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

To find \( \frac{dy}{dx} \), we will implicitly differentiate the equation \( 40x^{1/3}y^{2/3} = 480 \). First, we simplify the equation: \[ x^{1/3}y^{2/3} = 12 \] Now, we differentiate both sides with respect to \( x \): \[ \frac{d}{dx}(x^{1/3}y^{2/3}) = 0 \] Using the product rule: \[ \frac{1}{3}x^{-2/3}y^{2/3} + x^{1/3} \cdot \frac{2}{3}y^{-1/3}\frac{dy}{dx} = 0 \] Rearranging gives: \[ \frac{2}{3}x^{1/3}y^{-1/3}\frac{dy}{dx} = -\frac{1}{3}x^{-2/3}y^{2/3} \] Now we solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = -\frac{y^{2/3} \cdot x^{-2/3}}{2 \cdot x^{1/3} \cdot y^{-1/3}} \cdot 3 \] This simplifies to: \[ \frac{dy}{dx} = -\frac{y^{2/3} \cdot 3}{2x^{1/3} \cdot y^{-1/3} \cdot x^{2/3}} = -\frac{3y^{1}}{2x^{1}} \] Now we substitute \( x = 64 \) and \( y = 27 \): \[ \frac{dy}{dx} = -\frac{3 \cdot 27}{2 \cdot 64} = -\frac{81}{128} \] Thus, the derivative evaluated at the point \( (64, 27) \) is: \[ \frac{dy}{dx} = -\frac{81}{128} \] ### Interpretation The value \( -\frac{81}{128} \) indicates that if the manufacturer increases spending on labor \( x \) by 1 thousand dollars, the amount spent on capital \( y \) must decrease by approximately \( 0.6328 \) thousand dollars (or \$632.80) to maintain the same production level of cars. This negative relationship reflects the trade-off between labor and capital investments in production under the given constraints.