🧠:<think>Alright, so I have this problem about a property manager and a tycoon trying to reduce energy costs by 15% across their portfolio of three buildings. Let me try to figure this out step by step.First, the portfolio consists of three buildings: A, B, and C. Their monthly energy costs are 50,000, 70,000, and 80,000 respectively. The tycoon wants a 15% reduction in total energy costs over the next year. The manager is implementing different energy-saving measures for each building.Problem 1: Determine if the energy-saving measures will achieve the 15% reduction target.Okay, so I need to calculate the total current energy cost, then calculate the savings from each building, subtract those savings, and see if the total cost reduction meets or exceeds 15%.Let me start by calculating the current total monthly energy cost. That's straightforward:Building A: 50,000Building B: 70,000Building C: 80,000Total monthly cost = 50,000 + 70,000 + 80,000 = 200,000So, the total annual cost currently is 200,000 * 12 = 2,400,000.The target is a 15% reduction. So, the target savings would be 15% of 2,400,000.15% of 2,400,000 is 0.15 * 2,400,000 = 360,000.So, the total energy cost after savings should be 2,400,000 - 360,000 = 2,040,000.Now, let's calculate the savings from each building.Building A: 10% reduction on 50,000 per month.10% of 50,000 is 0.10 * 50,000 = 5,000 per month.So, annual savings for A: 5,000 * 12 = 60,000.Building B: 20% reduction on 70,000 per month.20% of 70,000 is 0.20 * 70,000 = 14,000 per month.Annual savings for B: 14,000 * 12 = 168,000.Building C: 25% reduction on 80,000 per month.25% of 80,000 is 0.25 * 80,000 = 20,000 per month.Annual savings for C: 20,000 * 12 = 240,000.Total savings from all buildings: 60,000 + 168,000 + 240,000 = 468,000.Wait, that's more than the target savings of 360,000. So, does that mean the target is achieved?But let me double-check. Maybe I should calculate the new total cost after savings and see the percentage reduction.Calculating the new monthly costs after savings:Building A: 50,000 - 5,000 = 45,000Building B: 70,000 - 14,000 = 56,000Building C: 80,000 - 20,000 = 60,000New total monthly cost: 45,000 + 56,000 + 60,000 = 161,000Annual cost: 161,000 * 12 = 1,932,000Original annual cost was 2,400,000. So, the reduction is 2,400,000 - 1,932,000 = 468,000.Percentage reduction: (468,000 / 2,400,000) * 100 = 19.5%19.5% is more than the 15% target. So, yes, the measures achieve the target.Wait, but let me think again. The problem says "over the next year." So, are we considering the savings over a year? Yes, because the measures are implemented and savings are annual.So, the first part is done. The total savings exceed the target.Problem 2: After implementing the measures, the energy price is expected to rise by 5%. So, we need to adjust the total energy cost after savings and then see if the original target is still met.Hmm. So, first, after the savings, the total annual cost is 1,932,000. But then, energy prices rise by 5%, so the cost will increase by 5%.Wait, is the 5% increase applied to the original cost or the reduced cost?I think it's applied to the reduced cost because the savings have already been implemented, and then the price increases.So, the new cost after price increase would be 1,932,000 * 1.05.Let me calculate that.1,932,000 * 1.05 = ?1,932,000 * 1 = 1,932,0001,932,000 * 0.05 = 96,600Total: 1,932,000 + 96,600 = 2,028,600Now, compare this adjusted total cost to the original target.Original target was a 15% reduction from 2,400,000, which was 2,040,000.But after the price increase, the total cost is 2,028,600, which is higher than 2,040,000.Wait, no. Wait, 2,028,600 is less than 2,400,000, but is it a 15% reduction?Wait, let's calculate the percentage reduction from the original.Original cost: 2,400,000Adjusted cost after savings and price increase: 2,028,600Reduction amount: 2,400,000 - 2,028,600 = 371,400Percentage reduction: (371,400 / 2,400,000) * 100 ≈ 15.475%So, approximately 15.475% reduction, which is just over 15%. So, the target is still met.Wait, but let me