🧠:<think>First, I need to set up a system of linear equations based on the information provided. The total number of tickets sold should be 200, and the total revenue from these tickets should be 1,200. This gives me two equations:x + y = 2005x + 8y = 1200Next, I'll solve this system to find the values of x and y. I can use the substitution method by expressing y in terms of x from the first equation and then substituting it into the second equation. This will allow me to solve for x first and then find y.After finding the initial solution, I need to incorporate the observed ratio of ticket sales, which is 3 bake sale tickets for every 2 craft workshop tickets. This means that y = (2/3)x. I'll substitute this new relationship into the original equations to find the adjusted number of tickets that meet both the fundraising goal and the observed ratio.</think>

🧠:<think>Okay, so I have this problem about a multi-level parking structure, and I need to figure out the coefficients a, b, and c for the quadratic equation that models the number of parking spaces on each level. Then, I also need to calculate the potential daily revenue based on the given rate and utilization. Let me break this down step by step.First, the problem states that the number of parking spaces on the n-th level is given by S_n = an² + bn + c. There are 5 levels, and the total number of spaces is 1,250. Additionally, we know the number of spaces on the 1st, 3rd, and 5th levels: 150, 250, and 350 respectively.So, let's write down what we know:- For level 1 (n=1): S₁ = a(1)² + b(1) + c = a + b + c = 150- For level 3 (n=3): S₃ = a(3)² + b(3) + c = 9a + 3b + c = 250- For level 5 (n=5): S₅ = a(5)² + b(5) + c = 25a + 5b + c = 350Also, the total number of spaces across all 5 levels is 1,250. So, S₁ + S₂ + S₃ + S₄ + S₅ = 1,250.But we don't know S₂ and S₄. However, since we have a quadratic model, we can express S₂ and S₄ in terms of a, b, and c as well.So, let me write equations for S₂ and S₄:- For level 2 (n=2): S₂ = a(2)² + b(2) + c = 4a + 2b + c- For level 4 (n=4): S₄ = a(4)² + b(4) + c = 16a + 4b + cTherefore, the total number of spaces is:S₁ + S₂ + S₃ + S₄ + S₅ = (a + b + c) + (4a + 2b + c) + (9a + 3b + c) + (16a + 4b + c) + (25a + 5b + c) = 1,250Let me compute the sum of the coefficients for a, b, and c:For a: 1 + 4 + 9 + 16 + 25 = 55aFor b: 1 + 2 + 3 + 4 + 5 = 15bFor c: 1 + 1 + 1 + 1 + 1 = 5cSo, 55a + 15b + 5c = 1,250Now, let's summarize the equations we have:1. a + b + c = 150 (from S₁)2. 9a + 3b + c = 250 (from S₃)3. 25a + 5b + c = 350 (from S₅)4. 55a + 15b + 5c = 1,250 (from the total)Hmm, so we have four equations here, but actually, the fourth equation is dependent on the first three because it's the sum. So, maybe we can use the first three equations to solve for a, b, c, and then verify with the fourth equation.Let me write the first three equations:Equation 1: a + b + c = 150Equation 2: 9a + 3b + c = 250Equation 3: 25a + 5b + c = 350Let me subtract Equation 1 from Equation 2 to eliminate c:Equation 2 - Equation 1: (9a - a) + (3b - b) + (c - c) = 250 - 150Which simplifies to:8a + 2b = 100Let me call this Equation 4: 8a + 2b = 100Similarly, subtract Equation 2 from Equation 3:Equation 3 - Equation 2: (25a - 9a) + (5b - 3b) + (c - c) = 350 - 250Which simplifies to:16a + 2b = 100Let me call this Equation 5: 16a + 2b = 100Now, we have Equation 4: 8a + 2b = 100 and Equation 5: 16a + 2b = 100Subtract Equation 4 from Equation 5:(16a - 8a) + (2b - 2b) = 100 - 100Which simplifies to:8a = 0So, 8a = 0 => a = 0Wait, a is zero? That would mean the quadratic term is zero, so the model is actually linear, not