Problem 36 - 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. $$U=f(\ell, g)=32 \ell^{2 / 3} g^{1 / 3} \text { subject to } 4 \ell+2 g=12$$

Step-by-Step Solution

Answer

Answer: The values of l and g that maximize the utility function are l = 2 and g = 2, with a maximum utility of 64.

Step 1: Formulate the Lagrangian function

The Lagrangian function \(L(\ell, g, \lambda)\) is defined as the utility function plus a multiplier (Lagrange multiplier, \(\lambda\)) times the constraint. In this case, it is given by: $$L(\ell, g, \lambda) = 32\ell^{2/3}g^{1/3} - \lambda(4\ell + 2g - 12)$$

Step 2: Find the partial derivatives

Next, we'll find the partial derivatives of the Lagrangian function with respect to each variable, \(\ell\), \(g\), and \(\lambda\). This will give us: $$\frac{\partial L}{\partial \ell} = \frac{64}{3}\ell^{-1/3}g^{1/3} - 4\lambda$$ $$\frac{\partial L}{\partial g} = \frac{32}{3}\ell^{2/3}g^{-2/3} - 2\lambda$$ $$\frac{\partial L}{\partial \lambda} = 4\ell + 2g - 12$$

Step 3: Set the partial derivatives equal to zero

Now, we will set each partial derivative equal to zero to find the critical points: $$\frac{64}{3}\ell^{-1/3}g^{1/3} - 4\lambda = 0$$ $$\frac{32}{3}\ell^{2/3}g^{-2/3} - 2\lambda = 0$$ $$4\ell + 2g - 12 = 0$$

Step 4: Determine the feasible critical points

Using the constraint, we can solve for one variable in terms of the other: $$4\ell + 2g = 12 \Rightarrow g = 6 - 2\ell$$ Now, substitute this expression for \(g\) into the first two partial derivative equations and solve for \(\ell\) and \(g\): $$\frac{64}{3}\ell^{-1/3}(6 - 2\ell)^{1/3} - 4\lambda = 0$$ $$\frac{32}{3}\ell^{2/3}(6 - 2\ell)^{-2/3} - 2\lambda = 0$$ By solving these equations simultaneously, we find the critical point \((\ell, g, \lambda) = (2, 2, 8)\).

Step 5: Find the value of the utility function at the optimal point

We found that the optimal point is \((\ell, g) = (2, 2)\). Now, we will plug these values into the utility function to find the value of the utility at this point: $$U(2, 2) = 32(2^{2/3})(2^{1/3}) = 32(2^{1}) = 32(2) = 64$$ Thus, the value of the utility function at the optimal point is 64. The values of \(\ell\) and \(g\) that maximize the utility function, subject to the given constraint, are \(\ell = 2\) and \(g = 2\).

Key Concepts

Utility Function

A utility function is a mathematical representation used in economics and consumer theory to describe an individual's preference for goods or combinations of goods. It measures the level of satisfaction or happiness a consumer receives from consuming certain bundles of goods. In this problem, the utility function is given as:

\[ U = f(\ell, g) = 32\ell^{2/3}g^{1/3} \]
This function suggests that utility depends on two variables: \( \ell \) and \( g \). Here, \( \ell \) and \( g \) could represent quantities of different goods. The idea is to find the optimal mix of these goods that yields maximum utility.

Key characteristics of utility functions include:

• Non-satiation: More is typically better; higher values for goods lead to higher utility.
• Substitutability: Goods can often be substituted for one another up to a certain point, as shown by different exponents in the function.
• Cardinal Utility: While utility can be measured numerically, the numbers don't necessarily have inherent meaning—only the ordering matters.

Partial Derivatives

Partial derivatives are a fundamental tool in multivariable calculus, used to understand how a function changes as each variable changes while all other variables are held constant. In the context of optimization problems, they help determine how changes in input variables affect the overall function value.

For the Lagrangian function \( L(\ell, g, \lambda) = 32\ell^{2/3}g^{1/3} - \lambda(4\ell + 2g - 12) \), we compute partial derivatives with respect to each variable. This gives us insights into how changes in \( \ell \), \( g \), and the multiplier \( \lambda \) influence the function:

• Partial derivative with respect to \( \ell \): This tells us how the Lagrangian changes with changes in \( \ell \), and is given by:
  \[ \frac{\partial L}{\partial \ell} = \frac{64}{3}\ell^{-1/3}g^{1/3} - 4\lambda \]
• Partial derivative with respect to \( g \): This measures the impact of changes in \( g \) on the Lagrangian:
  \[ \frac{\partial L}{\partial g} = \frac{32}{3}\ell^{2/3}g^{-2/3} - 2\lambda \]
• Partial derivative with respect to \( \lambda \): It captures the effect of changes in \( \lambda \) (the constraint's influence) and is:
  \[ \frac{\partial L}{\partial \lambda} = 4\ell + 2g - 12 \]

Setting these derivatives to zero helps find critical points where the utility function reaches a maximum or minimum, considering constraints.

Critical Points

Critical points are values in the domain of a function where its derivative is zero or undefined. These points are significant because they often correspond to locations of maxima, minima, or saddle points of the function. In optimization problems like this one, identifying critical points can help us determine the optimal values for the variables that maximize or minimize the function given certain constraints.

To find critical points in our context, you solve the system of equations formed by setting the partial derivatives of the Lagrangian to zero. For instance:

• \( \frac{64}{3}\ell^{-1/3}g^{1/3} - 4\lambda = 0 \)
• \( \frac{32}{3}\ell^{2/3}g^{-2/3} - 2\lambda = 0 \)
• \( 4\ell + 2g - 12 = 0 \)

Solving these three equations simultaneously helps pinpoint the values of \( \ell \), \( g \), and \( \lambda \) that satisfy both the utility function and its constraints. As calculated, the critical point \( \ell = 2, g = 2, \lambda = 8 \) provides the maximum utility within the specified constraints. After plugging these values into the original utility function, you find that the utility at this point is 64, which is the maximum utility achievable under the given circumstances.