check the calculations again.First, the total cost after savings: 1,932,000Then, a 5% increase on that: 1,932,000 * 1.05 = 2,028,600Original total cost: 2,400,000So, the difference is 2,400,000 - 2,028,600 = 371,400Percentage reduction: (371,400 / 2,400,000) * 100 = 15.475%So, yes, it's a 15.475% reduction, which is just above 15%. So, the target is still achieved.But wait, is the 5% increase applied to the original cost or the reduced cost? The problem says "the average energy price is expected to rise by 5% in the next year." So, I think it's applied to the cost after the savings. Because the savings are implemented first, then the price increases.Alternatively, if the price increase was applied before the savings, it would be different. But I think the order is savings first, then price increase.So, yes, the adjusted total cost is 2,028,600, which is a 15.475% reduction from the original 2,400,000. So, the target is still met.But let me think again. Maybe the price increase affects the savings? Or is the savings based on the original energy cost, regardless of price changes?Wait, the savings are based on the current energy costs, which are before the price increase. So, if the price increases, the savings percentages would still apply to the original costs, but the actual dollar savings might be different.Wait, no. The savings are implemented, which reduces the energy usage, so the cost is based on the reduced usage. Then, the price increases, so the cost per unit goes up, but the usage is already reduced.Wait, maybe I need to model it differently.Let me think in terms of energy usage and price.Suppose each building's energy cost is a function of usage and price.Let me denote:For each building, energy cost = usage * price.The savings measures reduce the usage, so the new cost is (usage - reduction) * price.But if the price increases, it's (usage - reduction) * (price * 1.05).Alternatively, if the price increase is applied before the savings, it would be (usage * 1.05) - reduction, but that doesn't make much sense because the savings are based on the original usage.Wait, no. The savings are based on the current usage, so the reduction is a percentage of the current usage. So, if the price increases, the cost per unit goes up, but the usage is already reduced.So, the correct way is:Original cost: usage * priceAfter savings: (usage - reduction) * priceAfter price increase: (usage - reduction) * (price * 1.05)So, the total cost after both savings and price increase is 1.05 times the cost after savings.Which is what I did earlier.So, the total cost after savings is 1,932,000, then multiplied by 1.05 gives 2,028,600.Which is a 15.475% reduction from the original 2,400,000.So, the target is still met.But let me check if the problem expects the price increase to be applied before the savings. That would be a different calculation.If the price increases first, then the savings are applied to the increased cost.So, original cost: 200,000/monthPrice increases by 5%, so new cost before savings: 200,000 * 1.05 = 210,000/monthThen, apply the savings:Building A: 10% of 50,000 is 5,000, but now the cost is 50,000 * 1.05 = 52,500. So, 10% of 52,500 is 5,250.Similarly, Building B: original 70,000 becomes 73,500. 20% of 73,500 is 14,700.Building C: original 80,000 becomes 84,000. 25% of 84,000 is 21,000.Then, total savings would be 5,250 + 14,700 + 21,000 = 40,950 per month.Total cost after savings: 210,000 - 40,950 = 169,050 per month.Annual cost: 169,050 * 12 = 2,028,600.Same as before.Wait, so regardless of the order, the result is the same? Because multiplication is commutative.Yes, because (A * B) * C = A * (B * C). So, whether you apply the price increase first or the savings first, the result is the same.So, the adjusted total cost is 2,028,600, which is a 15.475% reduction from the original 2,400,000.Therefore, the target is still met.Wait, but let me confirm with exact numbers.Original total annual cost: 2,400,000After savings and price increase: 2,028,600Difference: 2,400,000 - 2,028,600 = 371,400Percentage reduction: (371,400 / 2,400,000) * 100 = 15.475%So, approximately 15.48%, which is just over 15%. Therefore, the target is still achieved.So, summarizing:1. The energy-saving measures alone reduce the cost by 19.5%, which exceeds the 15% target.2. After a 5% energy price increase, the total cost reduction is approximately 15.48%, which still meets the 15% target.Therefore, both scenarios meet the target.</think>

