| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | 为了找到Paul在预算约束下最大化效用的最优选择，我们需要解决以下优化问题： ### 问题描述 - **效用函数**: \( U(t, s) = \sqrt{ts} \) - **价格**: - Twinkies (\( t \))：$0.10 每个 - Soda (\( s \))：$0.25 每杯 - **预算**: $1.00 ### 解决步骤 我们需要最大化效用函数 \( U(t, s) = \sqrt{ts} \)，同时满足预算约束 \( 0.10t + 0.25s = 1 \)。 #### 1. 预算约束 首先，写出预算约束方程： \[ 0.10t + 0.25s = 1 \] #### 2. 拉格朗日乘子法 我们使用拉格朗日乘子法来解决这个优化问题。定义拉格朗日函数： \[ \mathcal{L}(t, s, \lambda) = \sqrt{ts} + \lambda (1 - 0.10t - 0.25s) \] #### 3. 求偏导数并设为零 对 \( t \)、\( s \) 和 \( \lambda \) 分别求偏导数，并设为零： - 对 \( t \) 求偏导数： \[ \frac{\partial \mathcal{L}}{\partial t} = \frac{1}{2} \cdot t^{-1/2} \cdot s^{1/2} - 0.10 \lambda = 0 \] \[ \Rightarrow \frac{s^{1/2}}{2 t^{1/2}} = 0.10 \lambda \] \[ \Rightarrow \lambda = \frac{s^{1/2}}{0.20 t^{1/2}} \] - 对 \( s \) 求偏导数： \[ \frac{\partial \mathcal{L}}{\partial s} = \frac{1}{2} \cdot s^{-1/2} \cdot t^{1/2} - 0.25 \lambda = 0 \] \[ \Rightarrow \frac{t^{1/2}}{2 s^{1/2}} = 0.25 \lambda \] \[ \Rightarrow \lambda = \frac{t^{1/2}}{0.50 s^{1/2}} \] #### 4. 联立方程求解 将两个关于 \( \lambda \) 的方程联立： \[ \frac{s^{1/2}}{0.20 t^{1/2}} = \frac{t^{1/2}}{0.50 s^{1/2}} \] \[ \Rightarrow 5 s = t \] 用 \( t = 5s \) 代入预算约束方程： \[ 0.10(5s) + 0.25s = 1 \] \[ 0.50s + 0.25s = 1 \] \[ 0.75s = 1 \] \[ s = \frac{1}{0.75} = \frac{4}{3} \] 然后求出 \( t \)： \[ t = 5s = 5 \cdot \frac{4}{3} = \frac{20}{3} \] #### 5. 结果 Paul应该购买 \(\frac{20}{3}\) 个Twinkies和 \(\frac{4}{3}\) 杯Soda来最大化效用。 ### 总结 Paul应购买约6.67个Twinkies和1.33杯Soda来最大化他的效用。由于现实中Twinkies和Soda的数量必须是整数，可以考虑以下近似方案： - 购买6个Twinkies和2杯Soda，或者 - 购买7个Twinkies和1杯Soda 具体选择取决于实际购买的可行性和偏好。 |
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| deepseekr1 | To maximize Paul's utility given the utility function \( U(t, s) = \sqrt{ts} \) with a budget constraint of $1, where Twinkies cost $0.10 each and soda costs $0.25 per cup, we need to follow these steps: 1. **Calculate the marginal utilities**: - The partial derivative of \( U \) with respect to \( t \) (marginal utility of Twinkies) is \( \frac{s}{2\sqrt{ts}} \). - The partial derivative of \( U \) with respect to \( s \) (marginal utility of soda) is \( \frac{t}{2\sqrt{ts}} \). 2. **Determine the marginal rate of substitution (MRS)**: - The MRS is the ratio of the marginal utilities: \( \text{MRS} = \frac{MU_t}{MU_s} = \frac{s}{t} \). 3. **Set the MRS equal to the price ratio**: - The price ratio of Twinkies to soda is \( \frac{0.10}{0.25} = 0.4 \). - Therefore, \( \frac{s}{t} = 0.4 \) which simplifies to \( s = 0.4t \). 4. **Substitute \( s = 0.4t \) into the budget constraint**: - The budget constraint is \( 0.10t + 0.25s = 1.00 \). - Substituting \( s \) gives \( 0.10t + 0.25(0.4t) = 1.00 \). - Simplifying, we get \( 0.10t + 0.10t = 0.20t = 1.00 \). - Solving for \( t \), we find \( t = \frac{1.00}{0.20} = 5 \). 5. **Verify the solution**: - Substituting \( t = 5 \) back into \( s = 0.4t \) gives \( s = 0.4 \times 5 = 2 \). - The total cost is \( 5 \times 0.10 + 2 \times 0.25 = 0.50 + 0.50 = 1.00 \), which fits the budget constraint. - The utility \( U = \sqrt{5 \times 2} = \sqrt{10} \approx 3.16 \) is maximized. Thus, Paul should buy \boxed{5} Twinkies to maximize his utility. |
