🧠:<think>Alright, so I've got this problem about designing a luxury home with rare antiques. It's divided into two parts. Let me try to wrap my head around each part step by step.Starting with the first part: the professional wants to select a subset of antiques for a specific room. The goal is to maximize the aesthetic appeal while staying within a budget. Specifically, the total aesthetic score needs to be at least 400, and the total market value shouldn't exceed 300,000. Each antique has a unique aesthetic score ( a_i ) and market value ( v_i ), with ( i ) ranging from 1 to 10.Hmm, okay. So, this sounds like an optimization problem where we need to choose which antiques to include. Since we're dealing with yes/no decisions (include or exclude each antique), this is a binary integer programming problem. Let me recall the structure of such problems.In integer linear programming, we define decision variables, an objective function, and constraints. So, for each antique ( i ), we can define a binary variable ( x_i ) where ( x_i = 1 ) if we select the antique, and ( x_i = 0 ) otherwise.The objective is to maximize the total aesthetic score. So, the objective function would be the sum of ( a_i x_i ) for all ( i ) from 1 to 10. That makes sense because we want the highest possible aesthetic score.But we have constraints. The first constraint is that the total aesthetic score must be at least 400. So, the sum of ( a_i x_i ) should be greater than or equal to 400. The second constraint is about the budget: the total market value of the selected antiques shouldn't exceed 300,000. Since the market values ( v_i ) are in thousands of dollars, the sum of ( v_i x_i ) should be less than or equal to 300.Additionally, all ( x_i ) must be binary variables, either 0 or 1. So, putting it all together, the integer linear programming formulation would look something like this:Maximize ( sum_{i=1}^{10} a_i x_i )Subject to:1. ( sum_{i=1}^{10} a_i x_i geq 400 )2. ( sum_{i=1}^{10} v_i x_i leq 300 )3. ( x_i in {0, 1} ) for all ( i = 1, 2, ldots, 10 )Wait, but hold on. The objective is to maximize aesthetic appeal, but the first constraint is that the aesthetic score must be at least 400. So, actually, the objective is to maximize the aesthetic score, but it must be at least 400. So, the constraint is necessary to ensure that the selected antiques meet the minimum aesthetic requirement. The budget constraint is also necessary to ensure we don't exceed the investment limit.Is there anything else I need to consider? Maybe the number of antiques selected? The problem doesn't specify a limit on the number, just the aesthetic and budget constraints. So, I think the formulation is correct as is.Moving on to the second part: after selecting the antiques for the specific room, the remaining antiques need to be evenly distributed into 3 different sections of the house. Each section should have roughly the same total aesthetic score. So, this is a problem of partitioning the remaining antiques into three groups with balanced aesthetic scores.This sounds like a combinatorial optimization problem, specifically a partitioning problem. The goal is to divide the remaining antiques into three subsets such that the total aesthetic score of each subset is as equal as possible.Let me think about how to model this. Let's denote the remaining antiques as ( R ), which is the set of antiques not selected in the first part. So, ( R = { i | x_i = 0 } ). For each antique in ( R ), we need to assign it to one of the three sections, say section 1, 2, or 3.Let me define another set of decision variables. For each antique ( i ) in ( R ), let ( y_{i,j} ) be a binary variable where ( y_{i,j} = 1 ) if antique ( i ) is assigned to section ( j ), and 0 otherwise. Here, ( j ) can