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.
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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.