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| gpto1 | To solve this problem, we need to determine the quantities of Twinkies (t) and soda (s) that maximize Paul's utility given his budget constraint. **Given:** - Utility function: \( U(t, s) = \sqrt{ts} \) - Price of Twinkies (P\_t): \$0.10 each - Price of soda (P\_s): \$0.25 per cup - Budget (I): \$1.00 **Objective:** Maximize \( U(t, s) \) subject to the budget constraint. ### **Step 1: Set Up the Budget Constraint** The budget constraint represents all combinations of Twinkies and soda that Paul can purchase with his \$1.00. \[ P_t \cdot t + P_s \cdot s = I \\ 0.10t + 0.25s = 1 \] ### **Step 2: Calculate the Marginal Utilities** First, find the marginal utilities (MU) with respect to Twinkies and soda. - Marginal Utility of Twinkies: \[ MU_t = \frac{\partial U}{\partial t} = \frac{1}{2}(t^{-1/2}s^{1/2}) = \frac{1}{2}\left(\frac{s}{t}\right)^{1/2} \] - Marginal Utility of Soda: \[ MU_s = \frac{\partial U}{\partial s} = \frac{1}{2}(t^{1/2}s^{-1/2}) = \frac{1}{2}\left(\frac{t}{s}\right)^{1/2} \] ### **Step 3: Compute the Marginal Rate of Substitution (MRS)** The Marginal Rate of Substitution is the rate at which Paul is willing to substitute Twinkies for soda while maintaining the same level of utility. \[ MRS = \frac{MU_t}{MU_s} = \frac{\left(\frac{1}{2}\left(\frac{s}{t}\right)^{1/2}\right)}{\left(\frac{1}{2}\left(\frac{t}{s}\right)^{1/2}\right)} = \frac{s}{t} \] ### **Step 4: Apply the Utility Maximization Condition** At the utility-maximizing bundle, the MRS equals the ratio of the prices: \[ MRS = \frac{P_t}{P_s} \] Substitute the known prices: \[ \frac{s}{t} = \frac{0.10}{0.25} = 0.4 \] This gives us a relationship between s and t: \[ s = 0.4t \] ### **Step 5: Substitute Back into the Budget Constraint** Now, substitute \( s = 0.4t \) into the budget constraint: \[ 0.10t + 0.25(0.4t) = 1 \\ 0.10t + 0.10t = 1 \\ 0.20t = 1 \\ t = \frac{1}{0.20} \\ t = 5 \] ### **Step 6: Calculate the Quantity of Soda** Using \( s = 0.4t \): \[ s = 0.4 \times 5 = 2 \] ### **Conclusion** To maximize his utility, Paul should purchase: - **5 Twinkies** - **2 cups of soda** **Answer:** **5 Twinkies** |
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| GPT-4o | To maximize Paul's utility given his budget constraint, we need to solve the utility maximization problem. Let's denote the number of Twinkies Paul buys as \( t \) and the number of sodas as \( s \). The utility function is given by \( U(t, s) = \sqrt{ts} \). Given the prices: - Twinkies cost $0.10 each. - Soda costs $0.25 per cup. Paul's budget is $1. Therefore, the budget constraint can be written as: \[ 0.10t + 0.25s = 1 \] We need to maximize \( U(t, s) = \sqrt{ts} \) subject to the budget constraint. First, let's express \( s \) in terms of \( t \) using the budget constraint: \[ 0.10t + 0.25s = 1 \] \[ 0.25s = 1 - 0.10t \] \[ s = \frac{1 - 0.10t}{0.25} \] \[ s = \frac{1 - 0.10t}{0.25} = 4 - 0.4t \] Now, substitute this expression for \( s \) into the utility function: \[ U(t, s) = \sqrt{t(4 - 0.4t)} \] \[ U(t) = \sqrt{4t - 0.4t^2} \] To find the maximum utility, we need to take the derivative of \( U(t) \) with respect to \( t \) and set it to zero: \[ U(t) = \sqrt{4t - 0.4t^2} \] Let \( f(t) = 4t - 0.4t^2 \). Then we have: \[ U(t) = \sqrt{f(t)} \] The critical points occur where the derivative of \( f(t) \) is zero: \[ f(t) = 4t - 0.4t^2 \] \[ f'(t) = 4 - 0.8t \] \[ 4 - 0.8t = 0 \] \[ 0.8t = 4 \] \[ t = \frac{4}{0.8} = 5 \] We need to check whether \( t = 5 \) satisfies the budget constraint: \[ s = 4 - 0.4t \] \[ s = 4 - 0.4(5) \] \[ s = 4 - 2 = 2 \] Now check if the budget constraint is satisfied: \[ 0.10(5) + 0.25(2) = 0.50 + 0.50 = 1 \] The budget constraint is satisfied. Therefore, Paul should buy 5 Twinkies and 2 sodas to maximize his utility. |
