Problem 38 - Calculus Early Transcendentals

Question

Maximizing utility functions Find the values of \(\ell\) and \(g\) with \(\ell \geq 0\) and \(g \geq 0\) that maximize the following utility functions subject to the given constraints. Give the value of the utility function at the optimal point. $$U=f(\ell, g)=\ell^{1 / 6} g^{5 / 6} \text { subject to } 4 \ell+5 g=20$$

Step-by-Step Solution

Answer

Answer: The optimal values are \(g = \frac{4}{3}\) and \(\ell = \frac{20}{3}\), and the maximum value of the utility function is \(U_{\text{max}} \approx 2.3707\).

Step 1: 1. Understand the problem and create a Lagrangian function

Given a utility function \(U(\ell, g) = \ell^{1/6} g^{5/6}\) that we need to maximize, subject to the constraint \(4\ell + 5g = 20\), we first formulate a new function called the Lagrangian function as follows: $$L(\ell, g, \lambda) = U(\ell, g) + \lambda(20 - 4\ell - 5g)$$ Note that \(\lambda\) is the Lagrange multiplier.

Step 2: 2. Partial derivative w.r.t. \(\ell\), \(g\), and \(\lambda\)

The next step is to take the partial derivative of L with respect to \(\ell\), \(g\), and \(\lambda\), and set each derivative equal to 0: $$\frac{\partial L}{\partial \ell} = \frac{1}{6}\ell^{-5/6}g^{5/6} - 4\lambda = 0$$ $$\frac{\partial L}{\partial g} = \frac{5}{6}\ell^{1/6}g^{-1/6} - 5\lambda = 0$$ $$\frac{\partial L}{\partial \lambda} = 20 - 4\ell - 5g = 0$$

Step 3: 3. Solve for \(\ell\), \(g\), and \(\lambda\)

Now, we have a system of three equations and three unknown variables: \(\ell\), \(g\), and \(\lambda\). We need to solve this system. First, eliminate \(\lambda\) by dividing the first equation by the second equation: $$\frac{\frac{1}{6}\ell^{-5/6}g^{5/6} - 4\lambda}{\frac{5}{6}\ell^{1/6}g^{-1/6} - 5\lambda} = \frac{4}{5}$$ This equation simplifies to: $$\ell = 5g$$ Using the constraint equation, substitute \(\ell\) by \(5g\): $$4(5g) + 5g = 20 \Rightarrow g = \frac{4}{3}$$ Finally, substitute the value of \(g\) back into the equation \(\ell = 5g\): $$\ell = 5 \times \frac{4}{3} = \frac{20}{3}$$

Step 4: 4. Find the maximum utility function value

Now that we have the optimal values for \(\ell\) and \(g\), we can find the maximum value of the utility function. Plug these values back into the utility function: $$U_{\text{max}} = \left(\frac{20}{3}\right)^{1/6}\left(\frac{4}{3}\right)^{5/6}$$ Calculate \(U_{\text{max}}\): $$U_{\text{max}} \approx 2.3707$$ Thus, the optimal values are: \(g = \frac{4}{3}\), \(\ell = \frac{20}{3}\) and the maximum value of the utility function is \(U_{\text{max}} \approx 2.3707\).

Key Concepts

Utility Function

Utility functions are a fundamental concept in microeconomics and are used to represent consumer preferences. They assign a numerical value to a combination of goods or services, indicating the overall satisfaction or 'utility' a consumer derives from that combination. In this exercise, our utility function is given by \[U(\ell, g) = \ell^{1/6} g^{5/6}\]where \ell\ is one good and \g\ is another.
The exponents \(1/6\) and \(5/6\) signify the relative importance or preference for each good. A higher exponent implies a stronger preference for that good. The utility function describes how satisfaction changes as the amounts of each good change. By maximizing this function, we are essentially seeking the most preferable combination of goods \(\ell\) and \(g\) within given constraints.
Understanding this concept is crucial as it forms the basis for deciding how to allocate resources efficiently, given a constraint, such as a budget or supply limit.

Constrained Optimization

Constrained optimization is a method used to find the best solution to a problem that has specific restrictions. In economic contexts, it involves optimizing an objective, like maximizing utility, under constraints like income or resources. Here, our constraint is \[4\ell + 5g = 20\]which could represent a budget line where the cost of \(\ell\) and \(g\) must add up to 20 units. To solve such problems, we use the method of Lagrange multipliers. This involves introducing a Lagrange multiplier, \(\lambda\), which helps integrate the constraints into the optimization process. The method constructs a Lagrangian function:\[L(\ell, g, \lambda) = \ell^{1/6} g^{5/6} + \lambda (20 - 4\ell - 5g)\]By finding the maximum of this Lagrangian, while enforcing the constraint, we determine the optimal allocation that maximizes utility. This step ensures that we do not exceed the available resources while making decisions.

Partial Derivatives

Partial derivatives are a mathematical tool used to understand how a function changes as its variables change. They are essential in multivariable calculus when dealing with functions of several variables like utility functions. In this problem, partial derivatives help find the points at which the utility function is maximized under the given constraint.When we take the partial derivative of the Lagrangian \(L\) with respect to \(\ell\), \(g\), and \(\lambda\), we get:

• \(\frac{\partial L}{\partial \ell} = \frac{1}{6}\ell^{-5/6}g^{5/6} - 4\lambda = 0\)
• \(\frac{\partial L}{\partial g} = \frac{5}{6}\ell^{1/6}g^{-1/6} - 5\lambda = 0\)
• \(\frac{\partial L}{\partial \lambda} = 20 - 4\ell - 5g = 0\)

Each of these equations captures a condition needed to ensure maximum utility. Solving these gives the optimal values of \(\ell\), \(g\), and \(\lambda\). The partial derivatives reflect how small changes in \(\ell\) or \(g\) affect the utility, while ensuring the constraint is satisfied. This approach enables us to methodically arrive at the solution that provides the greatest utility under the limitations.