🧠:<think>Alright, so I need to figure out the probability of Jamie winning the lottery in both the original game and the one Alex proposed. Let me start with the first question.1. Original Lottery Game: - Jamie chooses 6 numbers out of 49. - The winning numbers are drawn randomly without replacement. I remember that in lotteries, the probability of winning is calculated using combinations because the order of the numbers doesn't matter. So, the total number of possible combinations is the number of ways to choose 6 numbers from 49. The formula for combinations is C(n, k) = n! / (k!(n - k)!), where n is the total number of items, and k is the number of items to choose. Let me compute that. So, C(49, 6) is 49! / (6! * (49 - 6)!). Calculating factorials can get really big, but I think I can simplify it. Alternatively, I can use the combination formula step by step: C(49, 6) = 49 × 48 × 47 × 46 × 45 × 44 / (6 × 5 × 4 × 3 × 2 × 1). Let me compute the numerator first: 49 × 48 = 2352 2352 × 47 = let's see, 2352 × 40 = 94,080 and 2352 × 7 = 16,464, so total is 94,080 + 16,464 = 110,544 110,544 × 46 = Hmm, 110,544 × 40 = 4,421,760 and 110,544 × 6 = 663,264, so total is 4,421,760 + 663,264 = 5,085,024 5,085,024 × 45 = Let's break it down: 5,085,024 × 40 = 203,400,960 and 5,085,024 × 5 = 25,425,120, so total is 203,400,960 + 25,425,120 = 228,826,080 228,826,080 × 44 = Hmm, 228,826,080 × 40 = 9,153,043,200 and 228,826,080 × 4 = 915,304,320, so total is 9,153,043,200 + 915,304,320 = 10,068,347,520 Now the denominator is 6! which is 720. So, C(49, 6) = 10,068,347,520 / 720. Let me divide that. 10,068,347,520 ÷ 720. Let's see: First, divide numerator and denominator by 10: 1,006,834,752 / 72. 1,006,834,752 ÷ 72. Let's do this step by step. 72 × 14,000,000 = 1,008,000,000, which is a bit more than 1,006,834,752. So, 14,000,000 - (1,008,000,000 - 1,006,834,752)/72. The difference is 1,008,000,000 - 1,006,834,752 = 1,165,248. So, 1,165,248 ÷ 72 = 16,200 approximately. Wait, 72 × 16,200 = 1,166,400, which is a bit more than 1,165,248. So, 16,200 - (1,166,400 - 1,165,248)/72 = 16,200 - 1,152/72 = 16,200 - 16 = 16,184. So, total is approximately 14,000,000 - 16,184 = 13,983,816. Wait, that seems off because I know the actual value of C(49,6) is 13,983,816. So, maybe my calculations were a bit messy, but I got the right number in the end. So, the total number of possible combinations is 13,983,816. Therefore, the probability of Jamie winning the jackpot is 1 divided by that number. So, probability P1 = 1 / 13,983,816 ≈ 0.0000000715 or about 7.15 in 100 million.2. Alex's Proposed Game: - Jamie chooses 4 numbers out of 30. - The winning numbers are drawn randomly without replacement. Similarly, the probability here is 1 divided by the number of combinations of 30 numbers taken 4 at a time. So, C(30, 4) = 30! / (4! * (30 - 4)!) = (30 × 29 × 28 × 27) / (4 × 3 × 2 × 1). Let me compute the numerator: 30 × 29 = 870, 870 × 28 = 24,360, 24,360 × 27 = let's see, 24,360 × 20 = 487,200 and 24,360 × 7 = 170,520, so total is 487,200 + 170,520 = 657,720. Denominator is 4! = 24. So, C(30, 4) = 657,720 / 24. Let me divide that. 657,720 ÷ 24. 24 × 27,000 = 648,000. Subtract that from 657,720: 657,720 - 648,000 = 9,720. Now, 24 × 405 = 9,720. So, total is 27,000 + 405 = 27,405. So, the total number of combinations is 27,405. Therefore, the probability P2 = 1 / 27,405 ≈ 0.00003649 or about 3.65 in 100,000. Now, to compare the two probabilities. Let's see how much better Alex's game is compared to the original. So, the original probability is approximately 7.15 in 100 million, which is 0.0000000715. Alex's game probability is approximately 0.00003649. To find the factor by which the probability increases, we can divide P2 by P1. So, factor = P2 / P1 = (1 / 27,405) / (1 / 13,983,816) = 13,983,816 / 27,405. Let me compute that division. 