🧠:<think>Alright, so I'm trying to help this indie filmmaker optimize their shooting schedule and budget for their documentary. It's about urban life and art, and they're filming in five different neighborhoods. They have a budget of 50,000 and 30 days to complete the filming. Let me start with the first problem. They need to decide how many days to allocate to each neighborhood to minimize the total cost. Each neighborhood has a cost per day, C_i, and a minimum number of days, D_i, required to capture its essence. The total days can't exceed 30. So, the goal is to find the number of days, d_i, for each neighborhood that minimizes the total cost while meeting the constraints.Hmm, okay, so this sounds like a linear programming problem. The objective function is the total cost, which is the sum of each neighborhood's cost per day multiplied by the number of days spent there. The constraints are that the total days can't exceed 30, and each neighborhood must have at least D_i days. Also, since you can't have negative days, each d_i has to be at least D_i.But wait, the problem mentions that d_i must be greater than or equal to D_i, so we don't need to worry about them being negative because D_i is a minimum. So, the constraints are:1. d_1 + d_2 + d_3 + d_4 + d_5 ≤ 302. d_i ≥ D_i for all iAnd the objective is to minimize the total cost, which is Σ (C_i * d_i).To solve this, I can set up the problem with variables d_1 to d_5, each with their lower bounds D_i. Then, I can use a linear programming solver to find the values of d_i that minimize the cost while satisfying the constraints.But wait, do we have specific values for C_i and D_i? The problem doesn't provide them, so maybe I need to assume some or perhaps it's a general formulation. Since the problem says "formulate and solve," I think I need to present the general form and then perhaps provide an example with hypothetical numbers.Let me think. If I had specific C_i and D_i, I could plug them into the model. For example, suppose:Neighborhood 1: C1 = 1000/day, D1 = 3 daysNeighborhood 2: C2 = 1500/day, D2 = 4 daysNeighborhood 3: C3 = 1200/day, D3 = 2 daysNeighborhood 4: C4 = 2000/day, D4 = 5 daysNeighborhood 5: C5 = 800/day, D5 = 1 dayThen, the minimum total days required would be 3 + 4 + 2 + 5 + 1 = 15 days. Since 15 ≤ 30, we have 15 extra days to allocate. To minimize cost, we should allocate the extra days to the neighborhood with the lowest cost per day, which is Neighborhood 5 at 800/day.So, we can add all 15 extra days to Neighborhood 5, making d5 = 16 days. The other neighborhoods would have their minimum days: d1=3, d2=4, d3=2, d4=5, d5=16. Total days = 30.Total cost would be 3*1000 + 4*1500 + 2*1200 + 5*2000 + 16*800 = 3000 + 6000 + 2400 + 10000 + 12800 = 34,200.But wait, is this the minimal cost? Yes, because we allocated extra days to the cheapest option.But in reality, the filmmaker might have different C_i and D_i. So, the general approach is:1. Calculate the sum of all D_i. If it's less than 30, allocate the remaining days to the neighborhoods with the lowest C_i to minimize cost.2. If the sum of D_i exceeds 30, it's impossible, but the problem states they have 30 days, so I assume sum D_i ≤ 30.So, the formulation is:Minimize Σ (C_i * d_i) for i=1 to 5Subject to:Σ d_i ≤ 30d_i ≥ D_i for all iAnd d_i are integers, I suppose, since you can't film a fraction of a day.But in linear programming, we can relax the integer constraint for simplicity, and then round up if necessary.Now, moving on to the second problem. After allocating days, the filmmaker wants to maximize the diversity score S, which is the sum over neighborhoods and time periods of w_{i,j} * h_{i,j}. Each neighborhood