quadratic. Hmm, that's interesting. Let me check my calculations.Equation 2 - Equation 1: 9a + 3b + c - (a + b + c) = 250 - 150Which is 8a + 2b = 100. That seems correct.Equation 3 - Equation 2: 25a + 5b + c - (9a + 3b + c) = 350 - 250Which is 16a + 2b = 100. That also seems correct.Then, subtracting Equation 4 from Equation 5: 16a + 2b - (8a + 2b) = 100 - 100Which is 8a = 0 => a = 0. Hmm, so a is zero.So, if a = 0, then plugging back into Equation 4: 8(0) + 2b = 100 => 2b = 100 => b = 50Then, from Equation 1: a + b + c = 150 => 0 + 50 + c = 150 => c = 100So, the coefficients are a = 0, b = 50, c = 100.Wait, but if a = 0, then the model is linear: S_n = 50n + 100.Let me check whether this satisfies the given conditions.For n=1: 50*1 + 100 = 150. Correct.For n=3: 50*3 + 100 = 150 + 100 = 250. Correct.For n=5: 50*5 + 100 = 250 + 100 = 350. Correct.Now, let's compute the total number of spaces:S₁ = 150S₂ = 50*2 + 100 = 100 + 100 = 200S₃ = 250S₄ = 50*4 + 100 = 200 + 100 = 300S₅ = 350Total = 150 + 200 + 250 + 300 + 350 = Let's compute:150 + 200 = 350350 + 250 = 600600 + 300 = 900900 + 350 = 1,250Yes, that's correct. So, the total is indeed 1,250.Therefore, the coefficients are a = 0, b = 50, c = 100.Wait, but the problem stated that the number of spaces follows a quadratic pattern. If a = 0, it's linear. So, is that acceptable? The problem says "quadratic pattern," but maybe it's just a quadratic function which could be degenerate into linear if a=0. So, perhaps it's acceptable.Alternatively, maybe I made a mistake in the equations.Wait, let me double-check the equations.Equation 1: n=1: a + b + c = 150Equation 2: n=3: 9a + 3b + c = 250Equation 3: n=5: 25a + 5b + c = 350Subtracting Equation 1 from Equation 2: 8a + 2b = 100Subtracting Equation 2 from Equation 3: 16a + 2b = 100Subtracting these two equations: 8a = 0 => a=0So, unless I made a mistake in the initial setup, a is zero. Therefore, the model is linear.So, perhaps the problem allows for a quadratic model which can be linear if a=0.So, moving forward, a=0, b=50, c=100.Now, part 2: Calculate the potential revenue generated per day if the parking rate is 5 per hour and each space is utilized for an average of 8 hours per day.So, first, we need to find the total number of spaces, which is 1,250.Each space is utilized for 8 hours per day, so the total number of hours per day is 1,250 * 8.Then, the revenue is total hours multiplied by 5 per hour.Let me compute that.Total spaces: 1,250Total hours per day: 1,250 * 8 = 10,000 hoursRevenue per day: 10,000 * 5 = 50,000Wait, that seems straightforward.But let me verify:Each space is used for 8 hours, so each space generates 8 * 5 = 40 per day.Total revenue: 1,250 * 40 = 1,250 * 401,250 * 40: 1,250 * 4 = 5,000; 5,000 * 10 = 50,000.Yes, 50,000 per day.So, the potential revenue is 50,000 per day.Wait, but let me think again: is it per space per hour? So, each space is rented out for 5 per hour, and each is used for 8 hours. So, per space revenue is 5*8 = 40 per day.Total revenue: 1,250 * 40 = 50,000.Yes, that's correct.Alternatively, if we think of it as total spaces * average utilization hours * rate.So, 1,250 * 8 * 5 = 1,250 * 40 = 50,000.Either way, same result.Therefore, the coefficients are a=0, b=50, c=100, and the daily revenue is 50,000.Final AnswerThe coefficients are ( a = boxed{0} ), ( b = boxed{50} ), and ( c = boxed{100} ). The potential daily revenue is ( boxed{50000} ) dollars.</think>

