   Chapter 14, Problem 62RE

Chapter
Section
Textbook Problem

Use Lagrange multipliers to find the maximum and minimum values of f subject to the given constraint(s).62. f(x, y, z) = x2 + 2y2 + 3z2;x + y + z =1, x − y + 2z = 2

To determine

To find: The extreme values of the function f(x,y,z)=x2+2y2+3z2 subject to the constraints x+y+z=1andxy+2z=2 by using Lagrange multipliers.

Explanation

Given:

The function is f(x,y,z)=x2+2y2+3z2 subject to the constraints x+y+z=1andxy+2z=2 .

Definition used:

“The Lagrange multipliers defined as f(x0,y0,z0)=λg(x0,y0,z0)+μh(x0,y0,z0) . This equation can be expressed as fx=λgx+μhx , fy=λgy+μhy , fz=λgz+μhz and g(x,y,z)=k , h(x,y,z)=c ”.

Calculation:

Let the function g(x,y,z)=x+y+z=1 and h(x,y,z)=xy+2z=2 .

The Lagrange multipliers f(x,y,z)=λg(x,y,z)+μh(x,y,z) is computed as follows.

f(x,y,z)=λg(x,y,z)+μh(x,y,z)fx,fy,fz=λgx,gy,gz+μhx,hy,hzfx(x2+2y2+3z2),fy(x2+2y2+3z2),fz(x2+2y2+3z2)=λgx(x+y+z),gy(x+y+z),gz(x+y+z)+μhx(xy+2z),hy(xy+2z),hz(xy+2z)2x,4y,6z=λ1,1,1+μ1,1,2

Thus, the value of f(x,y,z)=λg(x,y,z)+μh(x,y,z) is 2x,4y,6z=λ1,1,1+μ1,1,2 .

The reuslt, 2x,4y,6z=λ1,1,1+μ1,1,2 can be expressed as follows.

2x=λ+μ (1)

4y=λμ (2)

6z=λ+2μ (3)

From the equations (1), (2) and (3) and compute the values λandμ as follows

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