   Chapter 14.8, Problem 31E

Chapter
Section
Textbook Problem

Use Lagrange multipliers to give an alternate solution to the indicated exercise in Section 14.7.31. Exercise 4141. Find the shortest distance from the point (2, 0, -3) to die plane x + y + z = 1.

To determine

To find: The shortest distance from the point (2,0,3) to the plane x+y+z=1 by using Lagrange multipliers.

Explanation

Given:

The plane x+y+z=1.

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:

The shortest distance between the point (2,0,3) and (x,y,z).

d=(x2)2+(y0)2+(z+3)2d2=(x2)2+y2+(z+3)2

Thus, the minimum value of the function d2=f(x,y,z)=(x2)2+y2+(z+3)2 subject to the constraint g(x,y,z)=x+y+z=1.

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[(x2)2+y2+(z+3)2],fy[(x2)2+y2+(z+3)2],fz[(x2)2+y2+(z+3)2]=λgx(x+y+z),gy(x+y+z),gz(x+y+z)2(x2),2y,2(z+3)=λ1,1,1

Thus, the value of f(x,y,z)=λg(x,y,z) is 2(x2),2y,2(z+3)=λ1,1,1

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

Find the derivative. Simplify where possible. 38. f(t)=1+sinht1sinht

Single Variable Calculus: Early Transcendentals, Volume I

23-42 Find f. f(t)=12+sint

Calculus (MindTap Course List)

In Problems 7-34, perform the indicated operations and simplify. 14.

Mathematical Applications for the Management, Life, and Social Sciences

The area of the region bounded by , , and r = sec θ is:

Study Guide for Stewart's Multivariable Calculus, 8th

True or False: The graph in question 3 has a vertical asymptote at x = 1.

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