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| llama3.1-70B-Instruct | A classic problem in microeconomics! Let's break it down step by step: **Given:** * Utility function: U(t,s) = √(ts) * Price of Twinkies: $0.10 each * Price of soda: $0.25 per cup * Budget: $1 **Objective:** Maximize utility (U) subject to the budget constraint. **Step 1: Write down the budget constraint** Let t be the number of Twinkies and s be the number of cups of soda. The total cost of Twinkies and soda must not exceed $1: 0.10t + 0.25s ≤ 1 **Step 2: Express s in terms of t** From the budget constraint, we can express s in terms of t: s = (1 - 0.10t) / 0.25 s = 4 - 0.4t **Step 3: Substitute s into the utility function** Now, substitute the expression for s into the utility function: U(t,s) = √(ts) = √(t(4 - 0.4t)) = √(4t - 0.4t^2) **Step 4: Maximize utility** To maximize utility, we need to find the value of t that maximizes the expression inside the square root. Take the derivative of the expression with respect to t and set it equal to 0: d(4t - 0.4t^2)/dt = 0 4 - 0.8t = 0 t = 5 **Step 5: Check the second-order condition** To ensure that t = 5 is a maximum, we need to check the second-order condition: d^2(4t - 0.4t^2)/dt^2 = -0.8 < 0 This confirms that t = 5 is a maximum. **Answer:** Paul should buy 5 Twinkies to maximize his utility. |
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| llama3.1-405B-Instruct | To maximize utility, we need to find the values of t (Twinkies) and s (soda) that maximize the utility function U(t,s) = √(ts), subject to the budget constraint 0.10t + 0.25s = 1. First, let's rewrite the budget constraint as s = (1 - 0.10t) / 0.25. Now, substitute this expression for s into the utility function: U(t,s) = √(ts) = √(t(1 - 0.10t) / 0.25) To find the maximum utility, we need to find the value of t that maximizes this expression. We can do this by taking the derivative of U with respect to t and setting it equal to 0: dU/dt = d(√(t(1 - 0.10t) / 0.25))/dt = 0 Simplifying and solving for t, we get: t = 5 So, Paul should buy 5 Twinkies to maximize his utility. Note: This solution assumes that Paul can buy a fraction of a cup of soda, which may not be realistic. If we restrict s to be an integer, the solution would be slightly different. |
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| DeepSeek-Coder-V2 | To solve the problem of how many Twinkies Paul should buy to maximize his utility, we need to set up and solve a constrained optimization problem. Here are the steps: 1. **Define the Utility Function and Budget Constraint:** - Utility function: \( U(t, s) = \sqrt{ts} \) - Budget constraint: \( 0.10t + 0.25s = 1 \) 2. **Express the Budget Constraint in Terms of One Variable:** - Solve for \( s \) in terms of \( t \): \[ 0.25s = 1 - 0.10t \] \[ s = \frac{1 - 0.10t}{0.25} \] \[ s = 4 - 0.4t \] 3. **Substitute \( s \) into the Utility Function:** - Substitute \( s = 4 - 0.4t \) into \( U(t, s) \): \[ U(t) = \sqrt{t(4 - 0.4t)} \] \[ U(t) = \sqrt{4t - 0.4t^2} \] 4. **Maximize the Utility Function:** - To find the maximum value of \( U(t) \), we need to take the derivative of \( U(t) \) with respect to \( t \) and set it to zero: \[ U(t) = \sqrt{4t - 0.4t^2} \] \[ \frac{dU}{dt} = \frac{1}{2\sqrt{4t - 0.4t^2}} \cdot (4 - 0.8t) \] \[ \frac{dU}{dt} = \frac{4 - 0.8t}{2\sqrt{4t - 0.4t^2}} \] Set the derivative equal to zero: \[ \frac{4 - 0.8t}{2\sqrt{4t - 0.4t^2}} = 0 \] \[ 4 - 0.8t = 0 \] \[ 0.8t = 4 \] \[ t = 5 \] 5. **Verify the Solution:** - Substitute \( t = 5 \) back into the budget constraint to find \( s \): \[ 0.10(5) + 0.25s = 1 \] \[ 0.5 + 0.25s = 1 \] \[ 0.25s = 0.5 \] \[ s = 2 \] - Check the utility: \[ U(5, 2) = \sqrt{5 \cdot 2} = \sqrt{10} \] Thus, Paul should buy 5 Twinkies to maximize his utility. |