13,983,816 ÷ 27,405. Let me see, 27,405 × 500 = 13,702,500. Subtract that from 13,983,816: 13,983,816 - 13,702,500 = 281,316. Now, 27,405 × 10 = 274,050. Subtract that: 281,316 - 274,050 = 7,266. Now, 27,405 × 0.265 ≈ 7,266 (because 27,405 × 0.2 = 5,481, 27,405 × 0.06 = 1,644.3, so total ≈ 5,481 + 1,644.3 = 7,125.3, which is close to 7,266). So, total factor is approximately 500 + 10 + 0.265 ≈ 510.265. So, the probability in Alex's game is about 510 times higher than the original game. Wait, let me verify that division again because 27,405 × 510 = ? 27,405 × 500 = 13,702,500 27,405 × 10 = 274,050 So, 13,702,500 + 274,050 = 13,976,550 The numerator is 13,983,816, so the difference is 13,983,816 - 13,976,550 = 7,266. So, 7,266 / 27,405 ≈ 0.265. So, total factor is 510.265, which is approximately 510.27. So, the probability is about 510 times higher in Alex's game. Alternatively, we can compute it as 13,983,816 / 27,405 ≈ 510.27. So, the probability of winning in Alex's game is about 510 times higher than the original game. Therefore, Jamie's chances of winning are significantly better in Alex's proposed game. Let me just recap: - Original probability: 1 / 13,983,816 ≈ 7.15e-8 - Alex's game probability: 1 / 27,405 ≈ 3.65e-5 - Factor: 3.65e-5 / 7.15e-8 ≈ 510.27 So, yes, that seems correct. I think that's all for the calculations. I just need to make sure I didn't make any arithmetic errors, but the steps seem logical.</think>

🧠:<think>Okay, so I have this problem about designing an urban park to maximize biodiversity while minimizing the carbon footprint. It's divided into two parts. Let me start with the first part.1. Expressing Biodiversity Index B as a Function of θ and rFirst, the park has three sections: a forested area, a flower garden, and a grassland. The total area is 20,000 square meters.- Forested Area (A_f): It's modeled as a circular sector with central angle θ and radius r. The area of a circular sector is given by (1/2) * r² * θ. So, A_f = (1/2) * r² * θ.- Flower Garden (A_g): It's a rectangle adjacent to the forested area. The length is 2r and the width is r. So, A_g = length * width = 2r * r = 2r².- Grassland (A_gr): The remaining area in the park. Since the total area is 20,000, A_gr = 20,000 - A_f - A_g. Substituting the expressions for A_f and A_g, we get A_gr = 20,000 - (1/2)r²θ - 2r².So, the biodiversity index B is given by B = αA_f + βA_g + γA_gr. Plugging in the areas:B = α*(1/2 r² θ) + β*(2r²) + γ*(20,000 - (1/2 r² θ) - 2r²)Let me simplify that:B = (α/2) r² θ + 2β r² + γ*20,000 - (γ/2) r² θ - 2γ r²Combine like terms:- Terms with r² θ: (α/2 - γ/2) r² θ- Terms with r²: 2β r² - 2γ r²- Constant term: γ*20,000So,B = [(α - γ)/2] r² θ + [2(β - γ)] r² + 20,000γThat's B expressed as a function of θ and r.2. Maximizing Biodiversity Index BWe need to maximize B with respect to θ and r, subject to the total area constraint.Wait, but the total area is fixed at 20,000, so the constraint is already incorporated into A_gr. So, we just need to maximize B as a function of θ and r.But let me check if there are any other constraints. The problem mentions that θ is the central angle of the sector, so θ must be between 0 and 2π radians. Also, r must be positive, but it's limited by the total area.So, to maximize B, we can take partial derivatives with respect to θ and r, set them equal to zero, and solve for θ and r.Let's write B again:B(θ, r) = [(α - γ)/2] r² θ + [2(β - γ)] r² + 20,000γFirst, take the partial derivative with respect to θ:∂B/∂θ = [(α - γ)/2] r²Set this equal to zero for maximization:[(α - γ)/2] r² = 0Since r² is always positive (r > 0), this implies that (α - γ)/2 = 0, so α = γ.But α, β, γ are positive constants. If α = γ, then the coefficient of θ becomes zero, and B becomes independent of θ. That suggests that if α ≠ γ, we can't have a maximum with respect