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. $$U=f(\ell, g)=8 \ell^{4 / 5} g^{1 / 5} \text { subject to } 10 \ell+8 g=40$$

Short answer

Answer: The values of \(\ell\) and \(g\) that maximize the utility function are \(\ell \approx \frac{20}{21}\) and \(g \approx \frac{80}{21}\), and the maximum value of the utility function is approximately \(16.105\).

Step-by-step solution

Formulate the Lagrangian

First, we need to formulate the Lagrangian, which is a function that combines the utility function and the constraint. We do this by adding the constraint multiplied by a Lagrange multiplier (\(\lambda\)) to the utility function. The Lagrangian \(L(\ell, g, \lambda)\) is given by: $$L(\ell, g, \lambda) = 8 \ell^{4 / 5} g^{1 / 5} + \lambda (40 - 10\ell - 8g)$$

Take partial derivatives

Now, we need to take the partial derivatives of the Lagrangian with respect to each of the three variables: \(\ell\), \(g\), and \(\lambda\). Then, we will set these partial derivatives equal to zero and solve for \(\ell\), \(g\), and \(\lambda\). The partial derivatives are: $$\frac{\partial L}{\partial \ell} = \frac{32}{5} \ell^{-1/5} g^{1/5} - 10\lambda$$ $$\frac{\partial L}{\partial g} = \frac{8}{5} \ell^{4/5} g^{-4/5} - 8\lambda$$ $$\frac{\partial L}{\partial \lambda} = 40 - 10\ell - 8g$$

Solve for \(\ell\), \(g\), and \(\lambda\)

Setting the partial derivatives equal to zero, we get the following system of equations: $$\frac{32}{5} \ell^{-1/5} g^{1/5} - 10\lambda = 0$$ $$\frac{8}{5} \ell^{4/5} g^{-4/5} - 8\lambda = 0$$ $$40 - 10\ell - 8g = 0$$ Divide the first equation by the second equation to eliminate \(\lambda\): $$\frac{\frac{32}{5} \ell^{-1/5} g^{1/5}}{\frac{8}{5} \ell^{4/5} g^{-4/5}} = \frac{10\lambda}{8\lambda} \Rightarrow 4\ell = g$$ Substitute \(g=4\ell\) into the third equation: $$40 - 10\ell - 8(4\ell) = 0 \Rightarrow 42\ell = 40 \Rightarrow \ell = \frac{20}{21}$$ Now, substitute the value of \(\ell\) into \(g=4\ell\) to find the value of \(g\): $$g = 4 \left(\frac{20}{21}\right) = \frac{80}{21}$$

Find the maximum utility

Finally, we have the values \(\ell = \frac{20}{21}\) and \(g = \frac{80}{21}\). Plug these values back into the utility function to find the maximum value of the utility: $$U = 8 \left(\frac{20}{21}\right)^{4 / 5} \left(\frac{80}{21}\right)^{1 / 5} \approx 16.105$$ Thus, the values of \(\ell\) and \(g\) that maximize the utility function are \(\ell \approx \frac{20}{21}\) and \(g \approx \frac{80}{21}\), and the maximum value of the utility function is approximately \(16.105\).

Key concepts

Utility Maximization

Utility maximization is a fundamental concept in economics. It involves selecting levels of consumption that provide the most satisfaction or utility to the consumer, given their budget constraint. In the context of this problem, the utility function is given by\[ U = 8 \ell^{4/5} g^{1/5} \]where \( \ell \) and \( g \) represent the quantities of two goods. The goal is to find the optimal combination of \( \ell \) and \( g \) that maximizes utility subject to the constraint \( 10\ell + 8g = 40 \).

The constraint reflects the consumer's budget, limiting the total expenditure on the two goods. By maximizing the utility function under this constraint, we can determine the most preferred combination of goods the consumer can afford. This involves using a mathematical approach, specifically Lagrange multipliers, to find the point where the consumer gets the maximum satisfaction possible within their budget.

Constrained Optimization

Constrained optimization is the process of maximizing or minimizing an objective function subject to constraints. In economics, it often appears in problems where resources are limited. For instance, maximizing utility while being bounded by budget limitations, as shown by the constraint equation \(10\ell + 8g = 40\).

To solve such a problem, we formulate a Lagrangian. The Lagrangian integrates the utility function and the constraint using a Lagrange multiplier \(\lambda\). The Lagrangian for the given problem is:\[ L(\ell, g, \lambda) = 8 \ell^{4/5} g^{1/5} + \lambda (40 - 10\ell - 8g) \]The role of \(\lambda\) is to adjust the utility function to account for the constraint, essentially balancing the trade-offs between the two goods within the budget. By doing this, we find the optimal choice of goods within given constraints.

This method allows us to reformulate a constrained problem into a broader framework of optimization, solving for points where the constraint is efficiently met while maximizing utility.

Partial Derivatives

Partial derivatives are a crucial mathematical tool used to understand how a multivariable function changes in relation to one of its variables, holding the others constant. In the context of Lagrange multipliers and optimization, partial derivatives help define the characteristics of the utility function and the constraint.

After setting up the Lagrangian \(L(\ell, g, \lambda)\), we take partial derivatives with respect to \(\ell\), \(g\), and \(\lambda\):

• \( \frac{\partial L}{\partial \ell} = \frac{32}{5} \ell^{-1/5} g^{1/5} - 10\lambda \)
• \( \frac{\partial L}{\partial g} = \frac{8}{5} \ell^{4/5} g^{-4/5} - 8\lambda \)
• \( \frac{\partial L}{\partial \lambda} = 40 - 10\ell - 8g \)

Setting these partial derivatives equal to zero provides us with a system of equations. Solving this system helps identify the specific values of \(\ell\), \(g\), and \(\lambda\) that optimize the utility function. By efficiently navigating through each direction (axis of \(\ell\), \(g\) and respecting the constraint), we use these derivatives to pinpoint the maximum utility value within the constraint.