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| Qwen2-72B-Instruct | To maximize Paul's utility given his budget constraint, we need to find the optimal combination of Twinkies (t) and sodas (s) that he can purchase with his $1. The utility function is given by U(t,s) = \sqrt{ts}, and the price of Twinkies is $0.10 each, and the price of soda is $0.25 per cup. Paul's budget is $1. First, let's express the budget constraint in terms of t and s: 0.10t + 0.25s = 1 We need to solve this equation for one of the variables, let's solve for s: s = (1 - 0.10t) / 0.25 s = 4 - 0.4t Now we will substitute this expression for s into the utility function: U(t, 4 - 0.4t) = \sqrt{t(4 - 0.4t)} To find the maximum of this function, we need to take the derivative with respect to t and set it equal to zero: dU/dt = (1/2)(4 - 0.4t)^(-1/2)(4 - 0.8t) = 0 Solving this equation for t, we get: 4 - 0.8t = 0 0.8t = 4 t = 4 / 0.8 t = 5 So Paul should buy 5 Twinkies to maximize his utility. To find the number of sodas he can buy with the remaining budget, we substitute t = 5 back into the budget constraint equation: 0.10(5) + 0.25s = 1 0.5 + 0.25s = 1 0.25s = 0.5 s = 0.5 / 0.25 s = 2 So Paul should buy 5 Twinkies and 2 sodas to maximize his utility given his budget constraint. |
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| baichuan4 | To find the number of Twinkies that maximizes Paul's utility, we need to formulate an optimization problem. We are given the utility function: \[U(t, s) = \sqrt{ts}\] And the budget constraint: \[0.10t + 0.25s \leq 1\] Where \(t\) is the number of Twinkies and \(s\) is the number of cups of soda. First, let's normalize the budget constraint to equal one by dividing all terms by 0.10: \[t + 2.5s \leq 10\] Now, we can express \(s\) in terms of \(t\): \[s \leq \frac{10 - t}{2.5}\] Substitute this into the utility function: \[U(t) = \sqrt{t\left(\frac{10 - t}{2.5}\right)}\] Simplify: \[U(t) = \sqrt{\frac{10t - t^2}{2.5}}\] To find the maximum utility, we differentiate \(U(t)\) with respect to \(t\) and set it equal to zero: \[U'(t) = \frac{1}{2}\left(\frac{10t - t^2}{2.5}\right)^{-1/2} \cdot \frac{d}{dt}(10t - t^2) \cdot \frac{1}{2.5} = 0\] \[\frac{1}{2}\left(\frac{10t - t^2}{2.5}\right)^{-1/2} \cdot (10 - 2t) \cdot \frac{1}{2.5} = 0\] The term \((10t - t^2)/2.5\) cannot be zero because it represents a quantity of soda which must be non-negative. Therefore, we need the term \(10 - 2t\) to be zero: \[10 - 2t = 0\] \[2t = 10\] \[t = 5\] So, Paul should buy 5 Twinkies to maximize his utility. Let's check if this satisfies the budget constraint: \[0.10 \times 5 + 0.25s \leq 1\] \[0.5 + 0.25s \leq 1\] \[0.25s \leq 0.5\] \[s \leq 2\] Therefore, Paul can also buy 2 cups of soda, which satisfies the budget constraint and maximizes his utility. The utility at this point is: \[U(5, 2) = \sqrt{5 \times 2} = \sqrt{10}\] In conclusion, to maximize his utility, Paul should buy 5 Twinkies and 2 cups of soda, which will give him a utility of \(\sqrt{10}\). |
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| doubao | The budget constraint is 0.10t + 0.25s = 1. We want to maximize U(t,s) = √(ts). From the budget constraint, we can express s in terms of t: s = (1 - 0.10t) / 0.25. Substituting this into the utility function gives U(t) = √[t * ((1 - 0.10t) / 0.25)]. To find the maximum, take the derivative of U(t) with respect to t, set it equal to 0, and solve for t. After doing the calculations, we find the value of t that maximizes utility. |
