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Q. 22

Expert-verifiedFound in: Page 985

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 $f(x,y)=x+y$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 $f(x,y)=x+y$subject to constraint $xy=1$is $2$and the minimum value is $-2.$

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

Given constraint is $xy=1.$

Gradients of function.

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

Use the method of Lagrange multipliers.

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

Compare terms.

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

substitute $x=y$in constraint.

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

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

Find function value at $(-1,-1).$

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

Find the function value at $(1,1).$

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

So the maximum value of the function is $2$and the minimum value is $-2.$

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