🧠:<think>Alright, so I have this problem about portfolio optimization and stochastic calculus. It's divided into two parts. Let me start with the first part, which is about mean-variance optimization. Okay, the portfolio consists of three stocks: A, B, and C. Their expected returns are 7%, 10%, and 5% respectively. The covariance matrix is given, which is a 3x3 matrix. I need to find the allocation weights w_A, w_B, and w_C that minimize the portfolio variance for a target expected return of 8%. The weights must sum to 1, and short-selling is allowed, so weights can be negative.First, I remember that in mean-variance optimization, we're trying to find the portfolio with the minimum variance for a given expected return. This is part of the efficient frontier concept. The general approach involves setting up a Lagrangian with constraints on the expected return and the sum of weights.Let me recall the formula for portfolio variance. It's given by:[sigma_p^2 = w_A^2 sigma_A^2 + w_B^2 sigma_B^2 + w_C^2 sigma_C^2 + 2w_Aw_B sigma_{AB} + 2w_Aw_C sigma_{AC} + 2w_Bw_C sigma_{BC}]But since we have the covariance matrix, it's easier to represent this in matrix form. The covariance matrix Σ is given, so the portfolio variance is:[sigma_p^2 = mathbf{w}^T Sigma mathbf{w}]Where w is the vector of weights [w_A, w_B, w_C]^T.Our goal is to minimize this variance subject to two constraints:1. The expected return of the portfolio equals 8%:[mu_p = w_A mu_A + w_B mu_B + w_C mu_C = 0.08]2. The sum of weights equals 1:[w_A + w_B + w_C = 1]To solve this, I need to use Lagrange multipliers. Let me set up the Lagrangian function:[mathcal{L} = mathbf{w}^T Sigma mathbf{w} - lambda_1 (w_A mu_A + w_B mu_B + w_C mu_C - 0.08) - lambda_2 (w_A + w_B + w_C - 1)]Where λ₁ and λ₂ are the Lagrange multipliers.To find the minimum, I need to take the partial derivatives of L with respect to each weight and set them equal to zero.Let me denote the gradient of L with respect to w as zero:[nabla mathcal{L} = 2 Sigma mathbf{w} - lambda_1 mu - lambda_2 mathbf{1} = 0]Where μ is the vector of expected returns [0.07, 0.10, 0.05]^T and 1 is a vector of ones [1,1,1]^T.So, we have:[2 Sigma mathbf{w} = lambda_1 mu + lambda_2 mathbf{1}]This gives us three equations (since it's a 3x3 system) plus the two constraints, making a total of five equations to solve for the three weights and two Lagrange multipliers.Let me write out the equations explicitly.First, let's compute 2Σ:[2Sigma = begin{bmatrix}0.08 & 0.02 & 0.04 0.02 & 0.18 & 0.06 0.04 & 0.06 & 0.12end{bmatrix}]So, the gradient equations are:1. 0.08 w_A + 0.02 w_B + 0.04 w_C = λ₁ * 0.07 + λ₂2. 0.02 w_A + 0.18 w_B + 0.06 w_C = λ₁ * 0.10 + λ₂3. 0.04 w_A + 0.06 w_B + 0.12 w_C = λ₁ * 0.05 + λ₂And the constraints:4. 0.07 w_A + 0.10 w_B + 0.05 w_C = 0.085. w_A + w_B + w_C = 1So, now we have five equations. Let me denote them as Eq1, Eq2, Eq3, Eq4, Eq5.I need to solve for w_A, w_B, w_C, λ₁, λ₂.This seems a bit involved, but maybe I can express λ₁ and λ₂ from the first three equations and then substitute into the constraints.Let me rearrange each equation to express λ₁ and λ₂.From Eq1:0.08 w_A + 0.02 w_B + 0.04 w_C = 0.07 λ₁ + λ₂Similarly, Eq2:0.02 w_A + 0.18 w_B + 0.06 w_C = 0.10 λ₁ + λ₂And Eq3:0.04 w_A + 0.06 w_B + 0.12 w_C = 0.05 λ₁ + λ₂Let me subtract Eq1 from Eq2:(0.02 w_A + 0.18 w_B + 0.06 w_C) - (0.08 w_A + 0.02 w_B + 0.04 w_C) = (0.10 λ₁ + λ₂) - (0.07 λ₁ + λ₂)Simplify:(-0.06 w_A + 0.16 w_B + 0.02 w_C) = 0.03 λ₁Similarly, subtract Eq1 from Eq3:(0.04 w_A + 0.06 w_B + 0.12 w_C) - (0.08 w_A + 0.02 w_B + 0.04 w_C) = (0.05 λ₁ + λ₂) - (0.07 λ₁ + λ₂)Simplify:(-0.04 w_A + 0.04 w_B + 0.08 w_C) = -0.02 λ₁So now, I have two new equations:6. -0.06 w_A + 0.16 w_B + 0.02 w_C = 0.03 λ₁7. -0.04 w_A + 0.04 w_B + 0.08 w_C = -0.02 λ₁Let me denote these as Eq6 and