be 1, 2, or 3.The objective is to minimize the difference between the maximum and minimum total aesthetic scores across the three sections. Alternatively, we can aim to have each section's total aesthetic score as close as possible to the average.First, let's calculate the total aesthetic score of the remaining antiques. Let ( A = sum_{i in R} a_i ). Then, the target for each section would be ( A / 3 ). However, since we can't have fractions of aesthetic scores, we aim for each section to be as close as possible to this target.So, the problem becomes assigning each antique in ( R ) to one of the three sections such that the total aesthetic score of each section is balanced.To formulate this, we can define variables for the total aesthetic score of each section:( S_j = sum_{i in R} a_i y_{i,j} ) for ( j = 1, 2, 3 )Our goal is to minimize the maximum deviation from the target ( A / 3 ). Alternatively, we can aim to minimize the maximum ( S_j ) while ensuring that the minimum ( S_j ) is as high as possible.But in combinatorial optimization, it's often easier to set up constraints that enforce the balance. One approach is to ensure that the difference between any two sections is minimized. However, this can be tricky because it's a non-linear objective.Alternatively, we can set up constraints that each section's total aesthetic score is at least ( lfloor A / 3 rfloor ) and at most ( lceil A / 3 rceil ). But this might not always be possible, especially if ( A ) isn't perfectly divisible by 3.Another approach is to use an objective function that minimizes the variance or the maximum difference between sections. However, variance can be complex to model in integer programming.Perhaps a more straightforward way is to minimize the maximum ( S_j ) while ensuring that each section has at least ( lfloor A / 3 rfloor ). But I'm not sure if that's the best approach.Wait, maybe we can model it as a multi-objective problem where we try to balance the sections. But since it's combinatorial, perhaps we can use a constraint-based approach.Let me think. For each section, the total aesthetic score should be as close as possible to ( A / 3 ). So, we can define variables for each section's total and then set constraints that each section's total is within a certain range.But without knowing the exact values, it's hard to set precise constraints. Alternatively, we can use a minimax approach, where we minimize the maximum deviation from the target.Let me try to define the problem formally.Let ( R ) be the set of remaining antiques after selecting for the specific room.Define ( y_{i,j} ) as before.Define ( S_j = sum_{i in R} a_i y_{i,j} ) for ( j = 1, 2, 3 ).We need to assign each ( i in R ) to exactly one ( j ), so:For each ( i in R ), ( sum_{j=1}^{3} y_{i,j} = 1 )Also, for each ( j ), ( S_j geq L ) and ( S_j leq U ), where ( L ) and ( U ) are lower and upper bounds on the aesthetic scores of each section.But since we want them to be as equal as possible, we can set ( L = lfloor A / 3 rfloor ) and ( U = lceil A / 3 rceil ). However, this might not always be feasible, especially if the total ( A ) isn't divisible by 3 or if the individual antiques have scores that make balancing difficult.Alternatively, we can define a variable ( D ) representing the maximum difference between any two sections, and aim to minimize ( D ). So, the constraints would be:For all ( j, k ), ( |S_j - S_k| leq D )But this is non-linear because of the absolute value. To linearize it, we can write:For all ( j, k ), ( S_j - S_k leq D ) and ( S_k - S_j leq D )But this adds a lot of constraints, especially since ( j ) and ( k ) range over 3 sections, leading to 6 constraints.Alternatively, we can set ( D ) as the maximum deviation from the target ( A / 3 ). So, for each section ( j ):( S_j geq A / 3 - D )( S_j leq A / 3 + D )Then, minimize ( D ).This approach might be more manageable.So, putting it all together, the combinatorial optimization problem would be:Minimize ( D )Subject to:1. For each ( i in R ), ( sum_{j=1}^{3} y_{i,j} = 1 )2. For each ( j ), ( sum_{i in R} a_i y_{i,j} geq A / 3 - D )3. For each ( j ), ( sum_{i in R} a_i y_{i,j} leq A / 3 + D )4. ( y_{i,j} in {0, 1} ) for all ( i in R ), ( j = 1, 2, 3 )5. ( D geq 0 )This way, we're minimizing the maximum deviation ( D ) from the target aesthetic score per section.But wait, since ( A ) is the total aesthetic score of the remaining antiques, ( A ) is fixed once we've selected the antiques for the specific room. So, ( A ) is known, and we can compute ( A / 3 ) as the target.However, since ( A ) might not be an integer, and the aesthetic scores ( a_i ) are integers, the sections might not perfectly balance. Hence, ( D ) will account for the minimal possible imbalance.Alternatively, if we don't want to introduce a new variable ( D ), we could set up constraints that each section's total is within a certain range, but that might not be as flexible.Another thought: sometimes, in partitioning problems, people use the concept of "bin packing," where the goal is to distribute items into bins with capacities. Here, it's similar but instead of capacities, we're trying to balance the totals.But in this case, we don't have capacities; we have a target for each bin (section). So, it's more like a "multi-way number partitioning" problem, which is known to be NP-hard. So, exact solutions might be computationally intensive, but for 10 antiques, it's manageable.Wait, but in the first part, we're selecting a subset of 10 antiques, so the remaining could be up to 10 as well, but depending on how many are selected in the first part. But since the first part is a selection, the remaining could be from 0 to 10, but likely, since we need to have a total aesthetic score of at least 400, the remaining might have a lower total.But regardless, the second part is about distributing the remaining into 3 sections, so the exact number isn't specified, but it's a combinatorial problem.So, to summarize, the second part is a multi-way partitioning problem where we need to divide the remaining antiques into 3 sections with balanced aesthetic scores. The constraints are that each antique is assigned to exactly one section, and the total aesthetic scores of the sections are as equal as possible.Therefore, the formulation would involve decision variables for assignment, constraints for each antique being assigned exactly once, and constraints to balance the totals, possibly with an objective to minimize the maximum difference or maximize the minimum total.I think the key constraints are:1. Each antique is assigned to exactly one section.2. The total aesthetic score of each section is as close as possible to ( A / 3 ).So, the combinatorial optimization problem can be formulated with the objective of minimizing the maximum deviation from the target aesthetic score per section, subject to the assignment constraints.I think that's about it. Let me just recap:1. First part: Integer linear program to select antiques for the room, maximizing aesthetic score while meeting the minimum aesthetic and budget constraints.2. Second part: Combinatorial optimization to distribute remaining antiques into 3 sections with balanced aesthetic scores, ensuring each section is as close as possible in total aesthetic value.I think I've covered the necessary formulations and constraints for both parts. Hopefully, that makes sense!</think>

