# Math Help - Extrema

1. ## Extrema

Hello,

yesterday we had a test of calculus and one question was:

Find and classify the extreme of
h(x,y)=(1+1/(x))*(1+1/(y))*(1/(x)+1/(y))

I have absolutely no idea how to solve this thing. Can someone help me?

Greets,
Leslon

2. Originally Posted by Leslon
Hello,

yesterday we had a test of calculus and one question was:

Find and classify the extreme of
h(x,y)=(1+1/(x))*(1+1/(y))*(1/(x)+1/(y))

I have absolutely no idea how to solve this thing. Can someone help me?

Greets,
Leslon
An extrema is a local minimum or maximum within a function. This is when $f'(x)=0$

You have $h(x,y) = \left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\righ t)\left(\frac{1}{y}+\frac{1}{x}\right)$

Let's first try to simplify this...

$h(x,y)=\left(1 + \frac{1}{y} + \frac{1}{x} + \frac{1}{xy}\right) \left(\frac{1}{y}+\frac{1}{x}\right)$

$h(x,y)=\frac{1}{y} + \frac{1}{y^2} + \frac{1}{xy} + \frac{1}{xy^2} + \frac{1}{x}+\frac{1}{xy}+\frac{1}{x^2}+\frac{1}{x^ 2y}$

$h(x,y)=\frac{1}{y} + \frac{1}{y^2} + \frac{2}{xy} + \frac{1}{xy^2} + \frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^2y}$

Can you derive this?

3. Well if I derive this, I get:
-(1+1/y)*(1/x+1/y)/x^2-(1+1/x)*(1+1/y)/x^2 for dh(x,y)/dx

and

-(1+1/x)*(1/x+1/y)/y^2-(1+1/x)*(1+1/y)/y^2 for dh(x,y)/dy

and now I'm stuck..

What to do next??

PS: I've included a .tex file if necessary

4. To insert images on this site use 'manage attachments', not the tags

I know it works that way on some sites, but the way it works here is better.