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| minimax | To solve this problem, we need to maximize Paul's utility subject to his budget constraint. Let's denote the number of Twinkies Paul buys as \( t \) and the number of sodas as \( s \). The utility function is given by: \[ U(t,s) = \sqrt{ts} \] The budget constraint is: \[ 0.10t + 0.25s = 1 \] We need to maximize \( U(t,s) \) subject to the budget constraint. To do this, we can use the method of Lagrange multipliers or simply substitute the budget constraint into the utility function to express utility as a function of one variable. Let's solve for \( s \) in terms of \( t \) using the budget constraint: \[ 0.10t + 0.25s = 1 \] \[ 0.25s = 1 - 0.10t \] \[ s = \frac{1 - 0.10t}{0.25} \] \[ s = 4 - 0.4t \] Now we can express the utility function in terms of \( t \) only: \[ U(t) = \sqrt{t(4 - 0.4t)} \] To find the maximum utility, we need to take the derivative of \( U(t) \) with respect to \( t \) and set it equal to zero: \[ \frac{dU}{dt} = \frac{1}{2\sqrt{t(4 - 0.4t)}} \cdot (4 - 0.8t) \] Setting the derivative equal to zero to find the critical points: \[ \frac{1}{2\sqrt{t(4 - 0.4t)}} \cdot (4 - 0.8t) = 0 \] Since the denominator cannot be zero (as \( t \) must be positive), we only need to consider the numerator: \[ 4 - 0.8t = 0 \] \[ 0.8t = 4 \] \[ t = \frac{4}{0.8} \] \[ t = 5 \] Paul can buy up to 5 Twinkies with his $1, but we need to check if this is the maximum utility point or if it lies on the boundary of his budget constraint. To do this, we can check the utility at \( t = 5 \) and compare it with the utility at the boundary points (where he spends all his money on either Twinkies or soda). If Paul spends all his money on Twinkies (\( t = 10 \), since each costs $0.10), he cannot buy any soda, and his utility would be: \[ U(10,0) = \sqrt{10 \cdot 0} = 0 \] If Paul spends all his money on soda (\( s = 4 \), since each cup costs $0.25), he cannot buy any Twinkies, and his utility would be: \[ U(0,4) = \sqrt{0 \cdot 4} = 0 \] Clearly, the utility at \( t = 5 \) and \( s = 4 - 0.4 \cdot 5 = 2 \) will be higher than these boundary points. Let's calculate it: \[ U(5,2) = \sqrt{5 \cdot 2} = \sqrt{10} \] Since \( \sqrt{10} \) is greater than 0, Paul maximizes his utility by buying 5 Twinkies and 2 sodas, which gives him a utility of \( \sqrt{10} \). |
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| kimi | To solve this problem, we need to find the optimal number of Twinkies (t) and sodas (s) that Paul can buy with his $1 budget that maximizes his utility function U(t, s) = √(ts). Let's denote the cost of Twinkies as C_t = $0.10 and the cost of soda as C_s = $0.25. Paul's budget is B = $1. The total cost equation is: C_t * t + C_s * s = B Substituting the given values, we get: 0.10t + 0.25s = 1 Now, we need to express s in terms of t (or vice versa) to find the relationship between the two: 0.25s = 1 - 0.10t s = (1 - 0.10t) / 0.25 s = 4 - 0.4t Now, we substitute s in the utility function: U(t, s) = √(t * (4 - 0.4t)) To maximize the utility, we can take the derivative of U with respect to t and set it to zero: dU/dt = (1/2) * (t * (4 - 0.4t))^(-1/2) * (4 - 0.8t) Setting dU/dt = 0: 0 = (1/2) * (t * (4 - 0.4t))^(-1/2) * (4 - 0.8t) This equation is a bit complex to solve algebraically, so we can use a numerical method to find the value of t that maximizes the utility. However, we can also use some intuition here: since the cost of soda is higher than Twinkies, Paul should buy fewer sodas and more Twinkies to maximize his utility. Let's try to find the maximum utility by testing different combinations of t and s that fit within the budget. We can start by finding the maximum number of Twinkies Paul can buy without buying any sodas: 0.10t = 1 t = 10 If Paul buys 10 Twinkies, he will have no money left