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Expert-verified Found in: Page 985 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # In Exercises, find the maximum and minimum of the function f subject to the given constraint. In each case explain why the maximum and minimum must both exist.$f\left(x,y,z\right)=x+y+z\text{when}{x}^{2}+4{y}^{2}+16{z}^{2}=64$

The maximum value of the function is $2\sqrt{21}$and the minimum value is $-2\sqrt{21}$and both exist because the constraint is a bounded and closed ellipsoid.

See the step by step solution

## Step 1. Given information.

Given function is $f\left(x,y,z\right)=x+y+z.$

Given constraint is ${x}^{2}+4{y}^{2}+16{z}^{2}=64.$

## Step 2. critical points of the function.

$\nabla f\left(x,y,z\right)=i+j+k\phantom{\rule{0ex}{0ex}}\nabla g\left(x,y,z\right)=2xi+8yj+32zk$

Use the method of Lagrange multipliers.

$\nabla f\left(x,y,z\right)=\lambda \nabla g\left(x,y,z\right)\phantom{\rule{0ex}{0ex}}i+j+k=\lambda \left(2xi+8yj+32zk\right)\phantom{\rule{0ex}{0ex}}i+j+k=2\lambda xi+8\lambda yj+32\lambda zk$

Compare terms.

$1=2\lambda x⇒\lambda =\frac{1}{2x}\phantom{\rule{0ex}{0ex}}1=8\lambda y⇒\lambda =\frac{1}{8y}\phantom{\rule{0ex}{0ex}}1=32\lambda z⇒\lambda =\frac{1}{32z}\phantom{\rule{0ex}{0ex}}\mathrm{so}x=4y&z=\frac{y}{4}$

substitute $x=4y&z=\frac{y}{4}$in constraint.

${x}^{2}+4{y}^{2}+16{z}^{2}=64\phantom{\rule{0ex}{0ex}}{\left(4y\right)}^{2}+4{y}^{2}+16{\left(\frac{y}{4}\right)}^{2}=64\phantom{\rule{0ex}{0ex}}21{y}^{2}=64\phantom{\rule{0ex}{0ex}}y=±\frac{8}{\sqrt{21}}\phantom{\rule{0ex}{0ex}}x=±\frac{32}{\sqrt{21}}\phantom{\rule{0ex}{0ex}}z=±\frac{2}{\sqrt{21}}$

so critical points are $\left(\frac{-32}{\sqrt{21}},\frac{-8}{\sqrt{21}},\frac{-2}{\sqrt{21}}\right)&\left(\frac{32}{\sqrt{21}},\frac{8}{\sqrt{21}},\frac{2}{\sqrt{21}}\right).$

## Step 3. maximum and minimum of a function.

Find function value at $\left(\frac{-32}{\sqrt{21}},\frac{-8}{\sqrt{21}},\frac{-2}{\sqrt{21}}\right)&\left(\frac{32}{\sqrt{21}},\frac{8}{\sqrt{21}},\frac{2}{\sqrt{21}}\right).$

role="math" localid="1649896217514" $f\left(x,y,z\right)=x+y+z\phantom{\rule{0ex}{0ex}}f\left(\frac{-32}{\sqrt{21}},\frac{-8}{\sqrt{21}},\frac{-2}{\sqrt{21}}\right)=\left(\frac{-32}{\sqrt{21}}+\frac{-8}{\sqrt{21}}+\frac{-2}{\sqrt{21}}\right)\phantom{\rule{0ex}{0ex}}f\left(\frac{-32}{\sqrt{21}},\frac{-8}{\sqrt{21}},\frac{-2}{\sqrt{21}}\right)=\frac{-42}{\sqrt{21}}\phantom{\rule{0ex}{0ex}}f\left(\frac{-32}{\sqrt{21}},\frac{-8}{\sqrt{21}},\frac{-2}{\sqrt{21}}\right)=-2\sqrt{21}\phantom{\rule{0ex}{0ex}}\mathrm{and}f\left(\frac{32}{\sqrt{21}},\frac{8}{\sqrt{21}},\frac{2}{\sqrt{21}}\right)=2\sqrt{21}$

So the maximum value of the function is $2\sqrt{21}$and the minimum value is $-2\sqrt{21}.$

As constraint is bounded and closed ellipsoid so maximum and minimum both exist. ### Want to see more solutions like these? 