Eq7.Let me solve Eq6 and Eq7 for λ₁.From Eq6:0.03 λ₁ = -0.06 w_A + 0.16 w_B + 0.02 w_CSo,λ₁ = (-0.06 w_A + 0.16 w_B + 0.02 w_C) / 0.03Similarly, from Eq7:-0.02 λ₁ = -0.04 w_A + 0.04 w_B + 0.08 w_CSo,λ₁ = (0.04 w_A - 0.04 w_B - 0.08 w_C) / 0.02Simplify:λ₁ = 2 w_A - 2 w_B - 4 w_CSo, now we have two expressions for λ₁:From Eq6:λ₁ = (-0.06 w_A + 0.16 w_B + 0.02 w_C) / 0.03From Eq7:λ₁ = 2 w_A - 2 w_B - 4 w_CSet them equal:(-0.06 w_A + 0.16 w_B + 0.02 w_C) / 0.03 = 2 w_A - 2 w_B - 4 w_CMultiply both sides by 0.03:-0.06 w_A + 0.16 w_B + 0.02 w_C = 0.06 w_A - 0.06 w_B - 0.12 w_CBring all terms to the left:-0.06 w_A + 0.16 w_B + 0.02 w_C - 0.06 w_A + 0.06 w_B + 0.12 w_C = 0Combine like terms:(-0.06 - 0.06) w_A + (0.16 + 0.06) w_B + (0.02 + 0.12) w_C = 0Which is:-0.12 w_A + 0.22 w_B + 0.14 w_C = 0Let me write this as:-0.12 w_A + 0.22 w_B + 0.14 w_C = 0 --> Eq8Now, I have Eq8 and the two constraints Eq4 and Eq5.So, let's recap:Eq4: 0.07 w_A + 0.10 w_B + 0.05 w_C = 0.08Eq5: w_A + w_B + w_C = 1Eq8: -0.12 w_A + 0.22 w_B + 0.14 w_C = 0So, now I have three equations (Eq4, Eq5, Eq8) with three variables w_A, w_B, w_C.Let me write them again:1. 0.07 w_A + 0.10 w_B + 0.05 w_C = 0.082. w_A + w_B + w_C = 13. -0.12 w_A + 0.22 w_B + 0.14 w_C = 0Let me try to solve this system.First, from Eq5: w_C = 1 - w_A - w_BLet me substitute w_C into Eq4 and Eq8.Substitute into Eq4:0.07 w_A + 0.10 w_B + 0.05 (1 - w_A - w_B) = 0.08Simplify:0.07 w_A + 0.10 w_B + 0.05 - 0.05 w_A - 0.05 w_B = 0.08Combine like terms:(0.07 - 0.05) w_A + (0.10 - 0.05) w_B + 0.05 = 0.08Which is:0.02 w_A + 0.05 w_B + 0.05 = 0.08Subtract 0.05:0.02 w_A + 0.05 w_B = 0.03 --> Eq4aSimilarly, substitute w_C into Eq8:-0.12 w_A + 0.22 w_B + 0.14 (1 - w_A - w_B) = 0Simplify:-0.12 w_A + 0.22 w_B + 0.14 - 0.14 w_A - 0.14 w_B = 0Combine like terms:(-0.12 - 0.14) w_A + (0.22 - 0.14) w_B + 0.14 = 0Which is:-0.26 w_A + 0.08 w_B + 0.14 = 0Subtract 0.14:-0.26 w_A + 0.08 w_B = -0.14 --> Eq8aNow, we have two equations:Eq4a: 0.02 w_A + 0.05 w_B = 0.03Eq8a: -0.26 w_A + 0.08 w_B = -0.14Let me write them as:1. 0.02 w_A + 0.05 w_B = 0.032. -0.26 w_A + 0.08 w_B = -0.14Let me solve this system.First, let me multiply Eq4a by 13 to make the coefficients of w_A similar to Eq8a.Multiply Eq4a by 13:0.26 w_A + 0.65 w_B = 0.39 --> Eq4bNow, add Eq4b and Eq8a:(0.26 w_A - 0.26 w_A) + (0.65 w_B + 0.08 w_B) = 0.39 - 0.14Simplify:0 + 0.73 w_B = 0.25So,w_B = 0.25 / 0.73 ≈ 0.3425Now, substitute w_B ≈ 0.3425 into Eq4a:0.02 w_A + 0.05 * 0.3425 = 0.03Calculate 0.05 * 0.3425 ≈ 0.017125So,0.02 w_A + 0.017125 = 0.03Subtract 0.017125:0.02 w_A ≈ 0.012875So,w_A ≈ 0.012875 / 0.02 ≈ 0.64375Now, from Eq5: w_C = 1 - w_A - w_B ≈ 1 - 0.64375 - 0.3425 ≈ 1 - 0.98625 ≈ 0.01375So, approximately:w_A ≈ 0.64375w_B ≈ 0.3425w_C ≈ 0.01375Let me check if these satisfy Eq8a:-0.26 * 0.64375 + 0.08 * 0.3425 ≈ -0.1674 + 0.0274 ≈ -0.14, which matches Eq8a.Similarly, check Eq4a:0.02 * 0.64375 + 0.05 * 0.3425 ≈ 0.012875 + 0.017125 ≈ 0.03, which is correct.So, the weights are approximately:w_A ≈ 0.64375w_B ≈ 0.3425w_C ≈ 0.01375But let me verify if these weights give the correct expected return.Compute μ_p = 0.07 * 0.64375 + 0.10 * 0.3425 + 0.05 * 0.01375Calculate each term:0.07 * 0.64375 ≈ 0.04506250.10 * 0.3425 ≈ 0.034250.05 * 0.01375 ≈ 0.0006875Sum ≈ 0.0450625 + 0.03425 + 0.0006875 ≈ 0.0799999 ≈ 0.08, which is correct.Also, check the sum of weights: 0.64375 + 0.3425 + 0.01375 ≈ 1, which is correct.So, these are the weights.But let me express them more precisely.From w_B = 0.25 / 0.730.25 / 0.73 ≈ 0.3424657534From w_A = (0.03 - 0.05 w_B) / 0.02= (0.03 - 0.05 * 0.3424657534) / 0.02= (0.03 - 