for sodas, and the utility would be: U(10, 0) = √(10 * 0) = 0 Now let's try to buy one less Twinkie and see if we can afford one soda: 0.10(9) + 0.25(1) = 0.90 + 0.25 = 1.15, which is over the budget. So, let's try one less Twinkie again and buy one soda: 0.10(8) + 0.25(1) = 0.80 + 0.25 = 1.05, which is still over the budget. We continue this process until we find a combination that fits within the budget: 0.10(7) + 0.25(1) = 0.70 + 0.25 = 0.95, which is under the budget. So, Paul can buy 7 Twinkies and 1 soda for a total cost of $0.95, which leaves him with $0.05. The utility for this combination is: U(7, 1) = √(7 * 1) = √7 ≈ 2.64575 Since the utility function is continuous and differentiable, we can be confident that this combination is close to the maximum utility given the budget constraints. However, to find the exact maximum, we would need to use calculus or numerical methods to solve the optimization problem. In conclusion, Paul should buy 7 Twinkies and 1 soda to maximize his utility under the given budget constraints. |
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| wenxinyiyan | To solve Paul's problem, we need to maximize his utility function U(t, s) = √(ts) under the constraint that the total cost of the Twinkies and soda does not exceed $1. The cost of each Twinkie is $0.10, and the cost of each cup of soda is $0.25. Let's set up the budget constraint: 0.1t + 0.25s ≤ 1 Paul's goal is to maximize U(t, s) = √(ts). We can rewrite this as: Maximize U^2 = ts Since the utility function is concave, we can use the method of Lagrange multipliers to solve this optimization problem. However, given the simplicity of the problem, we can also solve it directly by expressing s in terms of t using the budget constraint and then substituting it into the utility function. Let's express s in terms of t: s = (1 - 0.1t) / 0.25 Now, substitute this expression for s into the utility function: U^2 = t * (1 - 0.1t) / 0.25 To find the maximum utility, we take the derivative of U^2 with respect to t and set it to zero: d(U^2) / dt = (1 - 0.2t) / 0.25 = 0 Solving for t: 1 - 0.2t = 0 t = 5 So, Paul should buy 5 Twinkies to maximize his utility. Note that this solution assumes that Paul can only buy whole numbers of Twinkies and cups of soda, and it does not account for the possibility of buying fractional amounts. However, since the problem is framed in terms of a third-grade student's lunch choices, it is reasonable to assume that Paul can only buy whole numbers of each item. |
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| zhipu | To determine the number of Twinkies Paul should buy to maximize his utility, we need to maximize the utility function U(t, s) = √(ts) subject to his budget constraint. **Step 1: Define the budget constraint.** Paul's mom gives him $1.00 to spend on Twinkies and soda. If Twinkies cost $0.10 each and soda costs $0.25 per cup, we can write the budget constraint as: \[ 0.10t + 0.25s = 1 \] **Step 2: Solve the budget constraint for one of the goods.** Let's solve for soda (s) in terms of Twinkies (t): \[ 0.25s = 1 - 0.10t \] \[ s = \frac{1 - 0.10t}{0.25} \] \[ s = 4 - 0.4t \] **Step 3: Substitute the constraint into the utility function.** Now, substitute the expression for s into the utility function: \[ U(t, s) = \sqrt{ts} = \sqrt{t(4 - 0.4t)} \] **Step 4: Maximize the utility function.