🧠:<think>First, I'll tackle Sub-problem 1. The sushi platter has 5 types of sushi, each with a fish to rice to seaweed ratio of 3:2:1. The chef has 3 kg of fish. I need to determine the total amounts of rice and seaweed required.Since the ratio is 3:2:1, for every 3 parts of fish, there are 2 parts of rice and 1 part of seaweed. The total number of parts is 3 + 2 + 1 = 6 parts.Given that the chef has 3 kg of fish, which corresponds to 3 parts, each part is equal to 1 kg. Therefore, rice is 2 parts × 1 kg/part = 2 kg, and seaweed is 1 part × 1 kg/part = 1 kg.Since there are 5 types of sushi, the total amounts needed are:- Fish: 3 kg × 5 = 15 kg- Rice: 2 kg × 5 = 10 kg- Seaweed: 1 kg × 5 = 5 kgNext, for Sub-problem 2, the pasta sauce combines miso and tomato sauce in a 2:3 ratio, totaling 2 liters. Additionally, there's a 10% special ingredient blend of ginger, garlic, and basil in the ratio 1:1:2.First, calculate the total sauce without the special blend:- Miso: (2/5) × 2 liters = 0.8 liters- Tomato sauce: (3/5) × 2 liters = 1.2 litersThe special blend is 10% of 2 liters, which is 0.2 liters. Within this blend, the ratio is 1:1:2, totaling 4 parts.- Ginger: (1/4) × 0.2 liters = 0.05 liters- Garlic: (1/4) × 0.2 liters = 0.05 liters- Basil: (2/4) × 0.2 liters = 0.1 litersFinally, adjust the main sauce components to account for the special blend:- Miso: 0.8 liters - 0.05 liters = 0.75 liters- Tomato sauce: 1.2 liters - 0.05 liters = 1.15 litersSo, the chef should use:- Miso: 0.75 liters- Tomato sauce: 1.15 liters- Ginger: 0.05 liters- Garlic: 0.05 liters- Basil: 0.1 liters</think>

🧠:<think>Okay, so I need to perform a linear regression analysis on the given data points to find the best-fit line, which is y = mx + b. Then, using that equation, I have to predict the client satisfaction rating when 30 tickets are resolved. Hmm, let me think about how to approach this step by step.First, I remember that linear regression involves finding the line that best fits the data points by minimizing the sum of the squares of the vertical distances between the observed points and the regression line. To do this, I need to calculate the slope (m) and the y-intercept (b) of the line.The formula for the slope m is:m = (n * Σ(xy) - Σx * Σy) / (n * Σx² - (Σx)²)And the formula for the y-intercept b is:b = (Σy - m * Σx) / nWhere n is the number of data points.Looking at the data points provided:(12, 7.5), (15, 8.0), (10, 6.5), (20, 9.0), (18, 8.5), (25, 9.5)So, n = 6.I think I need to compute several sums: Σx, Σy, Σxy, and Σx².Let me list out the x and y values separately:x: 12, 15, 10, 20, 18, 25y: 7.5, 8.0, 6.5, 9.0, 8.5, 9.5First, let me calculate Σx:12 + 15 + 10 + 20 + 18 + 2512 + 15 is 27, plus 10 is 37, plus 20 is 57, plus 18 is 75, plus 25 is 100. So Σx = 100.Next, Σy:7.5 + 8.0 + 6.5 + 9.0 + 8.5 + 9.57.5 + 8.0 is 15.5, plus 6.5 is 22, plus 9.0 is 31, plus 8.5 is 39.5, plus 9.5 is 49. So Σy = 49.Now, Σxy. I need to multiply each x by its corresponding y and sum them up.Let me compute each xy:12 * 7.5 = 9015 * 8.0 = 12010 * 6.5 = 6520 * 9.0 = 18018 * 8.5 = 15325 * 9.5 = 237.5Now, adding these up:90 + 120 = 210210 + 65 = 275275 + 180 = 455455 + 153 = 608608 + 237.5 = 845.5So Σxy = 845.5Next, Σx². I need to square each x and sum them up.12² = 14415² = 22510² = 10020² = 40018² = 32425² = 625Adding these up:144 + 225 = 369369 + 100 = 469469 + 400 = 869869 + 324 = 11931193 + 625 = 1818So Σx² = 1818Now, plugging these into the formula for m:m = (n * Σxy - Σx * Σy) / (n * Σx² - (Σx)²)Plugging in the numbers:n = 6, Σxy = 845.5, Σx = 100, Σy = 49, Σx² = 1818So numerator = 6 * 845.5 - 100 * 49Let me compute 6 * 845.5 first:845.5 * 6: 800*6=4800, 45.5*6=273, so total is 4800 + 273 = 5073Then, 100 * 49 = 4900So numerator = 5073 - 4900 = 173Denominator = 6 * 1818 - (100)^2Compute 6 * 1818:1818 * 6: 1000*6=6000, 800*6=4800, 18*6=108, so 6000 + 4800 = 10800 + 108 = 10908Then, (100)^2 = 10000So denominator = 10908 - 10000 = 908Therefore, m = 173 / 908Let me compute that division:173 ÷ 908Well, 908 goes into 173 zero times. So we can write it as 0.1905 approximately.Wait, let me do it more accurately.908 * 0.19 = 908 * 0.1 + 908 * 0.09 = 90.8 + 81.72 = 172.52That's very close to 173. So 0.19 gives 172.52, which is just 0.48 less than 173.So 0.19 + (0.48 / 908) ≈ 0.19 + 0.00053 ≈ 0.19053So m ≈ 0.1905So approximately 0.1905.Now, compute b:b = (Σy - m * Σx) / nΣy = 49, m ≈ 0.1905, Σx = 100, n = 6So compute m * Σx: 0.1905 * 100 = 19.05Then, Σy - m * Σx = 49 - 19.05 = 29.95Now, divide by n: 29.95 / 6 ≈ 4.9917So approximately 4.9917So, rounding off, m ≈ 0.1905 and b ≈ 4.9917Therefore, the regression line is y = 0.1905x + 4.9917But let me check my calculations again because sometimes when dealing with decimal points, it's easy to make a mistake.Wait, let me verify the numerator and denominator again.Numerator: 6 * 845.5 = 5073Σx * Σy = 100 * 49 = 4900So numerator = 5073 - 4900 = 173. Correct.Denominator: 6 * 1818 = 10908(Σx)^2 = 100^2 = 10000Denominator = 10908 - 10000 = 908. Correct.So m = 173 / 908 ≈ 0.1905. Correct.Then, b = (49 - 0.1905 * 100)/6 = (49 - 19.05)/6 = 29.95 /6 ≈ 4.9917. Correct.So, the equation is y ≈ 0.1905x + 4.9917Alternatively, we can write it as y ≈ 0.1905x + 4.99Now, for the second part, predicting y when x = 30.So plug x = 30 into the equation:y = 0.1905 * 30 + 4.9917Compute 0.1905 * 30:0.1905 * 30 = 5.715Then, 5.715 + 4.9917 ≈ 10.7067Wait, but the satisfaction rating is on a scale of 1 to 10. So 10.7067 is above 10. That doesn't make sense because the maximum is 10.Hmm, maybe my calculations are off? Or perhaps the regression line is extrapolating beyond the data range, leading to a prediction beyond the scale.Looking back at the data, the maximum x is 25, and the corresponding y is 9.5. So when x increases beyond 25, the model predicts y beyond 9.5, which is possible, but in reality, satisfaction might not go beyond 10.But since the question asks to predict using the regression line, regardless of the scale, I think we have to go with the calculation.So y ≈ 10.7067, which is approximately 10.71.But since the scale is 1 to 10, maybe we cap it at 10? Or perhaps the model is just predicting beyond the scale.I think in the context of the question, we should just provide the predicted value as per the regression line, even if it exceeds 10.Alternatively, maybe I made a mistake in the calculation.Wait, let me recalculate the prediction:m ≈ 0.1905b ≈ 4.9917So y = 0.1905 * 30 + 4.99170.1905 * 30: 0.1905 * 10 = 1.905, so times 3 is 5.7155.715 + 4.9917 = 10.7067Yes, that's correct.Alternatively, maybe I should use more precise values for m and b instead of rounding early.Let me compute m more precisely.m = 173 / 908Let me compute 173 divided by 908.908 goes into 173 zero times. So 0.Compute 1730 divided by 908: 908 * 1 = 908, subtract from 1730: 1730 - 908 = 822. Bring down a zero: 8220.908 goes into 8220 how many times? 908*9=8172, which is less than 8220. 908*9=8172, subtract: 8220 - 8172=48. Bring down a zero: 480.908 goes into 480 zero times. Bring down another zero: 4800.908*5=4540, subtract: 4800 - 4540=260. Bring down a zero: 2600.908*2=1816, subtract: 2600 - 1816=784. Bring down a zero: 7840.908*8=7264, subtract: 7840 - 7264=576. Bring down a zero: 5760.908*6=5448, subtract: 5760 - 5448=312. Bring down a zero: 3120.908*3=2724, subtract: 3120 - 2724=396. Bring down a zero: 3960.908*4=3632, subtract: 3960 - 3632=328. Bring down a zero: 3280.908*3=2724, subtract: 3280 - 2724=556. Bring down a zero: 5560.908*6=5448, subtract: 5560 - 5448=112. Bring down a zero: 1120.908*1=908, subtract: 1120 - 908=212. Bring down a zero: 2120.908*2=1816, subtract: 2120 - 1816=304. Bring down a zero: 3040.908*3=2724, subtract: 