to θ unless α = γ.Wait, that seems odd. Maybe I made a mistake.Wait, no, because B is linear in θ. So, if the coefficient of θ is positive, B increases as θ increases, so θ should be as large as possible (θ = 2π). If the coefficient is negative, B decreases as θ increases, so θ should be as small as possible (θ = 0). If the coefficient is zero, θ doesn't matter.So, let's analyze the coefficient of θ:Coefficient = (α - γ)/2If α > γ, then the coefficient is positive, so to maximize B, set θ as large as possible, which is 2π.If α < γ, the coefficient is negative, so set θ as small as possible, which is 0.If α = γ, θ doesn't affect B, so any θ is fine.Now, moving on to r.Take the partial derivative of B with respect to r:∂B/∂r = [(α - γ)/2] * 2r θ + [2(β - γ)] * 2rSimplify:= (α - γ) r θ + 4(β - γ) rSet this equal to zero:(α - γ) r θ + 4(β - γ) r = 0Factor out r:r [ (α - γ) θ + 4(β - γ) ] = 0Since r > 0, we can divide both sides by r:(α - γ) θ + 4(β - γ) = 0So,(α - γ) θ = -4(β - γ)θ = [ -4(β - γ) ] / (α - γ )But θ must be between 0 and 2π.So, let's consider the cases:Case 1: α ≠ γFrom earlier, if α > γ, θ = 2πIf α < γ, θ = 0But from the derivative, we have θ = [ -4(β - γ) ] / (α - γ )This suggests that θ is determined by this equation, but it must also satisfy the earlier condition based on the sign of (α - γ).Wait, this is getting a bit tangled. Maybe I should approach it differently.Alternatively, since B is linear in θ, the maximum occurs at the endpoints of θ. So, regardless of the derivative with respect to r, θ is either 0 or 2π.But then, when we take the derivative with respect to r, we have to consider θ as either 0 or 2π.Wait, perhaps I should consider both variables together.Let me think again.We have B(θ, r) = [(α - γ)/2] r² θ + [2(β - γ)] r² + 20,000γWe can treat this as a function of two variables, θ and r, with θ ∈ [0, 2π] and r > 0, but also subject to the area constraint, which is already incorporated.But since B is linear in θ, the maximum will occur at θ = 0 or θ = 2π, depending on the sign of the coefficient of θ.Similarly, for r, we can find the optimal r given θ.So, let's proceed step by step.First, determine θ:If (α - γ) > 0, then increasing θ increases B, so set θ = 2π.If (α - γ) < 0, then increasing θ decreases B, so set θ = 0.If (α - γ) = 0, θ doesn't affect B, so any θ is fine, but we still need to maximize B with respect to r.Now, for each case, find the optimal r.Case 1: α > γ (θ = 2π)Then, B becomes:B = [(α - γ)/2] r² * 2π + [2(β - γ)] r² + 20,000γSimplify:= (α - γ) π r² + 2(β - γ) r² + 20,000γCombine terms:= [ (α - γ) π + 2(β - γ) ] r² + 20,000γNow, to maximize B with respect to r, we can take the derivative:dB/dr = 2 [ (α - γ) π + 2(β - γ) ] rSet derivative to zero:2 [ (α - γ) π + 2(β - γ) ] r = 0Since r > 0, this implies that [ (α - γ) π + 2(β - γ) ] = 0So,(α - γ) π + 2(β - γ) = 0Solve for r:Wait, no, this is the condition for the derivative to be zero, but since we're maximizing, we need to check if the coefficient is positive or negative.If [ (α - γ) π + 2(β - γ) ] > 0, then B increases with r, so r should be as large as possible.But r is limited by the total area.Wait, the total area is fixed at 20,000, so r can't be arbitrary. Let's express r in terms of the total area.From the total area:A_f + A_g + A_gr = 20,000But A_gr = 20,000 - A_f - A_gSo, we can express r in terms of θ.But in this case, θ is fixed at 2π.So, A_f = (1/2) r² * 2π = π r²A_g = 2r²So, total area:π r² + 2r² + A_gr = 20,000But A_gr = 20,000 - π r² - 2r²So, the total area is fixed, so r is determined by the areas.Wait, but we need to express r in terms of the total area.Wait, no, the total area is fixed, so r is constrained by the areas of the sectors and rectangles.But in this case, since θ is fixed at 2π, the forested area is a full circle, so A_f = π r².A_g = 2r²So, total area:π r² + 