Problem 374 - Calculus Volume 3

Question

Maximize \(U(x, y)=8 x^{4 / 5} y^{1 / 5} ; 4 x+2 y=12\).

Step-by-Step Solution

Answer

Maximize at \( x = 1.5 \), \( y = 3 \).

Step 1: Identify the Objective Function & Constraint

The given problem involves maximizing the utility function \( U(x, y) = 8x^{4/5}y^{1/5} \) subject to the constraint \( 4x + 2y = 12 \). The objective function \( U(x, y) \) is what we want to maximize, while the constraint limits the feasible values of \( x \) and \( y \).

Step 2: Convert the Constraint Equation for Substitution

Solve the constraint equation for one variable in terms of the other. In this case, solve for \( y \) in terms of \( x \): \[ y = 6 - 2x. \] This solution allows us to express \( U(x, y) \) with one variable, which simplifies the problem.

Step 3: Substitute the Constraint into the Objective

Substitute \( y = 6 - 2x \) into the utility function: \[ U(x, y) = 8x^{4/5}(6 - 2x)^{1/5}. \] Now the utility function is expressed in terms of \( x \) only, which makes it easier to apply calculus for maximization.

Step 4: Differentiate the Utility Function with Respect to x

Find the derivative of \( U(x) = 8x^{4/5}(6 - 2x)^{1/5} \) with respect to \( x \). Use the product rule and chain rule: \[ \frac{dU}{dx} = 8 \left[ \frac{4}{5}x^{-1/5}(6 - 2x)^{1/5} - \frac{2}{5}x^{4/5}(6 - 2x)^{-4/5} \right]. \]

Step 5: Find the Critical Points

Set the derivative equal to zero to find critical points: \[ \frac{dU}{dx} = 0. \] Solve for \( x \) which involves equating \( \frac{4}{5}x^{-1/5}(6 - 2x)^{1/5} \) and \( \frac{2}{5}x^{4/5}(6 - 2x)^{-4/5} \). This simplification gives \( 2x = 6 - 2x \), leading to \( x = 1.5 \).

Step 6: Verify the Maximization

Substitute \( x = 1.5 \) back into the constraint equation \( y = 6 - 2x = 6 - 2(1.5) = 3 \) and verify that this point provides a maximum by checking the second derivative or analyzing endpoints if necessary.

Key Concepts

Constrained Optimization

Constrained optimization is a mathematical approach used to find the maximum or minimum value of a function subject to certain restrictions or constraints. In many real-world problems, decisions are made based on mathematical models that must honor some constraints. For instance:

• Budget limits in production planning.
• Resource constraints in scheduling problems.
• Time restrictions in logistics and transportation.

In this problem, the goal is to maximize the utility function \( U(x, y) = 8x^{4/5}y^{1/5} \) under the constraint\( 4x + 2y = 12 \). This constraint restricts values of\( x \) and \( y \) to lie on the line defined by \( 4x + 2y =12 \) in a two-dimensional space. Identifying constraints andobjective functions is crucial in constrained optimization since theydetermine the feasible solutions for the problem.Exploring technologies such as Lagrange multipliers can be particularly useful for solving such problems.

Utility Function

A utility function in economics and mathematical optimization represents a measure of preferences over a set of goods or outcomes. It quantifies satisfaction or happiness, allowing comparisons between different scenarios.

• In this exercise, the utility function is\( U(x, y) = 8x^{4/5}y^{1/5} \), representing the satisfaction levelfrom consuming quantities \( x \) and \( y \).
• The exponents \( 4/5 \) and \( 1/5 \) indicate the diminishingreturns of utility as more \( x \) or \( y \) is consumed.

This utility function shows the principle of trade-off, where thedifferentiation between goods \( x \) and \( y \) is clear through howmarginal utilities behave. Such functions allow calculative decisionsto pinpoint optimal consumption points given limited resources.

Calculus Problem Solving

Calculus-based problem solving is a method of utilizing differentiation and other calculus concepts to solve optimization problems. It plays a key role in both identifying optimal points and addressing the constraints effectively.To solve the given problem:

• The constraint equation \( 4x + 2y = 12 \) was solved for \( y \), simplifying the optimization task to a single-variable problem.
• Lagrange multipliers provide a compact method for handling constraints efficiently, but this exercise emphasizes transforming the constraints to ease differentiation.
• The utility function's derivative with respect to \( x \) is found using rules like the product and chain rule.

Techniques like these demonstrate calculus's power in identifying feasible, optimal solutions to complex problems under prescribed conditions.

Critical Points Analysis

Critical points analysis is a vital step in the optimization process. It helps to identify potential maximum or minimum points of a function.In this exercise:

• After substituting and differentiating, the critical points are determined by setting the derivative \( \frac{dU}{dx} \) equal to zero.
• This led to finding \( x = 1.5 \) as a critical point. By substituting it back, \( y = 3 \) is found.
• Further verification involves evaluating the second derivative or considering endpoint behavior to confirm a maximum or minimum.

Examining critical points ensures that solutions not only solve the mathematical constraints but cater to identifying points that fulfill the desired extremum condition, i.e., maximum utility in this scenario.