈ 16 + 0.9511 ≈ 16.9511t=5: 25 - 6.25 + sin(π/2) = 18.75 + 1 = 19.75t=10: 50 - 25 + sin(π) = 25 + 0 = 25t=15: 75 - 56.25 + sin(3π/2) = 18.75 -1 = 17.75t=20: 100 - 100 + sin(2π) = 0 + 0 = 0So, f(t) does give fractional wins, which sum up to 332.5. So, the answer is 332.5.But the problem might expect an integer. Maybe I need to round it? Or perhaps the function is designed such that the total is an integer. Wait, 332.5 is exactly halfway between 332 and 333. Maybe the problem expects 332.5, or perhaps it's a typo and should be 332 or 333. But since it's a mathematical function, I think 332.5 is acceptable.So, I think the total number of wins is 332.5.**Problem 2: Expected Total Number of Goals**Now, the second problem is about the number of goals scored by Penang F.C. in each match, which follows a Poisson distribution with mean ( lambda(t) = frac{6}{sqrt{t}} ). I need to compute the expected total number of goals over the entire 20 seasons.First, let's understand what's given.Each season t, they played 30 matches (from the first part: "played an average of 30 matches per season"). So, for each season t, the number of goals per match is Poisson with mean ( lambda(t) = frac{6}{sqrt{t}} ).Therefore, the expected number of goals in one match is ( lambda(t) ), and since there are 30 matches per season, the expected total goals per season is ( 30 times lambda(t) ).Therefore, the expected total goals over 20 seasons is the sum over t=1 to 20 of ( 30 times lambda(t) ).So, mathematically:[E[text{Total Goals}] = sum_{t=1}^{20} 30 times lambda(t) = 30 sum_{t=1}^{20} frac{6}{sqrt{t}} = 180 sum_{t=1}^{20} frac{1}{sqrt{t}}]So, I need to compute ( sum_{t=1}^{20} frac{1}{sqrt{t}} ) and then multiply by 180.Let me compute ( sum_{t=1}^{20} frac{1}{sqrt{t}} ).This is the sum of reciprocals of square roots from 1 to 20.I don't think there's a simple closed-form formula for this sum, so I'll have to compute it numerically.Let me list out the terms and compute each one:t=1: 1/√1 = 1t=2: 1/√2 ≈ 0.7071t=3: 1/√3 ≈ 0.5774t=4: 1/2 = 0.5t=5: 1/√5 ≈ 0.4472t=6: 1/√6 ≈ 0.4082t=7: 1/√7 ≈ 0.37796t=8: 1/√8 ≈ 0.3536t=9: 1/3 ≈ 0.3333t=10: 1/√10 ≈ 0.3162t=11: 1/√11 ≈ 0.3015t=12: 1/√12 ≈ 0.2887t=13: 1/√13 ≈ 0.2774t=14: 1/√14 ≈ 0.2673t=15: 1/√15 ≈ 0.2582t=16: 1/4 = 0.25t=17: 1/√17 ≈ 0.2425t=18: 1/√18 ≈ 0.2357t=19: 1/√19 ≈ 0.2294t=20: 1/√20 ≈ 0.2236Now, let's compute each term and sum them up step by step.Let me list them with approximate values:1.0000,0.7071,0.5774,0.5000,0.4472,0.4082,0.37796,0.3536,0.3333,0.3162,0.3015,0.2887,0.2774,0.2673,0.2582,0.2500,0.2425,0.2357,0.2294,0.2236.Let me add them sequentially:Start with 1.0000+0.7071 = 1.7071+0.5774 = 2.2845+0.5000 = 2.7845+0.4472 = 3.2317+0.4082 = 3.6399+0.37796 ≈ 4.0179+0.3536 ≈ 4.3715+0.3333 ≈ 4.7048+0.3162 ≈ 5.0210+0.3015 ≈ 5.3225+0.2887 ≈ 5.6112+0.2774 ≈ 5.8886+0.2673 ≈ 6.1559+0.2582 ≈ 6.4141+0.2500 ≈ 6.6641+0.2425 ≈ 6.9066+0.2357 ≈ 7.1423+0.2294 ≈ 7.3717+0.2236 ≈ 7.5953So, the sum ( sum_{t=1}^{20} frac{1}{sqrt{t}} ≈ 7.5953 )Therefore, the expected total number of goals is:180 x 7.5953 ≈ 180 x 7.5953Let me compute that:First, 100 x 7.5953 = 759.5380 x 7.5953 = 607.624So, total is 759.53 + 607.624 = 1,367.154Approximately 1,367.15 goals.But let me check my addition:Wait, 180 x 7.5953:Compute 7.5953 x 180:Multiply 7.5953 by 100: 759.53Multiply 7.5953 by 80: 7.5953 x 80 = 607.624Add them together: 759.53 + 607.624 = 1,367.154Yes, that's correct.So, approximately 1,367.15 goals.But let me check if I added the sum correctly. I approximated the sum as 7.5953, but let me verify that.Wait, when I added up all the terms step by step, I got approximately 7.5953. Let me recount the addition to ensure I didn't make a mistake.Let me list the cumulative sum step by step:1.00001.70712.28452.78453.23173.63994.01794.37154.70485.02105.32255.61125.88866.15596.41416.66416.90667.14237.37177.5953Yes, that seems correct. So, the sum is approximately 7.5953.Therefore, 180 x 7.5953 ≈ 1,367.15.But let me see if I can get a more accurate sum by using more decimal places for each term.Alternatively, perhaps I can use a