** To maximize this function, we can take the derivative with respect to t and set it to zero: \[ \frac{dU}{dt} = \frac{d}{dt} \left( \sqrt{t(4 - 0.4t)} \right) \] Using the chain rule: \[ \frac{dU}{dt} = \frac{1}{2} \left( t(4 - 0.4t) \right)^{-\frac{1}{2}} \cdot \frac{d}{dt} \left( t(4 - 0.4t) \right) \] First, find the derivative of the inside function: \[ \frac{d}{dt} \left( t(4 - 0.4t) \right) = 4 - 0.8t \] Now, substitute back into the derivative of the utility function: \[ \frac{dU}{dt} = \frac{1}{2} \left( t(4 - 0.4t) \right)^{-\frac{1}{2}} \cdot (4 - 0.8t) \] Set the derivative equal to zero to find the critical points: \[ \frac{1}{2} \left( t(4 - 0.4t) \right)^{-\frac{1}{2}} \cdot (4 - 0.8t) = 0 \] Since the term \(\left( t(4 - 0.4t) \right)^{-\frac{1}{2}}\) is never zero (it is a square root), we focus on: \[ 4 - 0.8t = 0 \] \[ 0.8t = 4 \] \[ t = 5 \] **Step 5: Verify the second derivative (optional but good practice).** To ensure this critical point is a maximum, we can check the second derivative: \[ \frac{d^2U}{dt^2} = \frac{d}{dt} \left( \frac{1}{2} \left( t(4 - 0.4t) \right)^{-\frac{1}{2}} \cdot (4 - 0.8t) \right) \] However, for simplicity, we can verify by checking the endpoints and ensuring the critical point is within the budget constraint. **Step 6: Check the budget constraint.** If t = 5: \[ 0.10(5) + 0.25s = 1 \] \[ 0.50 + 0.25s = 1 \] \[ 0.25s = 0.50 \] \[ s = 2 \] This satisfies the budget constraint. **Conclusion:** Paul should buy 5 Twinkies to maximize his utility. This combination allows him to spend his entire budget and achieve the highest utility given the prices of Twinkies and soda. |
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| gpto1 | To solve this problem, we need to determine the quantities of Twinkies (t) and soda (s) that maximize Paul's utility given his budget constraint. **Given:** - Utility function: \( U(t, s) = \sqrt{ts} \) - Price of Twinkies (P\_t): \$0.10 each - Price of soda (P\_s): \$0.25 per cup - Budget (I): \$1.00 **Objective:** Maximize \( U(t, s) \) subject to the budget constraint. ### **Step 1: Set Up the Budget Constraint** The budget constraint represents all combinations of Twinkies and soda that Paul can purchase with his \$1.00. \[ P_t \cdot t + P_s \cdot s = I \\ 0.10t + 0.25s = 1 \] ### **Step 2: Calculate the Marginal Utilities** First, find the marginal utilities (MU) with respect to Twinkies and soda. - Marginal Utility of Twinkies: \[ MU_t = \frac{\partial U}{\partial t} = \frac{1}{2}(t^{-1/2}s^{1/2}) = \frac{1}{2}\left(\frac{s}{t}\right)^{1/2} \] - Marginal Utility of Soda: \[ MU_s = \frac{\partial U}{\partial s} = \frac{1}{2}(t^{1/2}s^{-1/2}) = \frac{1}{2}\left(\frac{t}{s}\right)^{1/2} \] ### **Step 3: Compute the Marginal Rate of Substitution (MRS)** The Marginal Rate of Substitution is the rate at which Paul is willing to substitute Twinkies for soda while maintaining the same level of utility. \[ MRS = \frac{MU_t}{MU_s} = \frac{\left(\frac{1}{2}\left(\frac{s}{t}\right)^{1/2}\right)}{\left(\frac{1}{2}\left(\frac{t}{s}\right)^{1/2}\right)} = \frac{s}{t} \] ### **Step 4: Apply the Utility Maximization Condition** At the utility-maximizing bundle, the MRS equals the ratio of the prices: \[ MRS = \frac{P_t}{P_s} \] Substitute the known prices: \[ \frac{s}{t} = \frac{0.10}{0.25} = 0.4 \] This gives us a relationship between s and t: \[ s = 0.4t \] ### **Step 5: Substitute Back into the Budget Constraint** Now, substitute \( s = 0.4t \) into the budget constraint: \[ 0.10t + 0.25(0.4t) = 1 \\ 0.10t + 0.10t = 1 \\ 0.20t = 1 \\ t = \frac{1}{0.20} \\ t = 5 \] ### **Step 6: Calculate the Quantity of Soda** Using \( s = 0.4t \): \[ s = 0.4 \times 5 = 2 \] ### **Conclusion** To maximize his utility, Paul should purchase: - **5 Twinkies** - **2 cups of soda** **Answer:** **5 Twinkies** |
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