0.0171232877) / 0.02= 0.0128767123 / 0.02 ≈ 0.643835615w_C = 1 - 0.643835615 - 0.3424657534 ≈ 1 - 0.9863013684 ≈ 0.0136986316So, more precisely:w_A ≈ 0.643835615w_B ≈ 0.3424657534w_C ≈ 0.0136986316To make it cleaner, maybe round to four decimal places:w_A ≈ 0.6438w_B ≈ 0.3425w_C ≈ 0.0137Alternatively, express as fractions.But perhaps it's better to present them as decimals.So, the weights are approximately:w_A ≈ 64.38%w_B ≈ 34.25%w_C ≈ 1.37%Wait, but the problem allows short-selling, so negative weights are possible. However, in this case, all weights are positive, which is fine.But let me cross-verify by plugging back into the original equations.Compute Eq1: 0.08 w_A + 0.02 w_B + 0.04 w_C= 0.08 * 0.6438 + 0.02 * 0.3425 + 0.04 * 0.0137≈ 0.051504 + 0.00685 + 0.000548 ≈ 0.058902Similarly, λ₁ * 0.07 + λ₂From Eq1: 0.058902 = 0.07 λ₁ + λ₂Similarly, Eq2: 0.02 w_A + 0.18 w_B + 0.06 w_C= 0.02 * 0.6438 + 0.18 * 0.3425 + 0.06 * 0.0137≈ 0.012876 + 0.06165 + 0.000822 ≈ 0.075348Which should equal 0.10 λ₁ + λ₂So, 0.075348 = 0.10 λ₁ + λ₂Subtract Eq1 from Eq2:0.075348 - 0.058902 = (0.10 λ₁ + λ₂) - (0.07 λ₁ + λ₂)0.016446 = 0.03 λ₁So, λ₁ ≈ 0.016446 / 0.03 ≈ 0.5482Then, from Eq1: 0.058902 = 0.07 * 0.5482 + λ₂Calculate 0.07 * 0.5482 ≈ 0.038374So, λ₂ ≈ 0.058902 - 0.038374 ≈ 0.020528Similarly, check Eq3:0.04 w_A + 0.06 w_B + 0.12 w_C= 0.04 * 0.6438 + 0.06 * 0.3425 + 0.12 * 0.0137≈ 0.025752 + 0.02055 + 0.001644 ≈ 0.047946Which should equal 0.05 λ₁ + λ₂= 0.05 * 0.5482 + 0.020528 ≈ 0.02741 + 0.020528 ≈ 0.047938Which is approximately equal, considering rounding errors. So, it checks out.Therefore, the weights are approximately:w_A ≈ 0.6438w_B ≈ 0.3425w_C ≈ 0.0137So, to summarize, the allocation weights that minimize the portfolio variance for a target expected return of 8% are approximately 64.38% in Stock A, 34.25% in Stock B, and 1.37% in Stock C.Now, moving on to part 2. The manager uses a GBM model for the stock prices. The initial prices are given: S_A(0)=100, S_B(0)=150, S_C(0)=200. The drift rates are equal to their expected returns, so μ_A=7%, μ_B=10%, μ_C=5%. The volatilities are the square roots of the diagonal elements of Σ.First, let's compute the volatilities.The covariance matrix Σ is:0.04, 0.01, 0.020.01, 0.09, 0.030.02, 0.03, 0.06So, the diagonal elements are 0.04, 0.09, 0.06.Therefore, the volatilities σ_A, σ_B, σ_C are sqrt(0.04)=0.2, sqrt(0.09)=0.3, sqrt(0.06)=sqrt(6)/10≈0.2449.So, σ_A=20%, σ_B=30%, σ_C≈24.49%.Now, the SDEs for GBM are:dS_A(t) = μ_A S_A(t) dt + σ_A S_A(t) dW_A(t)Similarly for S_B and S_C.So, writing them out:For Stock A:dS_A(t) = 0.07 S_A(t) dt + 0.2 S_A(t) dW_A(t)For Stock B:dS_B(t) = 0.10 S_B(t) dt + 0.3 S_B(t) dW_B(t)For Stock C:dS_C(t) = 0.05 S_C(t) dt + sqrt(0.06) S_C(t) dW_C(t)Alternatively, sqrt(0.06) is approximately 0.2449, so:dS_C(t) ≈ 0.05 S_C(t) dt + 0.2449 S_C(t) dW_C(t)Now, to find the expected stock prices at time T=1 year.For GBM, the expected value E[S(t)] is given by:E[S(t)] = S(0) * e^{μ t}Because the drift term is μ, and the expectation of the stochastic integral involving dW is zero.So, for each stock:E[S_A(1)] = 100 * e^{0.07 * 1} ≈ 100 * e^{0.07} ≈ 100 * 1.072508 ≈ 107.25Similarly,E[S_B(1)] = 150 * e^{0.10} ≈ 150 * 1.105171 ≈ 165.7756 ≈ 165.78E[S_C(1)] = 200 * e^{0.05} ≈ 200 * 1.051271 ≈ 210.2542 ≈ 210.25So, the expected stock prices at T=1 are approximately 107.25, 165.78, and 210.25 for A, B, and C respectively.Let me verify the calculations.Compute e^{0.07} ≈ 1.072508So, 100 * 1.072508 ≈ 107.2508 ≈ 107.25e^{0.10} ≈ 1.105171150 * 1.105171 ≈ 165.77565 ≈ 165.78e^{0.05} ≈ 1.051271200 * 1.051271 ≈ 210.2542 ≈ 210.25Yes, that seems correct.So, to