2r² + A_gr = 20,000But A_gr is the remaining area, so:A_gr = 20,000 - π r² - 2r²But we can also express B in terms of r.Wait, but in this case, since θ is fixed, we can express B as a function of r and then find the r that maximizes B.But earlier, we saw that B is linear in r², so if the coefficient is positive, B increases with r², so r should be as large as possible, but r is limited by the total area.Wait, let's see:From the total area:π r² + 2r² + A_gr = 20,000But A_gr = 20,000 - π r² - 2r²So, the maximum r is when A_gr is zero, but A_gr can't be negative, so:π r² + 2r² ≤ 20,000r² (π + 2) ≤ 20,000r² ≤ 20,000 / (π + 2)r ≤ sqrt(20,000 / (π + 2))So, r is bounded above by sqrt(20,000 / (π + 2)).But in our case, we have B as a function of r:B = [ (α - γ) π + 2(β - γ) ] r² + 20,000γIf [ (α - γ) π + 2(β - γ) ] > 0, then B increases with r², so to maximize B, set r as large as possible, i.e., r = sqrt(20,000 / (π + 2)).If [ (α - γ) π + 2(β - γ) ] < 0, then B decreases with r², so set r as small as possible, but r must be positive. However, r can't be zero because then the areas would be zero, which isn't practical. So, perhaps the minimum r is determined by other constraints, but since the problem doesn't specify, we might assume that r can be as small as needed, but in reality, it's limited by the practical size of the park.But since the problem doesn't specify, perhaps we can assume that r can be varied freely, so if the coefficient is negative, the maximum occurs at the smallest possible r, but since r can be made arbitrarily small, B would approach 20,000γ. However, this might not be practical, so perhaps we need to consider the feasible range of r.Wait, but in the case where [ (α - γ) π + 2(β - γ) ] < 0, the maximum B occurs at the smallest r, but r can't be zero because then A_f and A_g would be zero, and A_gr would be 20,000, but B would be 20,000γ, which might not be the maximum.Wait, perhaps I'm overcomplicating. Let's think differently.Since B is a quadratic function in r², and we have a constraint on r from the total area, we can express r in terms of the total area and substitute back into B.Wait, but in this case, with θ fixed at 2π, the total area is:π r² + 2r² + A_gr = 20,000But A_gr is part of B, so perhaps we can express B in terms of r and then find the optimal r.Alternatively, since B is linear in r², the maximum occurs at the boundary of r.So, if the coefficient of r² in B is positive, set r as large as possible; if negative, set r as small as possible.But let's compute the coefficient:Coefficient = (α - γ) π + 2(β - γ)If this is positive, maximize r; if negative, minimize r.But r is constrained by the total area:π r² + 2r² ≤ 20,000So, r² ≤ 20,000 / (π + 2)Thus, r_max = sqrt(20,000 / (π + 2))Similarly, r_min is theoretically approaching zero, but in practice, it's limited by the minimum size of the park sections.But since the problem doesn't specify, we can proceed with r_max and r_min as above.So, in Case 1 (α > γ, θ = 2π):If (α - γ) π + 2(β - γ) > 0, then set r = r_max.If (α - γ) π + 2(β - γ) < 0, set r = r_min (approaching zero).But if r approaches zero, A_f and A_g approach zero, and A_gr approaches 20,000, so B approaches 20,000γ.But we need to check if this is indeed the maximum.Wait, but if the coefficient is negative, then increasing r decreases B, so the maximum occurs at the smallest r, which would be r approaching zero, making B approach 20,000γ.But if r is zero, the park would only have the grassland, which might not be desired, but mathematically, it's a possibility.Similarly, in Case 2 (α < γ, θ = 0):Then, B becomes:B = [(α - γ)/2] r² * 0 + [2(β - γ)] r² + 20,000γ= 0 + 2(β - γ) r² + 20,000γ= 2(β - γ) r² + 20,000γAgain, this is linear in r².So, the coefficient is 2(β - γ).If β > γ, then B increases with r², so set r as large as possible.But with θ = 0, the