must have at least 2 hours in each time period (morning, afternoon, evening). The total filming time in each neighborhood can't exceed allocated days * 8 hours.So, for each neighborhood i, total hours h_i = h_{i,1} + h_{i,2} + h_{i,3} ≤ 8 * d_i.And for each i, h_{i,j} ≥ 2 for j=1,2,3.The objective is to maximize S = Σ_{i=1}^5 Σ_{j=1}^3 w_{i,j} h_{i,j}.This is another linear programming problem, with variables h_{i,j}.Constraints:1. For each i: h_{i,1} + h_{i,2} + h_{i,3} ≤ 8 * d_i2. For each i, j: h_{i,j} ≥ 23. h_{i,j} ≥ 0 (though the second constraint already covers this)Objective: Maximize Σ w_{i,j} h_{i,j}To solve this, we can use a linear programming approach. Again, without specific w_{i,j} and d_i, it's hard to solve numerically, but we can explain the method.Assuming we have the weights w_{i,j}, we can set up the problem. For each neighborhood, we need to distribute the available hours (8*d_i) across the three time periods, with each period getting at least 2 hours. To maximize the diversity score, we should allocate more hours to the time periods with higher weights in each neighborhood.For example, if in neighborhood 1, the weights are highest for the morning, then we should allocate as many hours as possible to morning, then afternoon, then evening, while respecting the 2-hour minimum and the total hours constraint.So, the steps would be:1. For each neighborhood i, calculate the total available hours: 8 * d_i.2. Subtract the minimum required hours: 3 * 2 = 6 hours.3. The remaining hours: 8*d_i - 6.4. Allocate these remaining hours to the time periods with the highest weights in that neighborhood.This way, we maximize the contribution to the diversity score.But again, without specific numbers, it's a general approach. If we had specific weights, we could compute the exact allocation.So, in summary, for the first problem, it's a linear program to minimize cost with day allocation, and for the second, another linear program to maximize diversity score with hour allocation, given the day allocations from the first problem.</think>

🧠:<think>Alright, so I'm trying to figure out this polynomial that models the Oregon Ducks' basketball performance over five seasons. The problem says it's a degree 4 polynomial, which is interesting because usually, when you have n data points, you can fit a polynomial of degree n-1. Here, we have 5 points, so a 4th-degree polynomial should fit perfectly. That makes sense.First, let me write down the given data points:1. P(1) = 21002. P(2) = 22503. P(3) = 24004. P(4) = 25505. P(5) = 2700Hmm, looking at these numbers, they seem to be increasing by 150 each season. Let me check:2250 - 2100 = 1502400 - 2250 = 1502550 - 2400 = 1502700 - 2550 = 150So, each season, the points increase by 150. That seems linear. Wait, but the polynomial is supposed to be degree 4. If the points increase linearly, wouldn't a linear polynomial (degree 1) suffice? Maybe there's something more here.But the problem says it's a degree 4 polynomial, so perhaps the differences are constant, but higher-order differences might not be. Let me think about finite differences.In a polynomial of degree n, the nth differences are constant. So for a 4th-degree polynomial, the 4th differences should be constant. Let me compute the finite differences for the given data.First, let's list the points:x | P(x)---|---1 | 21002 | 22503 | 24004 | 25505 | 2700First differences (ΔP):2250 - 2100 = 1502400 - 2250 = 1502550 - 2400 = 1502700 - 2550 = 150So, first differences are all 150. That suggests a linear polynomial, but since the problem states it's a 4th-degree, maybe the higher differences are zero? Let's check second differences:Δ²P = ΔP[i+1] - ΔP[i]But since all ΔP are 150, the second differences would be 0.Similarly, third differences would also be 0, and fourth differences would be 0.Wait, so if all higher differences beyond the first are zero, that would mean the polynomial is actually linear, not degree 4. But the problem says it's degree 4. That seems contradictory.Is there a mistake here? Or maybe the polynomial isn't just