recap:1. The optimal weights are approximately w_A=0.6438, w_B=0.3425, w_C=0.0137.2. The SDEs are as written, and the expected prices at T=1 are approximately 107.25, 165.78, and 210.25.I think that's it. Let me just make sure I didn't make any calculation errors.In part 1, solving the system of equations step by step, I arrived at the weights. The expected return and sum constraints are satisfied, and the Lagrangian conditions are approximately met, so it seems correct.In part 2, the SDEs are straightforward applications of GBM with given parameters. The expected values are computed correctly using the formula for GBM expectation.Yes, I think that's solid.Final Answer1. The optimal allocation weights are (boxed{w_A approx 0.6438}), (boxed{w_B approx 0.3425}), and (boxed{w_C approx 0.0137}).2. The expected stock prices at (T = 1) year are (boxed{E[S_A(1)] approx 107.25}), (boxed{E[S_B(1)] approx 165.78}), and (boxed{E[S_C(1)] approx 210.25}).</think>

🧠:<think>Alright, so I have this problem about a professional wrestler analyzing his training regimen using two functions: one for strength and one for stamina. I need to solve two sub-problems: part a is about finding the maximum and minimum strength values and when they occur within the first 12 weeks, and part b is about calculating the total accumulated stamina over the first 12 weeks by integrating the stamina function. Let me tackle each part step by step.Starting with part a: Determine the maximum and minimum values of the wrestler's strength ( S(t) ) and the weeks at which these values occur within the first 12 weeks of training.The strength function is given by ( S(t) = 50 + 20 sin(0.5t) ). Hmm, okay, so this is a sine function with some amplitude and a vertical shift. I remember that sine functions oscillate between -1 and 1, so when multiplied by 20, they'll oscillate between -20 and 20. Then, adding 50 shifts the entire function up by 50 units. So, the maximum strength should be 50 + 20 = 70, and the minimum should be 50 - 20 = 30. But wait, the question is about when these maxima and minima occur within the first 12 weeks. So, I need to find the specific weeks t where ( S(t) ) reaches 70 and 30.To find the times when the strength is maximum or minimum, I should find the critical points of the function ( S(t) ). Critical points occur where the derivative is zero or undefined. Since ( S(t) ) is a sine function, its derivative will be a cosine function, which is never undefined, so I just need to find where the derivative is zero.Let me compute the derivative of ( S(t) ):( S'(t) = d/dt [50 + 20 sin(0.5t)] )The derivative of 50 is 0, and the derivative of ( 20 sin(0.5t) ) is ( 20 times 0.5 cos(0.5t) ) by the chain rule. So,( S'(t) = 10 cos(0.5t) )Set this equal to zero to find critical points:( 10 cos(0.5t) = 0 )Divide both sides by 10:( cos(0.5t) = 0 )When does cosine equal zero? At ( pi/2 + kpi ) for any integer k. So,( 0.5t = pi/2 + kpi )Solving for t:( t = ( pi/2 + kpi ) / 0.5 = pi + 2kpi )So, t equals pi, 3pi, 5pi, etc. Since pi is approximately 3.1416, so 3.1416, 9.4248, 15.7079, etc. But we're only looking at the first 12 weeks, so t from 0 to 12.So, let's compute these t values:First critical point: t = pi ≈ 3.1416 weeksSecond critical point: t = 3pi ≈ 9.4248 weeksThird critical point would be t = 5pi ≈ 15.7079 weeks, which is beyond 12 weeks, so we can ignore that.So, within the first 12 weeks, the critical points are at approximately t ≈ 3.1416 and t ≈ 9.4248 weeks.Now, to determine whether these critical points correspond to maxima or minima, I can use the second derivative test or analyze the sign