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Math
CalculusMultivariable Calculus8th Edition

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Chapter 14.8, Problem 31E
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Multivariable Calculus

8th Edition
James Stewart
ISBN: 9781305266643

Solutions

Chapter
Section
Problem 1E
Problem 2E
Problem 3E
Problem 4E
Problem 5E
Problem 6E
Problem 7E
Problem 8E
Problem 9E
Problem 10E
Problem 11E
Problem 12E
Problem 13E
Problem 14E
Problem 15E
Problem 16E
Problem 17E
Problem 18E
Problem 19E
Problem 20E
Problem 21E
Problem 22E
Problem 23E
Problem 24E
Problem 25E
Problem 27E
Problem 28E
Problem 29E
Problem 30E
Problem 31E
Problem 32E
Problem 33E
Problem 34E
Problem 35E
Problem 36E
Problem 37E
Problem 38E
Problem 39E
Problem 40E
Problem 41E
Problem 42E
Problem 43E
Problem 44E
Problem 45E
Problem 46E
Problem 49E
Problem 50E
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Multivariable Calculus

8th Edition
James Stewart
ISBN: 9781305266643
Textbook Problem

Use Lagrange multipliers to give an alternate solution to the indicated exercise in Section 14.7.

31. Exercise 41

41. 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)=λ∇g(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=(x−2)2+(y−0)2+(z+3)2d2=(x−2)2+y2+(z+3)2

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

The Lagrange multipliers ∇f(x,y,z)=λ∇g(x,y,z) is computed as follows,

∇f(x,y,z)=λ∇g(x,y,z)〈fx,fy,fz〉=λ〈gx,gy,gz〉〈fx[(x−2)2+y2+(z+3)2],fy[(x−2)2+y2+(z+3)2],fz[(x−2)2+y2+(z+3)2]〉=λ〈gx(x+y+z),gy(x+y+z),gz(x+y+z)〉〈2(x−2),2y,2(z+3)〉=λ〈1,1,1〉

Thus, the value of ∇f(x,y,z)=λ∇g(x,y,z) is 〈2(x−2),2y,2(z+3)〉=λ〈1,1,1〉

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