3040 - 2724=316. Bring down a zero: 3160.908*3=2724, subtract: 3160 - 2724=436. Bring down a zero: 4360.908*4=3632, subtract: 4360 - 3632=728. Bring down a zero: 7280.908*8=7264, subtract: 7280 - 7264=16. Bring down a zero: 160.908 goes into 160 zero times. So we can stop here.So putting it all together, m ≈ 0.190534...So approximately 0.1905.Similarly, b was 29.95 / 6 ≈ 4.991666...So b ≈ 4.9917.So, using more precise values:y = 0.190534 * 30 + 4.991666Compute 0.190534 * 30:0.190534 * 30 = 5.71602Then, 5.71602 + 4.991666 ≈ 10.707686So approximately 10.7077, which is about 10.71.So, as per the regression line, the predicted satisfaction rating is approximately 10.71.But since the scale is 1 to 10, maybe we should cap it at 10? Or perhaps the model is just extrapolating beyond the observed data.In the context of the problem, it's a predictive model, and even though the satisfaction rating can't exceed 10, the model doesn't know that. So, strictly speaking, the prediction is 10.71.Alternatively, if we consider that the maximum rating is 10, we might cap it at 10, but the question doesn't specify that. It just asks to predict using the regression line.Therefore, I think the answer is approximately 10.71.But let me double-check all my calculations to make sure I didn't make any errors.First, Σx = 100, correct.Σy = 49, correct.Σxy = 845.5, correct.Σx² = 1818, correct.Numerator: 6*845.5 = 5073; 100*49=4900; 5073 - 4900=173, correct.Denominator: 6*1818=10908; 100²=10000; 10908 - 10000=908, correct.m=173/908≈0.1905, correct.b=(49 - 0.1905*100)/6=(49 -19.05)/6=29.95/6≈4.9917, correct.So, y=0.1905x +4.9917At x=30, y≈0.1905*30 +4.9917≈5.715 +4.9917≈10.7067≈10.71.Yes, all steps seem correct.So, the final answer is approximately 10.71.But since the question might expect rounding to two decimal places, 10.71 is fine.Alternatively, if we want to present it as a whole number, it would be 11, but since the scale is 1-10, 10.71 is more precise.I think 10.71 is acceptable.Final AnswerThe predicted client satisfaction rating when resolving 30 support tickets is boxed{10.71}.</think>

🧠:<think>Okay, so I have this problem about a military social worker who wants to maximize the well-being score (W) of service members and their families by allocating 100 hours across three support programs: Psychological Counseling (PC), Financial Assistance (FA), and Family Support Programs (FSP). The initial model is a linear regression, and then there's a part where we have to consider diminishing returns if any program exceeds 40 hours.Starting with part 1: I need to set up an optimization problem to maximize W given the constraint that the total hours add up to 100. The equation given is:[ W = 2.5 + 1.3 cdot H_{PC} + 0.9 cdot H_{FA} + 1.1 cdot H_{FSP} ]And the constraint is:[ H_{PC} + H_{FA} + H_{FSP} = 100 ]So, since this is a linear model, the coefficients for each H represent the marginal increase in W per hour spent on that program. The higher the coefficient, the more impact each hour has. So, to maximize W, we should allocate as many hours as possible to the program with the highest coefficient.Looking at the coefficients: PC has 1.3, FSP has 1.1, and FA has 0.9. So, PC is the most impactful, followed by FSP, then FA.Therefore, without any constraints, to maximize W, we should put all 100 hours into PC. But wait, is there any other constraint? The problem doesn't mention any minimum hours required for the other programs, so I think it's okay.But let me verify. If we set H_PC = 100, then H_FA and H_FSP would be 0. Plugging into the equation:W = 2.5 + 1.3*100 + 0.9*0 + 1.1*0 = 2.5 + 130 = 132.5Is that the maximum? Let's see if allocating some hours to FSP or FA could give a higher W. Suppose we take one hour from PC and give it to FSP. Then W would