changes of the first derivative.Let me compute the second derivative:( S''(t) = d/dt [10 cos(0.5t)] = -5 sin(0.5t) )At t ≈ 3.1416:Compute ( S''(3.1416) = -5 sin(0.5 * 3.1416) = -5 sin(1.5708) )Since sin(1.5708) is sin(pi/2) which is 1, so:( S''(3.1416) = -5 * 1 = -5 ), which is negative. Therefore, this critical point is a local maximum.At t ≈ 9.4248:Compute ( S''(9.4248) = -5 sin(0.5 * 9.4248) = -5 sin(4.7124) )4.7124 is approximately 3pi/2, and sin(3pi/2) is -1, so:( S''(9.4248) = -5 * (-1) = 5 ), which is positive. Therefore, this critical point is a local minimum.So, the maximum strength occurs at t ≈ 3.1416 weeks, and the minimum strength occurs at t ≈ 9.4248 weeks.But let me express these t values more precisely. Since t = pi and t = 3pi, which are exact values, so maybe I should write them as pi and 3pi weeks, but since the question asks for the weeks, it's probably okay to leave them in terms of pi or convert them to decimal.But let me check the exact values:pi ≈ 3.1416 weeks3pi ≈ 9.4248 weeksSo, within the first 12 weeks, these are the points where the strength is maximum and minimum.Now, let me compute the strength at these points to confirm.At t = pi:( S(pi) = 50 + 20 sin(0.5 * pi) = 50 + 20 sin(pi/2) = 50 + 20 * 1 = 70 )At t = 3pi:( S(3pi) = 50 + 20 sin(0.5 * 3pi) = 50 + 20 sin(3pi/2) = 50 + 20 * (-1) = 30 )So, that confirms the maximum strength is 70 at t ≈ 3.14 weeks, and the minimum strength is 30 at t ≈ 9.42 weeks.But wait, the problem says "within the first 12 weeks," so I should also check the endpoints, t=0 and t=12, to make sure that the maximum and minimum aren't actually at the endpoints.Compute S(0):( S(0) = 50 + 20 sin(0) = 50 + 0 = 50 )Compute S(12):( S(12) = 50 + 20 sin(0.5 * 12) = 50 + 20 sin(6) )Now, sin(6) is in radians. 6 radians is approximately 343.774 degrees, which is in the fourth quadrant. The sine of 6 radians is approximately -0.2794.So,( S(12) ≈ 50 + 20 * (-0.2794) ≈ 50 - 5.588 ≈ 44.412 )So, at t=12, the strength is approximately 44.41, which is higher than the minimum of 30 but lower than the maximum of 70. Therefore, the maximum and minimum within the first 12 weeks are indeed 70 and 30, occurring at approximately 3.14 weeks and 9.42 weeks, respectively.So, summarizing part a:Maximum strength: 70 at t ≈ 3.14 weeksMinimum strength: 30 at t ≈ 9.42 weeksMoving on to part b: Calculate the total accumulated stamina ( M(t) ) over the first 12 weeks by finding the definite integral of ( M(t) ) from ( t = 0 ) to ( t = 12 ).The stamina function is given by ( M(t) = 30 + 15 e^{-0.1t} ). So, we need to compute the integral from 0 to 12 of ( M(t) dt ).Let me write that out:Total stamina = ( int_{0}^{12} [30 + 15 e^{-0.1t}] dt )I can split this integral into two separate integrals:= ( int_{0}^{12} 30 dt + int_{0}^{12} 15 e^{-0.1t} dt )Compute each integral separately.First integral: ( int 30 dt ) is straightforward. The integral of a constant is the constant times t. So,( int_{0}^{12} 30 dt = 30t bigg|_{0}^{12} = 30*12 - 30*0 = 360 - 0 = 360 )Second integral: ( int 15 e^{-0.1t} dt ). Let me factor out the constant 15:= 15 ( int e^{-0.1t} dt )To integrate ( e^{-0.1t} ), I recall that the integral of ( e^{kt} dt ) is ( (1/k) e^{kt} + C ). So, here, k = -0.1, so:( int e^{-0.1t} dt = (1/(-0.1)) e^{-0.1t} + C = -10 e^{-0.1t} + C )Therefore, multiplying by 15:15 ( int e^{-0.1t} dt = 15 * (-10) e^{-0.1t} + C = -150 e^{-0.1t} + C )Now, evaluate from 0 to 12:= [ -150 e^{-0.1*12} ] - [ -150 e^{-0.1*0} ]Simplify each term:First term at t=12: -150 e^{-1.2}Second term at t=0: -150 e^{0} = -150 * 1 = -150So,= [ -150 e^{-1.2} ] - [ -150 ] = -150 e^{-1.2} + 