calculator to compute the sum more accurately.But since I don't have a calculator, I'll proceed with the approximate value.But let me check if I can compute the sum more accurately by considering more decimal places for each term.Let me recompute the sum with more precision:t=1: 1.000000t=2: 0.707107t=3: 0.577350t=4: 0.500000t=5: 0.447214t=6: 0.408248t=7: 0.377964t=8: 0.353553t=9: 0.333333t=10: 0.316228t=11: 0.301511t=12: 0.288675t=13: 0.277350t=14: 0.267261t=15: 0.258198t=16: 0.250000t=17: 0.242535t=18: 0.235702t=19: 0.229416t=20: 0.223607Now, let's add them more precisely:Start with 1.000000+0.707107 = 1.707107+0.577350 = 2.284457+0.500000 = 2.784457+0.447214 = 3.231671+0.408248 = 3.639919+0.377964 = 4.017883+0.353553 = 4.371436+0.333333 = 4.704769+0.316228 = 5.021000+0.301511 = 5.322511+0.288675 = 5.611186+0.277350 = 5.888536+0.267261 = 6.155797+0.258198 = 6.413995+0.250000 = 6.663995+0.242535 = 6.906530+0.235702 = 7.142232+0.229416 = 7.371648+0.223607 = 7.595255So, the sum is approximately 7.595255.Therefore, 180 x 7.595255 ≈ 180 x 7.595255.Compute 7.595255 x 180:First, 7 x 180 = 1,2600.595255 x 180 = ?Compute 0.5 x 180 = 900.095255 x 180 ≈ 17.1459So, total ≈ 90 + 17.1459 = 107.1459Therefore, total ≈ 1,260 + 107.1459 ≈ 1,367.1459So, approximately 1,367.15 goals.Therefore, the expected total number of goals is approximately 1,367.15.But since the number of goals should be an integer, but since we're dealing with expectations, it's acceptable to have a decimal. However, the problem might expect an exact value or a more precise decimal.Alternatively, perhaps I can express the sum in terms of the Riemann zeta function or harmonic series, but for n=20, it's just a finite sum, so numerical approximation is the way to go.Alternatively, I can use the integral approximation for the sum ( sum_{t=1}^{n} frac{1}{sqrt{t}} approx 2sqrt{n} + zeta(1/2) ), but that might complicate things. Since n=20 is small, the exact numerical sum is better.So, I think 1,367.15 is a good approximation. But let me check if I can compute it more accurately.Alternatively, perhaps I can use a calculator to compute the sum more precisely, but since I'm doing it manually, I'll stick with 7.595255.Therefore, 180 x 7.595255 ≈ 1,367.1459, which is approximately 1,367.15.But let me check if I can compute 7.595255 x 180 more accurately:7.595255 x 180:Breakdown:7 x 180 = 1,2600.595255 x 180:Compute 0.5 x 180 = 900.095255 x 180:Compute 0.09 x 180 = 16.20.005255 x 180 ≈ 0.9459So, 16.2 + 0.9459 ≈ 17.1459Therefore, 0.595255 x 180 ≈ 90 + 17.1459 ≈ 107.1459Therefore, total ≈ 1,260 + 107.1459 ≈ 1,367.1459So, 1,367.1459, which is approximately 1,367.15.Therefore, the expected total number of goals is approximately 1,367.15.But let me see if I can represent this as a fraction or a more precise decimal.Alternatively, perhaps I can leave it as 180 x sum, but since the sum is approximately 7.595255, multiplying by 180 gives approximately 1,367.15.Therefore, the expected total number of goals is approximately 1,367.15.But let me check if I can compute the sum more accurately by using more decimal places for each term.Wait, I already used more decimal places for each term, so the sum is accurate to about 7.595255.Therefore, 180 x 7.595255 ≈ 1,367.15.So, I think that's as precise as I can get without a calculator.Therefore, the expected total number of goals is approximately 1,367.15.But since the problem might expect an exact value, perhaps I can express it as 180 times the sum, but since the sum is irrational, it's better to present the approximate value.Alternatively, perhaps I can round it to the nearest whole number, which would be 1,367 goals.But let me check if 1,367.15 is closer to 1,367 or 1,368. Since 0.15 is less than 0.5, it rounds down to 1,367.But in the context of expected goals, it's acceptable to have a decimal, but if the problem expects an integer, then 1,367 is the answer.But the problem says "compute the expected total number of goals," so it's acceptable to have a decimal. Therefore, 1,367.15 is the expected total.But let me check if I can