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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 # When you use the method of Lagrange multipliers to find the maximum and minimum of $f\left(x,y\right)=x+y$subject to the constraint $xy=1,$you obtain two points. Is there a relative maximum at one of the points and a relative minimum at the other? Which is which?

The maximum value of the function $f\left(x,y\right)=x+y$subject to constraint $xy=1$is $2$and the minimum value is $-2.$

See the step by step solution

## Step 1. Given information.

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

Given constraint is $xy=1.$

## Step 2. critical points of the function.

$\nabla f\left(x,y\right)=\mathbf{i}+\mathbf{j}\phantom{\rule{0ex}{0ex}}\nabla \mathrm{g}\left(\mathrm{x},\mathrm{y}\right)=y\mathbf{i}+x\mathbf{j}$

Use the method of Lagrange multipliers.

role="math" localid="1649887252906" $\nabla f\left(x,y\right)=\lambda \nabla g\left(x,y\right)\phantom{\rule{0ex}{0ex}}\mathbf{i}+\mathbf{j}=\lambda \left(y\mathbf{i}+x\mathbf{j}\right)\phantom{\rule{0ex}{0ex}}\mathbf{i}+\mathbf{j}=\lambda y\mathbf{i}+\lambda x\mathbf{j}$

Compare terms.

role="math" localid="1649887083426" $\lambda y=1⇒y=\frac{1}{\lambda }\phantom{\rule{0ex}{0ex}}\lambda x=1⇒x=\frac{1}{\lambda }\phantom{\rule{0ex}{0ex}}\mathrm{so}\mathrm{x}=\mathrm{y}$

substitute $x=y$in constraint.

$xy=1\phantom{\rule{0ex}{0ex}}{y}^{2}=1\phantom{\rule{0ex}{0ex}}y=±1\phantom{\rule{0ex}{0ex}}x=±1$

so critical points are role="math" localid="1649887209373" $\left(-1,-1\right)&\left(1,1\right).$

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

Find function value at $\left(-1,-1\right).$

role="math" localid="1649887515378" $f\left(x,y\right)=x+y\phantom{\rule{0ex}{0ex}}f\left(-1,-1\right)=\left(-1\right)+\left(-1\right)\phantom{\rule{0ex}{0ex}}f\left(-1,-1\right)=-2$

Find the function value at $\left(1,1\right).$

$f\left(x,y\right)=x+y\phantom{\rule{0ex}{0ex}}f\left(1,1\right)=\left(1\right)+\left(1\right)\phantom{\rule{0ex}{0ex}}f\left(1,1\right)=2$

So the maximum value of the function is $2$and the minimum value is $-2.$ ### Want to see more solutions like these? 