a straight line but has higher-degree terms that cancel out in the given data points?Alternatively, perhaps the data points lie on a linear polynomial, but the model is a higher-degree polynomial that also passes through these points. That is, the polynomial is not unique unless we have more constraints.But in that case, how do we determine the specific 4th-degree polynomial? Because without more points or constraints, there are infinitely many 4th-degree polynomials passing through these five points.Wait, but the problem says "the polynomial P(x)" implying that it's uniquely determined. So maybe I need to think differently.Alternatively, perhaps the points are on a linear polynomial, and the 4th-degree polynomial is constructed in a way that it coincides with the linear one at these points but differs elsewhere. But without more information, how can we find it?Wait a second, maybe I can use the method of finite differences to construct the polynomial. Since the first differences are constant, that would suggest a linear polynomial, but since it's supposed to be degree 4, maybe the higher differences are zero, which would still make it linear.But that seems conflicting. Maybe the problem is designed such that the polynomial is linear, but for some reason, it's given as degree 4. Maybe I need to proceed as if it's a 4th-degree polynomial and find its coefficients.Let me set up the general form of a 4th-degree polynomial:P(x) = ax⁴ + bx³ + cx² + dx + eWe have five points, so we can set up five equations:1. P(1) = a(1)^4 + b(1)^3 + c(1)^2 + d(1) + e = a + b + c + d + e = 21002. P(2) = a(16) + b(8) + c(4) + d(2) + e = 16a + 8b + 4c + 2d + e = 22503. P(3) = a(81) + b(27) + c(9) + d(3) + e = 81a + 27b + 9c + 3d + e = 24004. P(4) = a(256) + b(64) + c(16) + d(4) + e = 256a + 64b + 16c + 4d + e = 25505. P(5) = a(625) + b(125) + c(25) + d(5) + e = 625a + 125b + 25c + 5d + e = 2700So, we have a system of five equations:1. a + b + c + d + e = 21002. 16a + 8b + 4c + 2d + e = 22503. 81a + 27b + 9c + 3d + e = 24004. 256a + 64b + 16c + 4d + e = 25505. 625a + 125b + 25c + 5d + e = 2700Now, I need to solve this system for a, b, c, d, e.This seems a bit tedious, but let's proceed step by step.First, let's subtract equation 1 from equation 2:Equation 2 - Equation 1:(16a - a) + (8b - b) + (4c - c) + (2d - d) + (e - e) = 2250 - 210015a + 7b + 3c + d = 150 --> Let's call this Equation 6.Similarly, subtract equation 2 from equation 3:Equation 3 - Equation 2:(81a - 16a) + (27b - 8b) + (9c - 4c) + (3d - 2d) + (e - e) = 2400 - 225065a + 19b + 5c + d = 150 --> Equation 7.Subtract equation 3 from equation 4:Equation 4 - Equation 3:(256a - 81a) + (64b - 27b) + (16c - 9c) + (4d - 3d) + (e - e) = 2550 - 2400175a + 37b + 7c + d = 150 --> Equation 8.Subtract equation 4 from equation 5:Equation 5 - Equation 4:(625a - 256a) + (125b - 64b) + (25c - 16c) + (5d - 4d) + (e - e) = 2700 - 2550369a + 61b + 9c + d = 150 --> Equation 9.Now, we have four new equations:6. 15a + 7b + 3c + d = 1507. 65a + 19b + 5c + d = 1508. 175a + 37b + 7c + d = 1509. 369a + 61b + 9c + d = 150Now, let's subtract equation 6 from equation 7:Equation 7 - Equation 6:(65a - 15a) + (19b - 7b) + (5c - 3c) + (d - d) = 150 - 15050a + 12b + 2c = 0 --> Equation 10.Similarly, subtract equation 7 from equation 8:Equation 8 - Equation 7:(175a - 65a) + (37b - 19b) + (7c - 5c) + (d - d) = 150 - 150110a + 18b + 2c = 0 --> Equation 11.Subtract equation 8 from equation 9:Equation 9 - Equation 8:(369a - 175a) + (61b - 37b) + (9c - 7c) + (d - d) = 150 - 150194a + 24b + 2c = 0 --> Equation 12.Now, we have three equations:10. 50a + 12b + 2c = 011. 110a + 18b + 2c = 012. 194a + 24b + 2c = 0Let's subtract equation 10 from equation 11:Equation 11 - Equation 10:(110a - 50a) + (18b - 12b) + (2c - 2c) = 0 - 060a + 6b = 0 --> Simplify by dividing by 6: 10a + b = 0 --> Equation 13.Similarly, subtract equation 11 from equation 12:Equation 12 - Equation 11:(194a - 110a) + (24b - 18b) + (2c - 2c) = 0 - 084a + 6b = 0 --> Simplify by dividing by 6: 14a + b = 0 --> Equation 14.Now, we have:13. 10a + b = 014. 