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17. f (1,y, z) = x subject to z? + y? + 2² – z = 1 %3D

17. f (1,y, z) = x subject to z? + y? + 2² – z = 1 %3D

College Algebra
College Algebra
10th Edition
ISBN:9781337282291
Author:Ron Larson
Publisher:Ron Larson
Chapter8: Sequences, Series,and Probability
Section: Chapter Questions
Problem 10CT: Sketch the region corresponding to the system of constraints. Then find the minimum and maximum...
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Section 12.9 - Lagrange Multipliers Number 17

17. f (x, y, 2) = x subject to r? + y? + 2 – z = 1
Transcribed Image Text:17. f (x, y, 2) = x subject to r? + y? + 2 – z = 1
15-24. Lagrange multipliers in three variables Use Lagrange multipliers to find the maximum and minimum values of f (when they exist) subject to the given constraint.
Transcribed Image Text:15-24. Lagrange multipliers in three variables Use Lagrange multipliers to find the maximum and minimum values of f (when they exist) subject to the given constraint.
Expert Solution
Step 1

Let h(x, y, z)=x2+y2+z2-z. We now have ∇f(x, y, z)=(1, 0, 0) and ∇h(x, y, z)=(2x, 2y, 2z-1). Let λ be the lagrnage's multiplier. Then we have ∇f(x, y, z)=λ ∇h(x, y, z) implying that (1, 0, 0) =λ(2x, 2y, 2z-1). 

We also have the relation x2+y2+z2-z=1.

We now have 

2λx=12λy=02λz-λ=0x2+y2+z2-z=1

Solving the above relations, we get y=0, x=1λ, z=12, and the relation x2+y2+z2-z=1.

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