Chapter 14.5, Problem 13E

### 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 z   +   y subject to the constraint x 2 +   y 2 +   z 2 =   1 .

To determine

To calculate: The maximum value of w=xz+y which is subjected to the constraint x2+y2+z2=1.

Explanation

Given Information:

The provided function is w=xz+y and it is subjected to the constraint x2+y2+z2=1.

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=xz+y.

The provided constraint is x2+y2+z2=1.

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)=xz+y and g(x,y,z)=x2+y2+z2âˆ’1.

Put the values of f(x,y,z)=xz+yÂ andÂ g(x,y,z)=x2+y2+z2âˆ’1 in the objective function, F(x,y,z,Î»)=f(x,y,z)+Î»g(x,y,z).

F(x,y,z,Î»)=xz+y+Î»(x2+y2+z2âˆ’1)

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.

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

The objective function is F(x,y,z,Î»)=xz+y+Î»(x2+y2+z2âˆ’1),

âˆ‚Fâˆ‚x=0z+Î»(2x)=02xÎ»=âˆ’zÎ»=âˆ’z2x â€¦

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