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The number of cars produced when \( x \) dollars is spent on labor and \( y \) dollars is spent on capital invested by a manufacturer can be modeled by the equation \( 40 x^{1 / 3} y^{2 / 3}=480 \). (Both \( x \) and \( y \) are measured in thousands of dollars.) a. Find \( \frac{d y}{d x} \) and evaluate at the point \( (64,27) \). b. Interpret the result of a.

Ask by Lawson Warren. in the United States
Mar 17,2025

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**a. Finding \(\frac{dy}{dx}\) and evaluating at \((64,27)\):** Given the equation: \[ 40x^{1/3}y^{2/3} = 480 \] Differentiate both sides with respect to \(x\): \[ \frac{dy}{dx} = -\frac{y}{2x} \] At the point \((64,27)\): \[ \frac{dy}{dx} = -\frac{27}{128} \] **b. Interpretation:** The negative value of \(\frac{dy}{dx} = -\frac{27}{128}\) indicates that increasing labor spending by \$1,000 requires decreasing capital spending by approximately \$210.94 to maintain the same production level.

Solución

**a. Finding \(\frac{dy}{dx}\) and evaluating at \((64,27)\):** We start with the equation \[ 40x^{1/3}y^{2/3} = 480. \] Differentiate both sides with respect to \(x\). First, write the equation implicitly as \[ 40x^{1/3}y^{2/3} - 480 = 0. \] Differentiate term by term. The derivative of \(40x^{1/3}y^{2/3}\) using the product rule is: \[ \frac{d}{dx}\bigl(40x^{1/3}y^{2/3}\bigr) = 40\left(\frac{d}{dx}\bigl(x^{1/3}\bigr)y^{2/3} + x^{1/3}\frac{d}{dx}\bigl(y^{2/3}\bigr)\right). \] Differentiate the factors: - \(\frac{d}{dx}\bigl(x^{1/3}\bigr) = \frac{1}{3}x^{-2/3}\). - \(\frac{d}{dx}\bigl(y^{2/3}\bigr)\) requires the chain rule: \(\frac{d}{dx}\bigl(y^{2/3}\bigr) = \frac{2}{3}y^{-1/3}\frac{dy}{dx}\). Thus, \[ 40\left[\frac{1}{3}x^{-2/3}y^{2/3} + x^{1/3}\frac{2}{3}y^{-1/3}\frac{dy}{dx}\right] = 0. \] Simplify by dividing both sides by 40 and multiplying by 3: \[ x^{-2/3}y^{2/3} + 2x^{1/3}y^{-1/3}\frac{dy}{dx} = 0. \] Solve for \(\frac{dy}{dx}\): \[ 2x^{1/3}y^{-1/3}\frac{dy}{dx} = -x^{-2/3}y^{2/3}, \] \[ \frac{dy}{dx} = -\frac{x^{-2/3}y^{2/3}}{2x^{1/3}y^{-1/3}}. \] Now, combine the exponents: - For \(x\): \(x^{-2/3}/x^{1/3} = x^{-2/3-1/3} = x^{-1}\), - For \(y\): \(y^{2/3}/y^{-1/3} = y^{2/3+1/3} = y\). Thus, we have \[ \frac{dy}{dx} = -\frac{y}{2x}. \] Evaluating at the point \((x,y) = (64,27)\): \[ \frac{dy}{dx}\bigg|_{(64,27)} = -\frac{27}{2(64)} = -\frac{27}{128}. \] **b. Interpretation of the result:** The derivative \[ \frac{dy}{dx} = -\frac{27}{128} \] at the point \((64,27)\) represents the rate at which capital spending (\(y\), in thousands of dollars) must change with respect to a change in labor spending (\(x\), in thousands of dollars) in order to keep the production level constant. A value of \(-\frac{27}{128}\) means that if the manufacturer increases labor spending by one thousand dollars, capital spending must decrease by approximately \(\frac{27}{128}\) thousand dollars (about \$210.94) in order to maintain the same level of production. This negative relationship indicates a trade-off between spending on labor and spending on capital along a production isoquant.

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To find \( \frac{d y}{d x} \), we start from the given equation: \[ 40 x^{1 / 3} y^{2 / 3} = 480. \] First, we can simplify this to: \[ x^{1 / 3} y^{2 / 3} = 12. \] Next, we can differentiate both sides implicitly with respect to \( x \): 1. Differentiate the left side using the product rule: - The derivative of \( x^{1/3} \) is \( \frac{1}{3} x^{-2/3} \). - The derivative of \( y^{2/3} \) is \( \frac{2}{3} y^{-1/3} \frac{dy}{dx} \). So we have: \[ \frac{1}{3} x^{-2/3} y^{2/3} + x^{1/3} \cdot \frac{2}{3} y^{-1/3} \frac{dy}{dx} = 0. \] 2. Rearranging gives us: \[ x^{1/3} \cdot \frac{2}{3} y^{-1/3} \frac{dy}{dx} = -\frac{1}{3} x^{-2/3} y^{2/3}. \] 3. Now solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = -\frac{1}{2} \cdot \frac{y}{x} \cdot \left( \frac{y^{2/3}}{x^{2/3}} \right). \] Now substituting \( (x, y) = (64, 27) \): - Calculating \( x^{1/3} = 4 \) and \( y^{2/3} = 9 \). - Plugging in: \[ \frac{dy}{dx} = -\frac{9}{8}. \] Thus, \( \frac{dy}{dx} \) evaluated at the point \( (64, 27) \) is \(-\frac{9}{8}\). **Interpretation:** This means that for every increase of \$1,000 spent on labor (\(x\)), there is a decrease of approximately \$1,125 (or \( -\frac{9}{8} \) thousand dollars) in the amount spent on capital (\(y\)) to maintain the same level of car production. Essentially, it tells us how the investment in labor and capital are linked in producing a consistent number of cars.

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