   Chapter 7.6, Problem 27E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Finding Distance In Exercises 23-28, find the minimum distance from the curve or surface to the given point. (Hint: Start by minimizing the square of the distance.) Plane : x   +   y   +   z =   1 ,   ( 2 ,   1 ,   1 ) Minimize  d 2 =   ( x −   2 ) 2 +   ( y   −   1 ) 2 +   ( z   − 1 ) 2

To determine

To calculate: The minimum distance from plane x+y+z=1 to the point (2,1,1) by minimizing d2=(x2)2+(y1)2+(z1)2.

Explanation

Given Information:

The distance function is d2=(x2)2+(y1)2+(z1)2 and constant function is (g)=x+y+z1=0.

Formula used:

If f(x,y,z) has a maximum or minimum subject to the constraint g(x,y,z)=0 then it will occur at one of the critical numbers of the function F defined by

Fx(x,y,z,λ)=0Fy(x,y,z,λ)=0Fz(x,y,z,λ)=0Fλ(x,y,z,λ)=0

Where λ is Lagrange multiplier.

Step1: Solve the above system of equations.

Step 2: Evaluate f at each solution point obtained in the in the first step. The greatest value yields the maximum of f subject to the constraint g(x,y,z) and the least value yields the minimum of f subject to the constraint g(x,y,z)=0

Calculation:

For the minimum distance the function f is defined as,

f(x,y,z,λ)=(z2)2+(y1)2+λ(x+y+z1)=0Fx(x,y,z,λ)=2(x2)+λ=0λ=2(x2)

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