decrease by 1.3 and increase by 1.1, so net decrease of 0.2. Similarly, giving it to FA would decrease by 1.3 and increase by 0.9, net decrease of 0.4. So, yes, keeping all hours in PC gives the highest W.Therefore, the optimal allocation is H_PC = 100, H_FA = 0, H_FSP = 0.Now, moving on to part 2: There's a diminishing return if any program exceeds 40 hours. The penalty is 0.05*(H_i - 40)^2 for H_i > 40. So, we need to adjust the model to include this penalty term.So, the new well-being score W becomes:[ W = 2.5 + 1.3 cdot H_{PC} + 0.9 cdot H_{FA} + 1.1 cdot H_{FSP} - 0.05 cdot (H_{PC} - 40)^2 - 0.05 cdot (H_{FA} - 40)^2 - 0.05 cdot (H_{FSP} - 40)^2 ]But only if each H_i exceeds 40. So, if H_i <= 40, the penalty is zero. So, we have to consider that in our optimization.So, the problem now is to maximize W, considering that if any H_i is more than 40, we subtract the penalty. So, the trade-off is between the marginal gain from allocating more hours to a program and the penalty for exceeding 40 hours.This complicates things because now it's a nonlinear optimization problem. Since we have three variables and a quadratic term, it might be a bit tricky.First, let's consider that each program can have a maximum of 40 hours without any penalty. So, if we allocate 40 hours to each, that's 120 hours, but we only have 100. So, we can't allocate 40 to all. So, we have to choose which programs to allocate more than 40 hours, considering the penalty.But wait, actually, the penalty is only applied if H_i exceeds 40. So, for each program, if it's above 40, we have a penalty. So, perhaps it's better to have multiple programs at 40 or below, but since the total is 100, we can have two programs at 40 and the third at 20, but that might not be optimal.Alternatively, maybe it's better to have one program above 40, and the others below, but considering the penalties.So, let's think about the marginal benefit of each additional hour beyond 40. For each program, the marginal benefit is the coefficient minus twice the penalty coefficient times (H_i - 40). Wait, no, the penalty is 0.05*(H_i -40)^2, so the derivative is -0.1*(H_i -40). So, the marginal cost of allocating an extra hour beyond 40 is 0.1*(H_i -40). So, the net marginal benefit is the coefficient minus 0.1*(H_i -40).So, for each program, beyond 40 hours, each additional hour gives a net benefit of (coefficient - 0.1*(H_i -40)). So, we need to find the point where this net benefit is zero or negative.But since we have limited hours, we might need to compare which program gives the highest net benefit when exceeding 40.Alternatively, perhaps it's better to set up the problem with Lagrange multipliers, considering the constraints.But this might get complicated. Let me try to structure it.Let me denote:For each program, if H_i <=40, the contribution is coefficient*H_i.If H_i >40, the contribution is coefficient*H_i - 0.05*(H_i -40)^2.So, the total W is:2.5 + sum over i of [ coefficient_i * H_i - 0.05*(H_i -40)^2 if H_i >40 else coefficient_i * H_i ]So, to model this, perhaps we can consider that for each program, beyond 40 hours, the marginal benefit decreases.So, for each program, the marginal benefit beyond 40 is:coefficient_i - 0.1*(H_i -40)We can set this equal to the opportunity cost, which is the difference in coefficients between programs.But this is getting a bit abstract. Maybe it's better to consider that beyond 40, the effective coefficient decreases.So, for PC, beyond 40, the effective coefficient is 1.3 - 0.1*(H_PC -40). Similarly for FSP: 1.1 - 0.1*(H_FSP -40), and FA: 0.9 - 0.1*(H_FA -40).But since FA has the lowest coefficient, it's unlikely to be allocated beyond 40, because the penalty would make it even worse. So, maybe only PC and FSP can be allocated beyond 40.So, let's