forested area is zero, so A_f = 0.A_g = 2r²A_gr = 20,000 - 0 - 2r² = 20,000 - 2r²So, the total area constraint is 2r² ≤ 20,000, so r² ≤ 10,000, r ≤ 100.Thus, r_max = 100.So, if β > γ, set r = 100.If β < γ, then B decreases with r², so set r as small as possible, approaching zero, making B approach 20,000γ.If β = γ, then B is constant, 20,000γ, regardless of r.Case 3: α = γThen, the coefficient of θ in B is zero, so θ doesn't affect B.Then, B = 2(β - γ) r² + 20,000γAgain, similar to Case 2.If β > γ, maximize r.But with θ arbitrary, but since A_f is (1/2) r² θ, and A_g is 2r², the total area is:(1/2) r² θ + 2r² + A_gr = 20,000But A_gr = 20,000 - (1/2) r² θ - 2r²But since θ can vary, we can choose θ to adjust the areas.But since B is independent of θ, we can set θ to any value, but to maximize B, we need to maximize r.But r is constrained by the total area.Wait, but if θ is arbitrary, we can adjust θ to make r as large as possible.Wait, no, because r is a variable, and θ is another variable. So, perhaps we can express θ in terms of r.From the total area:(1/2) r² θ + 2r² + A_gr = 20,000But A_gr is part of B, so perhaps we can express θ in terms of r and A_gr.But since B is independent of θ, we can set θ to any value, but to maximize B, we need to maximize r.But r is limited by the total area.Wait, let's express θ in terms of r and A_gr:(1/2) r² θ = 20,000 - 2r² - A_grBut since A_gr is part of B, and B is 2(β - γ) r² + 20,000γ, we can express θ as:θ = [20,000 - 2r² - A_gr] / ( (1/2) r² )But since A_gr = 20,000 - (1/2) r² θ - 2r², we can substitute back.But this seems circular.Alternatively, since B is independent of θ, we can choose θ to be any value, but to maximize r, we need to minimize the area taken by the forested area, which is (1/2) r² θ.So, to maximize r, set θ as small as possible, which is θ = 0.Thus, in this case, θ = 0, and then A_f = 0, A_g = 2r², A_gr = 20,000 - 2r².Then, B = 2(β - γ) r² + 20,000γSo, similar to Case 2.If β > γ, maximize r, which is r = 100.If β < γ, minimize r, approaching zero.If β = γ, B is constant.So, putting it all together:- If α > γ: - Set θ = 2π - If (α - γ) π + 2(β - γ) > 0, set r = sqrt(20,000 / (π + 2)) - Else, set r approaching zero- If α < γ: - Set θ = 0 - If β > γ, set r = 100 - Else, set r approaching zero- If α = γ: - Set θ = 0 - If β > γ, set r = 100 - Else, set r approaching zeroBut this seems a bit too involved. Maybe there's a better way.Alternatively, perhaps we can use Lagrange multipliers to maximize B subject to the total area constraint.Let me try that.We need to maximize B = αA_f + βA_g + γA_grSubject to A_f + A_g + A_gr = 20,000Express A_gr = 20,000 - A_f - A_gSo, B = αA_f + βA_g + γ(20,000 - A_f - A_g)= (α - γ)A_f + (β - γ)A_g + 20,000γNow, express A_f and A_g in terms of θ and r:A_f = (1/2) r² θA_g = 2r²So,B = (α - γ)(1/2 r² θ) + (β - γ)(2r²) + 20,000γ= [(α - γ)/2] r² θ + 2(β - γ) r² + 20,000γNow, we can set up the Lagrangian:L = [(α - γ)/2] r² θ + 2(β - γ) r² + 20,000γ - λ[(1/2) r² θ + 2r² + A_gr - 20,000]But since A_gr = 20,000 - A_f - A_g, the constraint is automatically satisfied, so maybe we don't need Lagrange multipliers here.Alternatively, since we have two variables θ and r, and the total area is fixed, we can express θ in terms of r or vice versa.From the total area:(1/2) r² θ + 2r² + A_gr = 20,000But A_gr = 20,000 - (1/2) r² θ - 2r²So, we can express θ in terms of r:(1/2) r² θ = 20,000 - 2r² - A_grBut A_gr is part of B, so perhaps we can express θ as:θ = [20,000 - 2r² - A_gr] / ( (1/2) r² )But this seems circular.Alternatively, express θ in terms of r:From total area:(1/2) r² θ + 2r² ≤ 20,000So,θ ≤ [20,000 - 2r²] / ( (1/2) r² ) = [20,000 - 2r²] / (0.5 r²) = [20,000 / (0.5 r²)] - [2r² / (0.5 r²)] = 40,000 / r² - 4But θ must be ≤ 2π, so:40,000 / r² - 4 ≤ 2πThus,40,000 / r² ≤ 2π + 4r² ≥ 40,000 / (2π + 4)r ≥ sqrt(40,000 / (2π + 4)) = sqrt(20,000 / (π + 2))Which is the same as r_max in Case 1.So, r cannot be less than sqrt(20,000 / (π + 2)) if θ is to be at least zero.Wait, but if r is smaller, θ can be smaller.Wait, perhaps I'm overcomplicating.Let me try to express B in terms of r only.From the total area:(1/2) r² θ + 2r² + A_gr = 20,000But A_gr = 20,000 - (1/2) r² θ - 2r²So, B = (α - γ)(1/2 r² θ) + (β - γ)(2r²) + γ(20,000 - (1/2) r² θ - 2r²)Simplify:= (α - γ)(1/2 r² θ) + (β - γ)(2r²) + 20,000γ - γ(1/2 r² θ) - γ(2r²)= [ (α - γ)/2 - γ/2 ] r² θ + [ 2(β - γ) - 2γ ] r² + 20,000γ= [ (α - 2γ)/2 ] r² θ + [ 2β - 4γ ] r² + 20,000γWait, that's different from earlier. Did I make a mistake?Wait, let's recompute:B = αA_f + βA_g + γA_gr= α*(1/2 r² θ) + β*(2r²) + γ*(20,000 - 1/2 r² θ - 2r²)= (α/2) r² θ + 2β r² + 20,000γ - (γ/2) r² θ - 2γ r²= [ (α/2 - γ/2) ] r² θ + [ 2β - 2γ ] r² + 20,000γ= [ (α - γ)/2 ] r² θ + 2(β - γ) r² + 20,000γYes, that's correct.So, B = [ (α - γ)/2 ] r² θ + 2(β - γ) r² + 20,000γNow, from the total area:(1/2) r² θ + 2r² ≤ 20,000We can express θ in terms of r:θ = [20,000 - 2r²] / ( (1/2) r² ) - [A_gr] / ( (1/2) r² )Wait, no, A_gr is part of the total area, so:(1/2) r² θ + 2r² + A_gr = 20,000But A_gr is part of B, so perhaps we can express θ as:θ = [20,000 - 2r² - A_gr] / ( (1/2) r² )But since A_gr is part of B, we can't directly express θ without knowing A_gr.Alternatively, since A_gr = 20,000 - (1/2) r² θ - 2r², we can substitute back into B:B = [ (α - γ)/2 ] r² θ + 2(β - γ) r² + γ(20,000 - (1/2) r² θ - 2r²)= [ (α - γ)/2 ] r² θ + 2(β - γ) r² + 20,000γ - (γ/2) r² θ - 2γ r²= [ (α - γ)/2 - γ/2 ] r² θ + [ 2(β - γ) - 2γ ] r² + 20,000γ= [ (α - 2γ)/2 ] r² θ + [ 2β - 4γ ] r² + 20,000γWait, that's the same as before.But perhaps we can express θ in terms of r from the total area:(1/2) r² θ = 20,000 - 2r² - A_grBut A_gr is part of B, so perhaps we can write θ as:θ = [20,000 - 2r² - A_gr] / ( (1/2) r² )But since A_gr is part of B, which is a function of r and θ, this seems circular.Alternatively, perhaps we can treat θ as a function of r, given the total area.From total area:(1/2) r² θ + 2r² + A_gr = 20,000But A_gr = 20,000 - (1/2) r² θ - 2r²So, we can express θ as:θ = [20,000 - 2r² - A_gr] / ( (1/2) r² )But since A_gr is part of B, which is a function of θ and r, we can't directly solve for θ in terms of r without knowing A_gr.This seems like a dead end.Perhaps a better approach is to consider that since B is linear in θ, the maximum occurs at θ = 0 or θ = 2π, as earlier.So, let's proceed with that.Case 1: α > γ, θ = 2πThen, B = [ (α - γ)/2 ] r² * 2π + 2(β - γ) r² + 20,000γ= (α - γ) π r² + 2(β - γ) r² + 20,000γ= [ (α - γ) π + 2(β - γ) ] r² + 20,000γNow, to maximize B, we need to maximize r², subject to the total area constraint.From total area:π r² + 2r² + A_gr = 20,000But A_gr = 20,000 - π r² - 2r²So, the maximum r is when A_gr = 0:π r² + 2r² = 20,000r² (π + 2) = 20,000r² = 20,000 / (π + 2)r = sqrt(20,000 / (π + 2)) ≈ sqrt(20,000 / 5.1416) ≈ sqrt(3891.3) ≈ 62.4 metersSo, r_max ≈ 62.4 metersThus, in this case, if [ (α - γ) π + 2(β - γ) ] > 0, set r = 62.4 metersIf [ (α - γ) π + 2(β - γ) ] < 0, set r approaching zeroCase 2: α < γ, θ = 0Then, B = 2(β - γ) r² + 20,000γFrom total area:A_f = 0, A_g = 2r², A_gr = 20,000 - 2r²So, r² ≤ 10,000, r ≤ 100 metersThus, if β > γ, set r = 100 metersIf β < γ, set r approaching zeroCase 3: α = γThen, B = 2(β - γ) r² + 20,000γSame as Case 2.So, summarizing:- If α > γ and [ (α - γ) π + 2(β - γ) ] > 0: - θ = 2π - r = sqrt(20,000 / (π + 2)) ≈ 62.4 meters- If α > γ and [ (α - γ) π + 2(β - γ) ] < 0: - θ = 2π - r approaching zero- If α < γ and β > γ: - θ = 0 - r = 100 meters- If α < γ and β < γ: - θ = 0 - r approaching zero- If α = γ: - θ = 0 - If β > γ, r = 100 meters - Else, r approaching zeroBut in reality, setting r approaching zero would mean the park is mostly grassland, which