14a + b = 0Subtract equation 13 from equation 14:(14a - 10a) + (b - b) = 0 - 04a = 0 --> So, a = 0.Plugging a = 0 into equation 13:10(0) + b = 0 --> b = 0.Now, with a = 0 and b = 0, let's go back to equation 10:50(0) + 12(0) + 2c = 0 --> 2c = 0 --> c = 0.So, a = b = c = 0.Now, let's go back to equation 6:15a + 7b + 3c + d = 150Plugging a = b = c = 0:0 + 0 + 0 + d = 150 --> d = 150.Now, go back to equation 1:a + b + c + d + e = 2100Plugging a = b = c = 0, d = 150:0 + 0 + 0 + 150 + e = 2100 --> e = 2100 - 150 = 1950.So, the polynomial is:P(x) = 0x⁴ + 0x³ + 0x² + 150x + 1950Simplify:P(x) = 150x + 1950.Wait, that's a linear polynomial, which makes sense because the points are linear. But the problem stated it's a 4th-degree polynomial. So, is there a mistake?Wait, but when we solved the system, we found that a, b, c are zero, which reduces the polynomial to a linear one. So, in this case, the 4th-degree polynomial that fits the data is actually a linear polynomial. That's because the data is perfectly linear, so the higher-degree coefficients are zero.Therefore, the polynomial is P(x) = 150x + 1950.Now, for the second part, predicting the 6th season:P(6) = 150*6 + 1950 = 900 + 1950 = 2850.So, the predicted points for the 6th season would be 2850.But wait, let me double-check the calculations because it's surprising that the higher coefficients are zero.Let me verify with the given points:For x=1: 150*1 + 1950 = 2100 ✔️x=2: 150*2 + 1950 = 2250 ✔️x=3: 150*3 + 1950 = 2400 ✔️x=4: 150*4 + 1950 = 2550 ✔️x=5: 150*5 + 1950 = 2700 ✔️Yes, all points fit the linear model. So, even though the problem says it's a 4th-degree polynomial, the data is linear, so the higher coefficients are zero.Therefore, the polynomial is linear, and the prediction for x=6 is 2850.</think>

🧠:<think>Okay, so I have this problem about a food and beverage vendor catering to a music festival. The festival lasts for 3 days, and the number of attendees each day follows a sinusoidal pattern. The equation given is:[ A(t) = 5000 + 3000 sinleft(frac{pi}{6}(t - 4)right) ]where ( t ) is the time in hours since the gates open each day, ranging from 0 to 12. The first part asks for the total number of attendee-hours over the entire 3-day festival. An attendee-hour is one attendee present for one hour. So, I think this means I need to calculate the integral of the number of attendees over each day and then multiply by 3 since it's a 3-day festival.Let me break it down. For each day, the number of attendees at time ( t ) is given by ( A(t) ). To find the total attendee-hours per day, I need to integrate ( A(t) ) from ( t = 0 ) to ( t = 12 ). Then, multiply that result by 3 for the three days.So, the formula for total attendee-hours would be:[ text{Total Attendee-Hours} = 3 times int_{0}^{12} A(t) , dt ]Substituting ( A(t) ):[ 3 times int_{0}^{12} left(5000 + 3000 sinleft(frac{pi}{6}(t - 4)right)right) dt ]I can split this integral into two parts:[ 3 times left( int_{0}^{12} 5000 , dt + int_{0}^{12} 3000 sinleft(frac{pi}{6}(t - 4)right) dt right) ]Calculating the first integral:[ int_{0}^{12} 5000 , dt = 5000 times (12 - 0) = 5000 times 12 = 60,000 ]So, that part is straightforward. Now, the second integral:[ int_{0}^{12} 3000 sinleft(frac{pi}{6}(t - 4)right) dt ]Let me make a substitution to simplify this integral. Let ( u = frac{pi}{6}(t - 4) ). Then, ( du = frac{pi}{6} dt ), which means ( dt = frac{6}{pi} du ).Changing the limits of integration accordingly. When ( t = 0 ):[ u = frac{pi}{6}(0 - 4) = frac{pi}{6}(-4) = -frac{2pi}{3} ]When ( t = 12 ):[ u = frac{pi}{6}(12 - 4) = frac{pi}{6}(8) = frac{4pi}{3} ]So, substituting into the integral:[ 3000 times int_{-2pi/3}^{4pi/3} sin(u) times frac{6}{pi} du ]Simplify the constants:[ 3000 times frac{6}{pi} times int_{-2pi/3}^{4pi/3} sin(u) du ]Calculating the integral of ( sin(u) ):[ int sin(u) du = -cos(u) + C ]So, evaluating from ( -2pi/3 ) to ( 4pi/3 ):[ -cos(4pi/3) + cos(-2pi/3) ]We know that ( cos(-theta) = cos(theta) ), so ( cos(-2pi/3) = cos(2pi/3) ).Calculating each term:( cos(4pi/3) ) is ( cos(pi + pi/3) = -cos(pi/3) = -0.5 )( cos(2pi/3) ) is ( cos(pi - pi/3) = -cos(pi/3) = -0.5 )So, substituting back:[ -(-0.5) + (-0.5) = 0.5 - 0.5 = 0 ]Wait, that can't be right. If the integral of the sine function over this interval is zero, that would mean the average number of attendees is just the constant term, 5000, over each day. Hmm, let me double-check my calculations.Wait, ( cos(4pi/3) = -0.5 ) and ( cos(-2pi/3) = cos(2pi/3) = -0.5 ). So:[ -cos(4pi/3) + cos(-2pi/3) = -(-0.5) + (-0.5) = 0.5 - 0.5 = 0 ]So, yes, the integral of the sine function over this interval is zero. That makes sense because the sine function is symmetric over its period, and the interval from ( -2pi/3 ) to ( 4pi/3 ) is exactly one full period of the sine function. Therefore, the positive and negative areas cancel out, resulting in zero.Therefore, the second integral is zero. So, the total attendee-hours per day is just 60,000. Therefore, over 3 days, it would be:[ 3 times 60,000 = 180,000 ]Wait, so the total attendee-hours is 180,000. That seems straightforward, but let me make sure I didn't make a mistake in the substitution.Alternatively, maybe I can think about the integral of the sine function over its period. Since the function ( A(t) ) is sinusoidal with a period. Let me check the period of the sine function in the equation.The general sine function is ( sin(B(t - C)) ), where the period is ( 2pi / B ). In this case, ( B = pi/6 ), so the period is ( 2pi / (pi/6) = 12 ) hours. So, the period is exactly 12 hours, which is the duration of each day. So, integrating over one full period, the integral of the sine function is zero, which confirms our earlier result.Therefore, the total attendee-hours is indeed 3 * 60,000 = 180,000.Moving on to the second part: determining the minimum value of supplies needed. Each attendee consumes 0.75 worth of supplies per hour. So, the total cost would be the total attendee-hours multiplied by 0.75.So, total supplies needed:[ 180,000 times 0.75 = 135,000 ]So, the vendor should prepare 135,000 worth of supplies.Wait, but let me think again. Is the minimum value based on the peak demand or just the total? The question says "to meet demand", so I think it's referring to the total consumption over the entire festival, not the peak hour. Because if it were peak hour, we would need to find the maximum number of attendees and multiply by 0.75 and then by 3 days or something. But the way it's phrased, "to cover the 3-day festival", so I think it's the total.But just to make sure, let me check. The first part is total attendee-hours, which is 180,000. Then, each attendee-hour consumes 0.75, so total supplies needed would be 180,000 * 0.75 = 135,000.Alternatively, if they were asking for peak demand, we would have to find the maximum number of attendees in any given hour, multiply by 0.75, and then perhaps by the number of hours or something. But the question says "to meet demand", and since they have to prepare enough supplies for the entire festival, it's more about the total consumption rather than the peak. So, I think 135,000 is correct.But just to be thorough, let me check what the maximum number of attendees is. The function is ( A(t) = 5000 + 3000 sin(...) ). The sine function varies between -1 and 1, so the maximum number of attendees is 5000 + 3000*1 = 8000, and the minimum is 5000 - 3000 = 2000.So, the peak hour has 8000 attendees. If we were to calculate the supplies needed for the peak hour, it would be 8000 * 0.75 = 6000 per hour. But since the festival is 12 hours each day, and 3 days, if they were to prepare for the peak, they would need 6000 * 12 * 3 = 216,000. But that seems like overkill because the total consumption is only 135,000. So, the question is asking for the minimum value needed to cover the festival, which is the total consumption, not the peak. So, 135,000 is the correct answer.Therefore, my answers are:1. 180,000 attendee-hours.2. 135,000 worth of supplies.Final Answer1. The total number of attendee-hours over the entire 3-day festival is boxed{180000}.2. The minimum value of supplies needed is boxed{135000} dollars.</think>