Problem 324 - Calculus Volume 1

Question

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 \(30 x^{1 / 3} y^{2 / 3}=360\) \((\mathrm{Both} x \text { and } y \text { are measured in thousands of dollars.) }\) a. Find \(\frac{d y}{d x}\) and evaluate at the point \((27,8) .\) b. Interpret the result of a.

Step-by-Step Solution

Answer

At \((27, 8)\), for every increase in labor, capital should decrease by \( \frac{4}{27} \) to keep the production constant.

Step 1: Identify the function

The given function to model the number of cars is \(30 x^{1/3} y^{2/3} = 360\). First, solve this equation to isolate one variable in terms of the other.

Step 2: Simplify the function

Divide both sides of the equation by 30 to simplify the function:\[ x^{1/3} y^{2/3} = 12 \].

Step 3: Implicit differentiation

To find \( \frac{dy}{dx} \), use implicit differentiation on \( x^{1/3} y^{2/3} = 12 \) with respect to \( x \):Apply the product rule: \( \frac{d}{dx}(x^{1/3}) \cdot y^{2/3} + x^{1/3} \cdot \frac{d}{dx}(y^{2/3}) = 0 \).

Step 4: Calculate derivatives

Find derivatives: \( \frac{d}{dx}(x^{1/3}) = \frac{1}{3}x^{-2/3} \)and \( \frac{d}{dx}(y^{2/3}) = \frac{2}{3}y^{-1/3} \frac{dy}{dx} \).

Step 5: Substitute and solve for \(\frac{dy}{dx}\)

Substitute derivatives into the product rule equation: \( \frac{1}{3}x^{-2/3}y^{2/3} + x^{1/3} \cdot \frac{2}{3}y^{-1/3} \frac{dy}{dx} = 0 \). Reorganize to solve for \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = -\frac{y}{2x} \].

Step 6: Evaluate \(\frac{dy}{dx}\) at \((27, 8)\)

Substitute \( x = 27 \) and \( y = 8 \) into \( \frac{dy}{dx} = -\frac{y}{2x} \): \[ \frac{dy}{dx} = -\frac{8}{2 \cdot 27} \].Simplify to get \( \frac{dy}{dx} = -\frac{4}{27} \).

Step 7: Interpret the result

The rate of change \( \frac{dy}{dx} \) indicates how the capital \( y \) should change with respect to labor \( x \) to maintain production at a constant level. At the point \((27, 8)\), for every increase in \( x \), \( y \) should decrease by \( \frac{4}{27} \) of that increase.

Key Concepts

Rate of Change

In mathematics, the rate of change between two variables is a crucial concept. It tells us how one variable behaves as the other changes, often in relation to a real-world scenario. In our exercise, we are interested in how the capital investment, represented by \( y \), changes in response to changes in labor cost, represented by \( x \). This is captured by the derivative \( \frac{dy}{dx} \), which indicates the rate at which \( y \) changes as \( x \) changes, while keeping the output (number of cars) constant.

When we evaluate this derivative at a specific point, such as \((27, 8)\), we are looking at a snapshot of how much \( y \) should change for a slight increase or decrease in \( x \). Here, we find \( \frac{dy}{dx} = -\frac{4}{27} \), meaning if labor costs increase slightly, the capital investment must decrease by \( \frac{4}{27} \times \) the change in labor cost in order to maintain the same level of production. Understanding this concept helps industries optimize resource allocation to achieve efficient production.

Product Rule

In calculus, the product rule is a fundamental tool used to differentiate functions where two or more terms are multiplied together. When we perform differentiation, every term within the product must be considered. Specifically, if \( u(x) \) and \( v(x) \) are two differentiable functions, the product rule is given by:

• \( \frac{d}{dx}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x) \)

In the context of our problem, the expression \( x^{1/3} y^{2/3} \) calls for the use of the product rule to find the derivatives of each component. We break it down by:

• Differentiating \( x^{1/3} \), treating \( y^{2/3} \) as a constant, yielding \( \frac{1}{3} x^{-2/3} y^{2/3} \).
• Then differentiating \( y^{2/3} \), considering \( x^{1/3} \) as constant, resulting in \( \frac{2}{3} x^{1/3} y^{-1/3} \frac{dy}{dx} \).

These derivatives are then added together to apply the product rule in our implicit differentiation process.

Implicit Function

An implicit function is where the relationship between variables is not expressed as a single variable function, like \( y = f(x) \). Instead, it involves a function where \( x \) and \( y \) are interdependent, like in our example \( x^{1/3} y^{2/3} = 12 \). This means there is no simple solution for one variable without involving the other.

Implicit differentiation is a technique applied in such cases. Here, we differentiate each variable while recognizing their dependence. This approach allows us to solve for \( \frac{dy}{dx} \) without first needing to solve one variable in terms of the other. In practical terms, functions like these reflect complex realities, such as economic models where variables are naturally intertwined and cannot be separated neatly. By using implicit differentiation, we can understand how changes in one aspect (like labor costs) affect another (like capital investment) without needing an explicit function for \( y \) in terms of \( x \).

In our case, the implicit function ensures we understand how investments should adjust to maintain equilibrium in production, a critical insight for strategic decision-making.