Chapter 14.5, Problem 18E

### Mathematical Applications for the ...

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

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 commodities is given by U =   x 2 y 3 , and the budget constraint is 10 x   +   15 y   = 250 . Find the values of x and y that maximize utility. Check by graphing the budget constraint with the indifference curve for maximum utility and with two other indifference curves.

To determine

To calculate: The values of x and y that maximize utility. Suppose that the utility function for two commodities is given by U=x2y3 and that the budget constraint is 10x+15y=250. Check by graphing the budget constraint with the indifference curve for maximum utility and with two other indifference curves.

Explanation

Given Information:

The provided utility function is U=x2y3 subject to the constraint 10x+15y=250.

Formula used:

According to the Lagrange multipliers steps 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 Î» to form the objective function F(x,y,Î»)=f(x,y)+Î»g(x,y).

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

Step 3: The critical points of F(x,y,Î») are the points that satisfy âˆ‚Fâˆ‚x=0, âˆ‚Fâˆ‚y=0, and âˆ‚Fâˆ‚Î»=0, that is, the points which make all the partial derivatives of zero.

For a function f(x,y), the partial derivative of f 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 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 fâ€²(x)=nxnâˆ’1.

Constant function rule for a constant c is such that, if f(x)=c then fâ€²(x)=0.

Coefficient rule for a constant c is such that, if f(x)=câ‹…u(x), where u(x) is a differentiable function of x, then fâ€²(x)=câ‹…uâ€²(x).

Calculation:

Consider the function, U=x2y3.

The provided constraint is 10x+15y=250.

According to the Lagrange multipliers method,

The objective function is F(x,y,Î»)=f(x,y)+Î»g(x,y).

Thus, f(x,y)=x2y3 and g(x,y)=10x+15yâˆ’250.

Substitute x2y3 for f(x,y) and 10x+15yâˆ’250 for g(x,y) in F(x,y,Î»)=f(x,y)+Î»g(x,y).

F(x,y,Î»)=x2y3+Î»(10x+15yâˆ’250)

Since, the critical points of F(x,y,Î») are the points that satisfy âˆ‚Fâˆ‚x=0, âˆ‚Fâˆ‚y=0, and âˆ‚Fâˆ‚Î»=0.

Recall that, for a function f(x,y), the partial derivative of f 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.

Use the power of x rule for derivatives, the constant function rule and the coefficient rule.

Thus,

âˆ‚Fâˆ‚x=0(2x)y3+Î»(10)=010Î»=âˆ’2xy3Î»=âˆ’xy35

And,

âˆ‚Fâˆ‚y=0x2(3y2)+Î»(15)=015Î»=âˆ’3x2y2Î»=âˆ’x2y25

And,

âˆ‚Fâˆ‚Î»=010x+15yâˆ’250=010x+15y=250

Solve the equations Î»=âˆ’xy35 and Î»=âˆ’x2y25.

Substitute âˆ’xy35 for Î» in Î»=âˆ’x2y25.

âˆ’xy35=âˆ’x2y25âˆ’5xy3=âˆ’5x2y2âˆ’5xy3+5x2y2=05xy2(âˆ’y+x)=0

Simplify it further,

xy2=0 or âˆ’y+x=0

That is, x=0, y=0 or x=y.

Now, calculate values of x and y.

Consider the equation x=0

Substitute 0 for x in 10x+15y=250.

10(0)+15y=2500+15y=25015y=250y=503

Consider the equation y=0

Substitute 0 for y in 10x+15y=250

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