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12. The number of cell phones 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 \( 60 x^{\frac{3}{4}} y^{\frac{1}{4}}=3240 \). a) Find \( \frac{d y}{d x} \) and evaluate at the point \( (81,16) \). b) Interpret the result of part a.

Ask by Frank Gonzalez. in the United States
Mar 06,2025

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**a)** The derivative \(\frac{dy}{dx}\) is \(-\frac{16}{27}\) when \(x = 81\) and \(y = 16\). **b)** This means that for every dollar increase in labor spending, capital spending must decrease by approximately 59 cents to keep the number of cell phones produced constant.

Solution

Below is the step‐by‐step solution in markdown format. --- ### Problem Statement We are given the production function \[ 60\, x^{\frac{3}{4}}\, y^{\frac{1}{4}} = 3240, \] where \(x\) is the dollar amount spent on labor and \(y\) is the dollars spent on capital. We need to: a) Find the derivative \(\frac{dy}{dx}\) and evaluate it at the point \((81, 16)\). b) Interpret the result from part (a). --- ### Part (a): Finding \(\frac{dy}{dx}\) #### Step 1. Differentiate Implicitly Starting with \[ 60\, x^{\frac{3}{4}}\, y^{\frac{1}{4}} = 3240, \] differentiate both sides with respect to \(x\). The right-hand side is constant (3240), so its derivative is 0. Differentiate the left-hand side using the product rule, where \(u(x) = x^{\frac{3}{4}}\) and \(v(x) = y^{\frac{1}{4}}\) (note that \(y\) is a function of \(x\)): \[ \frac{d}{dx}\left[60\, u(x)\, v(x)\right] = 60\left[u'(x)\, v(x) + u(x)\, v'(x)\right]. \] Compute the derivatives: - \(u(x) = x^{\frac{3}{4}}\) so \[ u'(x) = \frac{3}{4}x^{\frac{3}{4}-1} = \frac{3}{4}x^{-\frac{1}{4}}. \] - \(v(x) = y^{\frac{1}{4}}\) so by the chain rule, \[ v'(x) = \frac{1}{4}y^{\frac{1}{4}-1}\frac{dy}{dx} = \frac{1}{4}y^{-\frac{3}{4}}\frac{dy}{dx}. \] Plug these back: \[ 60\left[\frac{3}{4} x^{-\frac{1}{4}}\, y^{\frac{1}{4}} + x^{\frac{3}{4}}\cdot\frac{1}{4} y^{-\frac{3}{4}}\frac{dy}{dx}\right] = 0. \] #### Step 2. Simplify the Equation Divide both sides by 60: \[ \frac{3}{4} x^{-\frac{1}{4}}\, y^{\frac{1}{4}} + \frac{1}{4} x^{\frac{3}{4}}\, y^{-\frac{3}{4}}\frac{dy}{dx} = 0. \] Multiply the entire equation by 4 to eliminate the fractions: \[ 3\, x^{-\frac{1}{4}}\, y^{\frac{1}{4}} + x^{\frac{3}{4}}\, y^{-\frac{3}{4}}\frac{dy}{dx} = 0. \] #### Step 3. Solve for \(\frac{dy}{dx}\) Isolate the term with \(\frac{dy}{dx}\): \[ x^{\frac{3}{4}}\, y^{-\frac{3}{4}}\frac{dy}{dx} = -3\, x^{-\frac{1}{4}}\, y^{\frac{1}{4}}. \] Divide both sides by \(x^{\frac{3}{4}} y^{-\frac{3}{4}}\): \[ \frac{dy}{dx} = -3\, x^{-\frac{1}{4}}\, y^{\frac{1}{4}} \cdot \frac{1}{x^{\frac{3}{4}}\, y^{-\frac{3}{4}}}. \] Combine the \(x\) and \(y\) exponents: \[ x^{-\frac{1}{4}} \div x^{\frac{3}{4}} = x^{-\frac{1}{4} - \frac{3}{4}} = x^{-1}, \] \[ y^{\frac{1}{4}} \cdot y^{\frac{3}{4}} = y^{\frac{1}{4} + \frac{3}{4}} = y. \] Thus, we obtain: \[ \frac{dy}{dx} = -3\, \frac{y}{x}. \] #### Step 4. Evaluate at the Point \((81, 16)\) Substitute \(x = 81\) and \(y = 16\) into the expression: \[ \frac{dy}{dx}\Bigg|_{(81,16)} = -3\, \frac{16}{81} = -\frac{48}{81}. \] Simplify the fraction by dividing numerator and denominator by 3: \[ -\frac{48}{81} = -\frac{16}{27}. \] So, \[ \frac{dy}{dx}\Bigg|_{(81,16)} = -\frac{16}{27}. \] --- ### Part (b): Interpretation of the Result The derivative \(\frac{dy}{dx} = -\frac{16}{27}\) at the point \((81, 16)\) tells us how the spending on capital \(y\) must adjust with respect to spending on labor \(x\) to keep the production level (the number of cell phones produced) constant. - The negative sign indicates an inverse relationship: as spending on labor increases, spending on capital must decrease to hold production constant. - The magnitude \(\frac{16}{27}\) means that for every 1 dollar increase in labor spending, the capital spending must decrease by approximately \(\frac{16}{27}\) dollars (roughly 59 cents). In economic terms, this is the marginal rate of technical substitution (MRTS) between labor and capital at the given point. It quantifies the trade-off between labor and capital in the production process. --- ### Final Answers **a)** \[ \frac{dy}{dx} = -\frac{16}{27} \quad \text{at the point} \quad (81, 16). \] **b)** This result means that at the production level corresponding to \(x = 81\) dollars spent on labor and \(y = 16\) dollars spent on capital, an increase of 1 dollar in labor costs requires a reduction of \(\frac{16}{27}\) dollars in capital spending to maintain the constant level of cell phone production. This reflects the marginal rate of technical substitution between labor and capital.