consider that we might allocate more than 40 hours to PC and/or FSP, but not to FA.So, let's suppose we allocate H_PC = 40 + x, H_FSP = 40 + y, and H_FA = 100 - (40 +x) - (40 + y) = 20 -x - y.But since H_FA can't be negative, 20 -x - y >=0, so x + y <=20.Also, x and y must be >=0.So, our variables are x and y, with x + y <=20.Now, the total W is:2.5 + 1.3*(40 +x) + 0.9*(20 -x - y) + 1.1*(40 + y) - 0.05*x^2 - 0.05*y^2Let me compute this:First, expand the terms:1.3*(40 +x) = 52 + 1.3x0.9*(20 -x - y) = 18 - 0.9x - 0.9y1.1*(40 + y) = 44 + 1.1ySo, summing these up:52 + 1.3x + 18 - 0.9x - 0.9y + 44 + 1.1yCombine like terms:52 +18 +44 = 1141.3x -0.9x = 0.4x-0.9y +1.1y = 0.2ySo, total from the linear terms: 114 + 0.4x + 0.2yNow, subtract the penalties:-0.05x^2 -0.05y^2So, total W = 2.5 + 114 + 0.4x + 0.2y -0.05x^2 -0.05y^2Simplify:W = 116.5 + 0.4x + 0.2y -0.05x^2 -0.05y^2Now, we need to maximize this function subject to x + y <=20, x >=0, y >=0.This is a quadratic optimization problem with two variables.To find the maximum, we can take partial derivatives with respect to x and y, set them to zero, and solve.Compute partial derivative of W with respect to x:dW/dx = 0.4 - 0.1xSimilarly, partial derivative with respect to y:dW/dy = 0.2 - 0.1ySet these equal to zero:0.4 -0.1x =0 => x=40.2 -0.1y=0 => y=2So, the critical point is at x=4, y=2.Now, check if this point satisfies the constraint x + y <=20. 4+2=6 <=20, so yes.So, the maximum occurs at x=4, y=2.Therefore, the optimal allocation is:H_PC =40 +4=44H_FSP=40 +2=42H_FA=20 -4 -2=14So, let's compute W at this point:W=116.5 +0.4*4 +0.2*2 -0.05*(4)^2 -0.05*(2)^2Compute each term:0.4*4=1.60.2*2=0.40.05*16=0.80.05*4=0.2So,W=116.5 +1.6 +0.4 -0.8 -0.2Compute step by step:116.5 +1.6=118.1118.1 +0.4=118.5118.5 -0.8=117.7117.7 -0.2=117.5So, W=117.5Now, let's check if this is indeed the maximum. We can also check the boundaries.For example, if x=20, y=0:H_PC=60, H_FSP=40, H_FA=0Compute W:W=2.5 +1.3*60 +0.9*0 +1.1*40 -0.05*(60-40)^2 -0.05*(40-40)^2 -0.05*(0-40)^2Wait, but H_FA=0, which is less than 40, so no penalty for FA. Similarly, H_FSP=40, no penalty. H_PC=60, penalty is 0.05*(20)^2=0.05*400=20So,W=2.5 +78 +0 +44 -20=2.5+78+44-20=104.5Which is less than 117.5.Similarly, if x=0, y=20:H_PC=40, H_FSP=60, H_FA=0Penalty for FSP: 0.05*(20)^2=20W=2.5 +1.3*40 +0.9*0 +1.1*60 -20=2.5 +52 +0 +66 -20=100.5Still less than 117.5.What about x=0, y=0:H_PC=40, H_FSP=40, H_FA=20No penalties.W=2.5 +1.3*40 +0.9*20 +1.1*40=2.5 +52 +18 +44=116.5Which is less than 117.5.Another point: x=10, y=10:H_PC=50, H_FSP=50, H_FA=0Penalties: 0.05*(10)^2=5 for each, total penalty=10W=2.5 +1.3*50 +0.9*0 +1.1*50 -10=2.5 +65 +0 +55 -10=112.5Still less than 117.5.So, the maximum seems to be at x=4, y=2, giving W=117.5.Therefore, the optimal allocation is H_PC=44, H_FSP=42, H_FA=14.Wait, but let me double-check the calculations for W.Original expression:W = 116.5 +0.4x +0.2y -0.05x² -0.05y²At x=4, y=2:0.4*4=1.60.2*2=0.40.05*(4)^2=0.80.05*(2)^2=0.2So,116.5 +1.6 +0.4 -0.8 -0.2=116.5 +2 -1=117.5Yes, correct.Alternatively, let's compute W directly from the original equation:W=2.5 +1.3*44 +0.9*14 +1.1*42 -0.05*(44-40)^2 -0.05*(14-40)^2 -0.05*(42-40)^2Wait, but H_FA=14, which is less than 40, so no penalty for FA.So,W=2.5 +1.3*44 +0.9*14 +1.1*42 -0.05*(4)^2 -0.05*(2)^2Compute each term:1.3*44=57.20.9*14=12.61.1*42=46.20.05*16=0.80.05*4=0.2So,W=2.5 +57.2 +12.6 +46.2 -0.8 -0.2Compute step by step:2.5 +57.2=59.759.7 +12.6=72.372.3 +46.2=118.5118.5 -0.8=117.7117.7 -0.2=117.5Yes, same result.Therefore, the optimal allocation is H_PC=44, H_FSP=42, H_FA=14, giving W=117.5.So, to summarize:1. Without penalties, allocate all 100 hours to PC.2. With penalties, allocate 44 to PC, 42 to FSP, and 14 to FA.I think that's the solution.</think>