Chapter 4: Problem 374
Maximize \(U(x, y)=8 x^{4 / 5} y^{1 / 5} ; 4 x+2 y=12\).
Short Answer
Expert verified
Maximize at \( x = 1.5 \), \( y = 3 \).
Step by step solution
01
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 \( y \).
02
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.
03
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.
04
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: \[ \frac{dU}{dx} = 8 \left[ \frac{4}{5}x^{-1/5}(6 - 2x)^{1/5} - \frac{2}{5}x^{4/5}(6 - 2x)^{-4/5} \right]. \]
05
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 \( x = 1.5 \).
06
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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.
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.