150Factor out 150:= 150 (1 - e^{-1.2})Now, compute the numerical value.First, compute e^{-1.2}. I know that e^{-1} ≈ 0.3679, and e^{-1.2} is a bit less. Let me calculate it more accurately.Using a calculator, e^{-1.2} ≈ 0.3011942.So,1 - e^{-1.2} ≈ 1 - 0.3011942 ≈ 0.6988058Multiply by 150:150 * 0.6988058 ≈ 104.82087So, the second integral is approximately 104.82087.Therefore, total stamina is the sum of the two integrals:360 + 104.82087 ≈ 464.82087So, approximately 464.82.But let me write the exact expression before approximating:Total stamina = 360 + 150 (1 - e^{-1.2}) = 360 + 150 - 150 e^{-1.2} = 510 - 150 e^{-1.2}If I want to express it exactly, it's 510 - 150 e^{-1.2}, but if I compute the numerical value:As above, e^{-1.2} ≈ 0.3011942So,150 e^{-1.2} ≈ 150 * 0.3011942 ≈ 45.17913Therefore,510 - 45.17913 ≈ 464.82087So, approximately 464.82.But let me double-check my calculations to make sure I didn't make a mistake.First integral: 30t from 0 to 12 is 360. That seems correct.Second integral: 15 e^{-0.1t} integrated is -150 e^{-0.1t}. Evaluated from 0 to 12:At 12: -150 e^{-1.2}At 0: -150 e^{0} = -150So, subtracting: (-150 e^{-1.2}) - (-150) = -150 e^{-1.2} + 150 = 150 (1 - e^{-1.2})Yes, that's correct.Compute 150*(1 - e^{-1.2}):1 - e^{-1.2} ≈ 1 - 0.3011942 ≈ 0.6988058Multiply by 150: 0.6988058 * 150 ≈ 104.82087Add to 360: 360 + 104.82087 ≈ 464.82087So, approximately 464.82. Depending on how precise we need to be, maybe round to two decimal places: 464.82.Alternatively, if we need an exact expression, it's 510 - 150 e^{-1.2}, but since the question says "calculate the total accumulated stamina," it's probably expecting a numerical value.Alternatively, maybe we can write it as 150(3.4 - e^{-1.2}), but 510 - 150 e^{-1.2} is also acceptable.But let me compute 510 - 150 e^{-1.2} numerically:150 e^{-1.2} ≈ 150 * 0.3011942 ≈ 45.17913So,510 - 45.17913 ≈ 464.82087Yes, same result.So, the total accumulated stamina over the first 12 weeks is approximately 464.82.But let me check if I did the integral correctly. The integral of 30 is 30t, correct. The integral of 15 e^{-0.1t} is indeed -150 e^{-0.1t}, because the integral of e^{kt} is (1/k)e^{kt}, so here k = -0.1, so 1/k = -10, so 15 * (-10) = -150. Correct.Evaluated from 0 to 12: (-150 e^{-1.2}) - (-150 e^{0}) = -150 e^{-1.2} + 150. Correct.So, yes, the calculations seem correct.Therefore, the total accumulated stamina is approximately 464.82.But to make sure, let me compute e^{-1.2} more accurately.Using a calculator:e^{-1.2} = 1 / e^{1.2}Compute e^{1.2}:We know that e^1 = 2.718281828e^{0.2} ≈ 1.221402758So, e^{1.2} = e^1 * e^{0.2} ≈ 2.718281828 * 1.221402758 ≈Let me compute that:2.718281828 * 1.221402758First, 2 * 1.221402758 = 2.4428055160.7 * 1.221402758 ≈ 0.85498193060.018281828 * 1.221402758 ≈ approximately 0.02233Adding up: 2.442805516 + 0.8549819306 ≈ 3.297787447 + 0.02233 ≈ 3.320117447So, e^{1.2} ≈ 3.320117447Therefore, e^{-1.2} ≈ 1 / 3.320117447 ≈ 0.3011942So, that's consistent with my earlier calculation.Thus, 150 e^{-1.2} ≈ 150 * 0.3011942 ≈ 45.17913So, 510 - 45.17913 ≈ 464.82087So, 464.82087 is accurate to about 5 decimal places.Therefore, the total accumulated stamina is approximately 464.82.Alternatively, if we need to present it as an exact expression, it's 510 - 150 e^{-1.2}, but since the question says "calculate," likely expecting a numerical value.So, I think that's it.Final Answera) The maximum strength is boxed{70} at approximately week boxed{3.14} and the minimum strength is boxed{30} at approximately week boxed{9.42}.b) The total accumulated stamina over the first 12 weeks is boxed{464.82}.</think>