Chapter 14.5, Problem 14E

### 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

# Find the maximum value of w   = x 2 y z subject to the constraint 4 x + y   +   z =   4 , x   ≥ 0 , y   ≥ 0 ,  and  z ≥ 0 .

To determine

To calculate: The maximum value of w=x2yz which is subjected to the constraint 4x+y+z=4, x0, y0 and z0.

Explanation

Given Information:

The provided function is w=x2yz and it is subjected to the constraint 4x+y+z=4, xâ‰¥0, yâ‰¥0 and zâ‰¥0.

Formula used:

Lagrange Multipliers Method:

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

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

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

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

âˆ‚Fâˆ‚x=0, âˆ‚Fâˆ‚y=0, âˆ‚Fâˆ‚z=0 and âˆ‚Fâˆ‚Î»=0, that is, the points which make all the partial derivatives equal to zero.

For a function f(x,y,z), the partial derivative of f(x,y,z) with respect to y is calculated by taking the derivative of f(x,y,z) with respect to y and keeping the other variables x and z constant. The partial derivative of f(x,y,z) 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, w=x2yz.

The provided constraint is 4x+y+z=4, xâ‰¥0, yâ‰¥0 and zâ‰¥0.

According to the Lagrange multipliers method,

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

Here, f(x,y,z)=x2yz and g(x,y,z)=4x+y+zâˆ’4.

Put the values of f(x,y,z)=x2yzÂ andÂ g(x,y,z)=4x+y+zâˆ’4 in the objective function, F(x,y,z,Î»)=f(x,y,z)+Î»g(x,y,z).

F(x,y,z,Î»)=x2yz+Î»(4x+y+zâˆ’4)

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

âˆ‚Fâˆ‚x=0, âˆ‚Fâˆ‚y=0, âˆ‚Fâˆ‚z=0 and âˆ‚Fâˆ‚Î»=0.

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

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

## Additional Math Solutions

#### Read each scale:

Elementary Technical Mathematics

#### Evaluate the integral. 12(4x33x2+2x)dx

Single Variable Calculus: Early Transcendentals

#### True or False: If f(x) = f (âˆ’x) for all x then .

Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th

#### j Ă— (âˆ’k) = i âˆ’i j + k âˆ’j âˆ’ k

Study Guide for Stewart's Multivariable Calculus, 8th