   Chapter 7.6, Problem 28E ### 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.)Cone: z = x 2 + y 2 ,   ( 4 ,   0 ,   0 ) Minimize d 2 = ( x − 4 ) 2 + y 2 + z 2

To determine

To calculate: To determine maximum distance of cone, z=x2+y2 from point (4,0,0) that maximize d2=(x4)2+y2+z2

Explanation

Given Information:

The cone z=x2+y2 from point (4,0,0) that maximize d2=(x4)2+y2+z2

Formula used:

If f(x,y,z) has a maximum or minimum subject to the constrain g(x,y,z) then it will occur at

one of the critical numbers of the function F defines as,

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

Where, λ=Lograngemultiplier.

Step 1: Solve the above system of equations.

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

Calculation:

For the minimum distance as the function f is defined as,

F(x,y,z,λ)=(x4)2+y2+z2+λ[x2+y2z2]

Now, differentiate the above equation with respect to x as,

Fx(x,y,z,λ)=2(x4)2xλFx(x,y,z,λ)=02(x4)2xλ=02x(1λ)=8

Now, differentiate the above equation with respect to y as,

Fy(x,y,z,λ)=2y2yλ

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