might not be desirable, but mathematically, it's a possibility.Now, moving on to part 2.2. Feasible Range for θ and r Considering Carbon FootprintThe carbon footprint C is given by C = δA_f + εA_g + ζA_grWe need to ensure that C ≤ TExpress C in terms of θ and r:C = δ*(1/2 r² θ) + ε*(2r²) + ζ*(20,000 - 1/2 r² θ - 2r²)Simplify:= (δ/2) r² θ + 2ε r² + 20,000ζ - (ζ/2) r² θ - 2ζ r²= [ (δ/2 - ζ/2) ] r² θ + [ 2ε - 2ζ ] r² + 20,000ζ= [ (δ - ζ)/2 ] r² θ + 2(ε - ζ) r² + 20,000ζWe need C ≤ TSo,[ (δ - ζ)/2 ] r² θ + 2(ε - ζ) r² + 20,000ζ ≤ TWe need to find the feasible range of θ and r that satisfy this inequality while maximizing B.But since B is already a function of θ and r, and we have constraints on θ and r from part 1, we need to find the intersection of the feasible regions.This is getting quite complex, but perhaps we can approach it by considering the same cases as in part 1.Case 1: α > γ, θ = 2π, r = sqrt(20,000 / (π + 2)) ≈ 62.4 metersCheck if C ≤ T:C = [ (δ - ζ)/2 ] r² * 2π + 2(ε - ζ) r² + 20,000ζ= (δ - ζ) π r² + 2(ε - ζ) r² + 20,000ζ= [ (δ - ζ) π + 2(ε - ζ) ] r² + 20,000ζSubstitute r² = 20,000 / (π + 2):C = [ (δ - ζ) π + 2(ε - ζ) ] * (20,000 / (π + 2)) + 20,000ζ= 20,000 [ (δ - ζ) π + 2(ε - ζ) ] / (π + 2) + 20,000ζFactor out 20,000:= 20,000 [ ( (δ - ζ) π + 2(ε - ζ) ) / (π + 2) + ζ ]Simplify the expression inside the brackets:= [ (δ - ζ) π + 2(ε - ζ) + ζ(π + 2) ] / (π + 2)= [ δ π - ζ π + 2ε - 2ζ + ζ π + 2ζ ] / (π + 2)Simplify:- ζ π cancels with + ζ π- -2ζ + 2ζ cancelsSo,= (δ π + 2ε) / (π + 2)Thus, C = 20,000 (δ π + 2ε) / (π + 2)We need this to be ≤ TSo,20,000 (δ π + 2ε) / (π + 2) ≤ TIf this holds, then θ = 2π and r ≈ 62.4 meters is feasible.If not, we need to reduce r or θ.But since θ is fixed at 2π in this case, we might need to reduce r.But r is already at its maximum given θ = 2π, so if C exceeds T, we need to reduce r.But how?Alternatively, perhaps we can solve for r such that C = T.From C = [ (δ - ζ)/2 ] r² θ + 2(ε - ζ) r² + 20,000ζ ≤ TWith θ = 2π, we have:[ (δ - ζ)/2 ] r² * 2π + 2(ε - ζ) r² + 20,000ζ ≤ TSimplify:(δ - ζ) π r² + 2(ε - ζ) r² + 20,000ζ ≤ TFactor out r²:[ (δ - ζ) π + 2(ε - ζ) ] r² + 20,000ζ ≤ TSolve for r²:r² ≤ (T - 20,000ζ) / [ (δ - ζ) π + 2(ε - ζ) ]But we also have the total area constraint:π r² + 2r² ≤ 20,000So,r² ≤ 20,000 / (π + 2)Thus, the feasible r² is the minimum of the two:r² ≤ min( 20,000 / (π + 2), (T - 20,000ζ) / [ (δ - ζ) π + 2(ε - ζ) ] )But we need to ensure that (T - 20,000ζ) / [ (δ - ζ) π + 2(ε - ζ) ] is positive.So,(T - 20,000ζ) > 0 and [ (δ - ζ) π + 2(ε - ζ) ] > 0Or,(T - 20,000ζ) < 0 and [ (δ - ζ) π + 2(ε - ζ) ] < 0But since T is a threshold, it's likely that T > 20,000ζ, so the first condition holds.Thus,r² ≤ min( 20,000 / (π + 2), (T - 20,000ζ) / [ (δ - ζ) π + 2(ε - ζ) ] )If (T - 20,000ζ) / [ (δ - ζ) π + 2(ε - ζ) ] ≤ 20,000 / (π + 2), then r is limited by the carbon constraint.Otherwise, r is limited by the total area.Similarly, for other cases.This is getting quite involved, but I think the key takeaway is that the feasible range of θ and r depends on the constants δ, ε, ζ, and the threshold T.In summary, the values of θ and r that maximize B while satisfying C ≤ T would be:- If α > γ and [ (α - γ) π + 2(β - γ) ] > 0: - θ = 2π - r = sqrt( min( 20,000 / (π + 2), (T - 20,000ζ) / [ (δ - ζ) π + 2(ε - ζ) ] ) )- If α > γ and [ (α - γ) π + 2(β - γ) ] < 0: - θ = 2π - r approaching zero, but must satisfy C ≤ T- If α < γ and β > γ: - θ = 0 - r = min(100, sqrt( (T - 20,000ζ) / [ 2(ε - ζ) ] ) )- If α < γ and β < γ: - θ = 0 - r approaching zero, but must satisfy C ≤ T- If α = γ: - θ = 0 - If β > γ, r = min(100, sqrt( (T - 20,000ζ) / [ 2(ε - ζ) ] ) ) - Else, r approaching zeroBut this is a very high-level summary, and the exact feasible range would require solving for r in each case, considering both the biodiversity maximization and the carbon footprint constraint.I think I've covered the main steps, but it's quite complex, and I might have made some errors in the algebra. It's important to carefully check each step, especially when dealing with multiple constraints and variables.</think>

