   Chapter 14.5, Problem 15E Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

Solutions

Chapter
Section Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

Utility Suppose that the utility function for two com modifies is given by U   = x y 2 and that the budget constraint is 3 x   + 6 y =   18 . What values of x and y will maximize utility?

To determine

To calculate: The values of x and y which maximize the utility function for two commodities is U=xy2 and that the budget constraint is 3x+6y=18.

Explanation

Given Information:

The provided utility function is U=xy2 and it is subjected to the constraint 3x+6y=18.

Formula used:

Lagrange Multipliers Method:

According to the Lagrange multipliers method to obtain maxima or minima for a function z=f(x,y) subject to the constraint g(x,y)=0,

Step 1: Find the critical values of f(x,y) using the new variable λ to form the objective function F(x,y,λ)=f(x,y)+λg(x,y).

Step 2: The critical points of f(x,y) are the critical values of F(x,y,λ) which satisfies g(x,y)=0.

Step 3: The critical points of F(x,y,λ) are the points that satisfy:

Fx=0, Fy=0, and Fλ=0, that is, the points which make all the partial derivatives of zero.

For a function f(x,y), the partial derivative of f(x,y) with respect to y is calculated by taking the derivative of f(x,y) with respect to y and keeping the other variable x constant. The partial derivative of f(x,y) with respect to y is denoted by fy.

Power of x rule for a real number n is such that, if f(x)=xn then f(x)=nxn1.

Constant function rule for a constant c is such that, if f(x)=c then f(x)=0.

Coefficient rule for a constant c is such that, if f(x)=cu(x), where u(x) is a differentiable function of x, then f(x)=cu(x).

Calculation:

Consider the function, U=xy2.

The provided constraint is 3x+6y=18.

According to the Lagrange multipliers method,

The objective function is F(x,y,λ)=f(x,y)+λg(x,y).

Here, f(x,y)=xy2 and g(x,y)=3x+6y18.

Substitute xy2 for f(x,y) and 3x+6y18 for g(x,y) in F(x,y,λ)=f(x,y)+λg(x,y).

F(x,y,λ)=xy2+λ(3x+6y18)

Since, the critical points of F(x,y,λ) are the points that satisfy:

Fx=0, Fy=0, and Fλ=0.

Recall that, for a function f(x,y), the partial derivative of f(x,y) with respect to y is calculated by taking the derivative of f(x,y) with respect to y and keeping the other variable x constant

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