express it as a fraction.1,367.15 is equal to 1,367 + 0.15, which is 1,367 + 3/20, so 1,367 3/20.But perhaps it's better to present it as a decimal.Alternatively, perhaps I can use more precise decimal places.But given that the sum was approximated to 7.595255, multiplying by 180 gives 1,367.1459, which is approximately 1,367.15.Therefore, I think 1,367.15 is a good answer.But let me check if I can compute the sum more accurately.Alternatively, perhaps I can use the formula for the sum of reciprocals of square roots.The sum ( sum_{k=1}^{n} frac{1}{sqrt{k}} ) can be approximated by ( 2sqrt{n} + zeta(1/2) + frac{1}{2sqrt{n}} - frac{1}{24 n^{3/2}} + dots ), where ( zeta(1/2) ) is the Riemann zeta function at 1/2, which is approximately -1.4603545088.But for n=20, the approximation might not be very accurate, but let's try.Compute ( 2sqrt{20} + zeta(1/2) + frac{1}{2sqrt{20}} - frac{1}{24 times 20^{3/2}} )First, ( 2sqrt{20} ≈ 2 x 4.4721 ≈ 8.9442 )( zeta(1/2) ≈ -1.46035 )( frac{1}{2sqrt{20}} ≈ 0.1118 )( frac{1}{24 times 20^{3/2}} = frac{1}{24 x (20 x sqrt{20})} = frac{1}{24 x 20 x 4.4721} ≈ frac{1}{24 x 89.442} ≈ frac{1}{2146.608} ≈ 0.000466 )So, putting it all together:8.9442 - 1.46035 + 0.1118 - 0.000466 ≈8.9442 - 1.46035 = 7.483857.48385 + 0.1118 = 7.595657.59565 - 0.000466 ≈ 7.595184Which is very close to our manual sum of 7.595255. So, the approximation is quite accurate.Therefore, the sum is approximately 7.595184, which is very close to our manual calculation.Therefore, 180 x 7.595184 ≈ 180 x 7.595184 ≈ 1,367.13312So, approximately 1,367.13.Therefore, the expected total number of goals is approximately 1,367.13.But since we're dealing with expectations, it's acceptable to have a decimal. However, if we need to present it as a whole number, it would be 1,367 goals.But the problem doesn't specify, so I think 1,367.13 is acceptable.But let me check if I can compute it more precisely.Alternatively, perhaps I can use the exact sum I calculated earlier, which was 7.595255, leading to 1,367.1459, which is approximately 1,367.15.So, I think 1,367.15 is a good answer.But let me check if I can represent it as a fraction.0.15 is 3/20, so 1,367.15 = 1,367 + 3/20 = 1,367 3/20.But perhaps it's better to present it as a decimal.Therefore, the expected total number of goals is approximately 1,367.15.But let me check if I can compute the sum more accurately by using more precise values for each term.Wait, I already used more precise values for each term, so the sum is accurate to about 7.595255.Therefore, 180 x 7.595255 ≈ 1,367.1459, which is approximately 1,367.15.Therefore, I think that's as precise as I can get without a calculator.So, to summarize:Problem 1: Total wins = 332.5Problem 2: Expected total goals ≈ 1,367.15But let me check if I can present these answers in a more precise way.For Problem 1, 332.5 is exact, as the sum of the function f(t) over t=1 to 20 gives exactly 332.5.For Problem 2, the sum of 1/√t from t=1 to 20 is approximately 7.595255, leading to 180 x 7.595255 ≈ 1,367.1459, which is approximately 1,367.15.Therefore, the answers are:1. 332.52. Approximately 1,367.15But let me check if the problem expects an exact value for the second part.Wait, the problem says "compute the expected total number of goals," so it's acceptable to present it as a decimal. Alternatively, perhaps I can express it as a fraction.But 1,367.15 is equal to 1,367 + 0.15, which is 1,367 + 3/20, so 1,367 3/20, which is 27,343/20.But that's a bit messy, so decimal is better.Alternatively, perhaps I can present it as 1,367.15.Therefore, I think that's the answer.**Final Answer**1. The total number of wins over the 20 seasons is boxed{332.5}.2. The expected total number of goals scored over the entire 20 seasons is boxed{1367.15}.

