   Chapter 14.8, Problem 39E

Chapter
Section
Textbook Problem

Use Lagrange multipliers to give an alternate solution to the indicated exercise in Section 14.7.39. Exercise 4949. Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane x + 2y + 3z = 6.

To determine

To find: The volume of the largest rectangular box which lies in the first octant with three faces in the coordinate planes and its one of the vertex in the plane x+2y+3z=6 by using Lagrange multipliers.

Explanation

Given:

One of the vertex of the rectangular box lies in the plane x+2y+3z=6.

The rectangular box lies in the first octant.

Definition used:

“The Lagrange multipliers defined as f(x,y,z)=λg(x,y,z). This equation can be expressed as fx=λgx, fy=λgy,fz=λgz and g(x,y,z)=k”.

Calculation:

Let the volume of the rectangular box be V=f(x,y,z)=xyz where x>0,y>0,z>0.

Thus, the maximize function f(x,y,z)=xyz subject to the constraint g(x,y,z)=x+2y+3z=6.

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

f(x,y,z)=λg(x,y,z)fx,fy,fz=λgx,gy,gzfx(xyz),fy(xyz),fz(xyz)=λgx(x+2y+3z),gy(x+2y+3z),gz(x+2y+3z)yz,xz,xy=λ1,2,3

Thus, the value of f(x,y,z)=λg(x,y,z) is yz,xz,xy=λ1,2,3

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