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Q. 27

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.$f\left(x,y\right)=xy\mathrm{when}{x}^{2}+4{y}^{2}=16$

The maximum value of the function is $4$and the minimum value is $-4$and both exist as the constraint is a bounded and closed ellipse.

See the step by step solution

## Step 1. Given information.

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

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

## Step 2. critical points of the function.

$\nabla f\left(x,y\right)=yi+xj\phantom{\rule{0ex}{0ex}}\nabla g\left(x,y\right)=2xi+8zj$

Use the method of Lagrange multipliers.

$\nabla f\left(x,y\right)=\lambda \nabla g\left(x,y\right)\phantom{\rule{0ex}{0ex}}yi+xj=\lambda \left(2xi+8yj\right)\phantom{\rule{0ex}{0ex}}yi+xj=2\lambda xi+8\lambda yj$

Compare terms.

$y=2\lambda x⇒\lambda =\frac{y}{2x}\phantom{\rule{0ex}{0ex}}x=8\lambda y⇒\lambda =\frac{x}{8y}\phantom{\rule{0ex}{0ex}}\mathrm{so}{x}^{2}=4{y}^{2}$

substitute ${x}^{2}=4{y}^{2}$in constraint.

role="math" localid="1649893915369" ${x}^{2}+4{y}^{2}=16\phantom{\rule{0ex}{0ex}}4{y}^{2}+4{y}^{2}=16\phantom{\rule{0ex}{0ex}}8{y}^{2}=16\phantom{\rule{0ex}{0ex}}y=±\sqrt{2}\phantom{\rule{0ex}{0ex}}x=±2\sqrt{2}$

so critical points are role="math" localid="1649893960368" $\left(-2\sqrt{2},-\sqrt{2}\right),\left(2\sqrt{2},-\sqrt{2}\right),\left(-2\sqrt{2},\sqrt{2}\right),&\left(2\sqrt{2},\sqrt{2}\right).$

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

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

$f\left(x,y\right)=xy\phantom{\rule{0ex}{0ex}}f\left(-2\sqrt{2},-\sqrt{2}\right)=\left(-2\sqrt{2}\right)\left(-\sqrt{2}\right)\phantom{\rule{0ex}{0ex}}f\left(-2\sqrt{2},-\sqrt{2}\right)=4$

similarly,

$f\left(2\sqrt{2},-\sqrt{2}\right)=-4\phantom{\rule{0ex}{0ex}}f\left(-2\sqrt{2},\sqrt{2}\right)=-4\phantom{\rule{0ex}{0ex}}f\left(-2\sqrt{2},-\sqrt{2}\right)=4$

So the maximum value of the function is $4$and the minimum value is $-4.$

As constraint is bounded and closed ellipse so maximum and minimum both exist.