answer：Alright, so I have this problem about optimizing the transportation of iron ore from Kryvyi Rih to Germany. The consultant has identified three routes, each with different distances, fuel costs per kilometer, and fixed international tariffs. I need to figure out which route is the cheapest and then, considering that tariffs increase by 5% each year, calculate the total cost over three years for the optimal route.Let me start with the first part. I need to calculate the total transportation cost for each of the three routes. The total cost should include both the variable cost (which depends on distance and fuel cost) and the fixed international tariff.For each route, the total cost can be calculated as:Total Cost = (Distance × Fuel Cost per km) + Fixed TariffSo, for route 1:Total Cost 1 = (1500 km × 2 euros/km) + 300 eurosLet me compute that:1500 × 2 = 3000 euros3000 + 300 = 3300 eurosSo, route 1 costs 3300 euros.For route 2:Total Cost 2 = (1800 km × 1.8 euros/km) + 350 eurosCalculating that:1800 × 1.8. Hmm, 1800 × 1 = 1800, 1800 × 0.8 = 1440, so total is 1800 + 1440 = 3240 euros3240 + 350 = 3590 eurosSo, route 2 is 3590 euros.For route 3:Total Cost 3 = (2000 km × 1.7 euros/km) + 400 eurosCalculating:2000 × 1.7 = 3400 euros3400 + 400 = 3800 eurosSo, route 3 is 3800 euros.Comparing the three total costs: 3300, 3590, 3800. Clearly, route 1 is the cheapest. So, the optimal route is route 1.Now, moving on to the second part. The tariffs are subject to a 5% annual increase over three years. I need to calculate the total cost for route 1 over three years, considering the compounded increase in tariffs.First, let's understand what's happening. Each year, the fixed tariff for route 1 will increase by 5%. So, the tariff in year 1 is 300 euros, year 2 it's 300 × 1.05, year 3 it's 300 × (1.05)^2.But wait, is the fuel cost also subject to inflation? The problem statement says the tariffs are subject to a 5% annual increase, but it doesn't mention fuel costs. So, I think only the fixed tariffs increase each year, while the fuel cost per kilometer remains constant.So, each year, the total cost for route 1 will be:Year 1: (1500 × 2) + 300 = 3300 eurosYear 2: (1500 × 2) + (300 × 1.05) = 3000 + 315 = 3315 eurosYear 3: (1500 × 2) + (300 × 1.05^2) = 3000 + 330.75 = 3330.75 eurosWait, let me compute 1.05 squared: 1.05 × 1.05 = 1.1025. So, 300 × 1.1025 = 330.75 euros.So, total cost over three years is the sum of these three amounts.Total Cost = 3300 + 3315 + 3330.75Let me add them up:3300 + 3315 = 66156615 + 3330.75 = 9945.75 eurosSo, the total cost over three years is 9945.75 euros.But wait, let me double-check my calculations because sometimes it's easy to make a mistake with the exponents.Year 1: 300 eurosYear 2: 300 × 1.05 = 315 eurosYear 3: 300 × (1.05)^2 = 300 × 1.1025 = 330.75 eurosSo, the fixed tariffs over three years are 300, 315, 330.75.Adding these: 300 + 315 = 615, 615 + 330.75 = 945.75 eurosBut wait, the variable cost is 3000 euros each year, right? Because fuel cost is fixed. So, each year, the variable cost is 3000 euros, so over three years, that's 3000 × 3 = 9000 euros.Then, adding the fixed tariffs over three years: 945.75 euros.So, total cost is 9000 + 945.75 = 9945.75 euros.Yes, that's consistent with my earlier calculation.Therefore, the total cost over three years is 9945.75 euros.Wait, but let me think again: is the fuel cost per kilometer fixed each year? The problem says "the cost of fuel per kilometer for each route is..." and doesn't mention it changing, so I think it's fixed. So, variable cost is fixed each year, only the fixed tariff increases.Therefore, my calculation is correct.So, summarizing:1. The optimal route is route 1 with a total cost of 3300 euros per year.2. Over three years, considering the 5% annual increase in tariffs, the total cost is 9945.75 euros.I think that's it. Let me just make sure I didn't miss anything.Wait, the problem says "the total cost for the optimal route identified in sub-problem 1 over the 3-year period, considering the compounded increase in tariffs."So, I think I did that correctly. Each year, the tariff increases by 5%, so year 1: 300, year 2: 315, year 3: 330.75. Then, add up all three years' costs.Yes, that seems right.**Final Answer**1. The optimal route is route 1 with a total cost of boxed{3300} euros.2. The total cost over three years is boxed{9945.75} euros.