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To find \( \frac{dy}{dx} \), we first differentiate the given equation implicitly. Starting from: \[ 60 x^{\frac{3}{4}} y^{\frac{1}{4}} = 3240 \] We can rewrite it as: \[ x^{\frac{3}{4}} y^{\frac{1}{4}} = 54 \] Now, applying implicit differentiation with respect to \( x \): \[ \frac{3}{4} x^{-\frac{1}{4}} y^{\frac{1}{4}} + x^{\frac{3}{4}} \cdot \frac{1}{4} y^{-\frac{3}{4}} \frac{dy}{dx} = 0 \] Solving for \( \frac{dy}{dx} \): \[ \frac{1}{4} x^{\frac{3}{4}} y^{-\frac{3}{4}} \frac{dy}{dx} = -\frac{3}{4} x^{-\frac{1}{4}} y^{\frac{1}{4}} \] \[ \frac{dy}{dx} = -\frac{3 y^{\frac{1}{4}}}{3 x^{\frac{1}{4}} y^{\frac{3}{4}}} = -\frac{3}{3} \cdot \frac{y^{\frac{1}{4}}}{x^{\frac{1}{4}}} = -\frac{y^{\frac{1}{4}}}{x^{\frac{1}{4}}} \] Now, we evaluate \( \frac{dy}{dx} \) at the point \( (81, 16) \): Substituting \( x = 81 \) and \( y = 16 \): \[ \frac{dy}{dx} = -\frac{(16)^{\frac{1}{4}}}{(81)^{\frac{1}{4}}} = -\frac{2}{3} \] So, \( \frac{dy}{dx} = -\frac{2}{3} \). For interpretation, \( \frac{dy}{dx} = -\frac{2}{3} \) means that for each additional dollar spent on labor \( (x) \), the manufacturer must decrease the spending on capital \( (y) \) by \(\frac{2}{3}\) of a dollar to maintain the same level of cell phone production. This ratio illustrates the trade-off between labor and capital in the production process, highlighting the balance manufacturers need to achieve for optimal output.

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