🧠:<think>Alright, so I have this problem about evaluating an adaptive learning platform's algorithm. The project manager is using two metrics: Mean Squared Error (MSE) and Pearson correlation coefficient. I need to figure out both. Let me take it step by step.Starting with the first part: calculating the MSE. I remember that MSE is a common metric to measure the average squared difference between predicted and actual values. The formula for MSE is the average of the squared differences between each actual score and the predicted score. So, if I have actual scores ( A_{ij} ) and predicted scores ( P_{ij} ) for each student ( i ) and assessment ( j ), the MSE should be the sum of all squared differences divided by the total number of data points.Given that there are 1000 students and each takes 10 assessments, the total number of data points is 10,000. The project manager provided the sum of squared differences as 1,500,000. So, plugging into the formula, MSE would be 1,500,000 divided by 10,000. Let me write that down:MSE = (1/10,000) * ΣΣ(A_{ij} - P_{ij})²Given that ΣΣ(A_{ij} - P_{ij})² = 1,500,000, so:MSE = 1,500,000 / 10,000Calculating that, 1,500,000 divided by 10,000 is 150. So, the MSE is 150. That seems straightforward.Moving on to the second part: Pearson correlation coefficient. I know Pearson's r measures the linear correlation between two datasets. The formula involves covariance and the product of the standard deviations of the two datasets. The formula is:r = Cov(A, P) / (σ_A * σ_P)Where Cov(A, P) is the covariance between actual and predicted scores, σ_A is the standard deviation of actual scores, and σ_P is the standard deviation of predicted scores.The problem states that the covariance is 400, the standard deviation of actual scores is 50, and the standard deviation of predicted scores is 60. Plugging these into the formula:r = 400 / (50 * 60)Calculating the denominator: 50 multiplied by 60 is 3000. So,r = 400 / 3000Simplifying that, 400 divided by 3000 is 0.1333... So, approximately 0.1333. To express this as a fraction, 400/3000 reduces to 4/30, which is 2/15. So, 2/15 is approximately 0.1333.Wait, let me double-check that. 400 divided by 3000: both numerator and denominator can be divided by 100, so 4/30, which simplifies to 2/15. Yes, that's correct. So, 2/15 is about 0.1333.Is there anything else I need to consider for Pearson's r? I think that's it. The formula only requires covariance and the product of standard deviations. So, as long as those values are given correctly, this should be accurate.Just to recap:1. For MSE, total squared differences are 1,500,000 over 10,000 data points, so 150.2. For Pearson's r, covariance is 400, standard deviations 50 and 60, so 400/(50*60) = 400/3000 = 2/15 ≈ 0.1333.I think that's all. I don't see any mistakes in my calculations, so I feel confident about these results.Final Answer1. The Mean Squared Error (MSE) is boxed{150}.2. The Pearson correlation coefficient is boxed{dfrac{2}{15}}.</think>

🧠:<think>First, I need to calculate the displacement of the model in grams. The actual displacement of the USS Missouri is 58,000 tons, and since 1 ton equals 907,185 grams, I'll multiply these values to find the total displacement in grams.Next, I'll determine the center of gravity for the model. The actual ship's center of gravity is 40% of the way down from the top of its height, which is proportional to the draft of 28.5 feet. I'll convert the draft to centimeters using the scale of 1:150 and then calculate 40% of this height to find the position of the center of gravity from the base of the model.</think>