# maxima, minima, saddle point problem

• Sep 4th 2008, 09:02 PM
Dr Zoidburg
Find and classify all local maxima, minima and saddle points for the surface:

$
f(x,y)=x^3+y^3-3x-3y
$

last question for my assignment
thanks!
• Sep 4th 2008, 09:27 PM
Chris L T521
Quote:

Originally Posted by Dr Zoidburg
Find and classify all local maxima, minima and saddle points for the surface:

$
f(x,y)=x^3+y^3-3x-3y
$

last question for my assignment
thanks!

Find where $\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$

$\frac{\partial f}{\partial x}=3x^2-3$
$\frac{\partial f}{\partial y}=3y^2-3$

So we see that we have critical points when $3x^2-3=0$ and $3y^2-3=0$

This implies that $x^2=1$ and $y^2=1$

So we see that $x=\pm1$ and $y=\pm1$

So our critical points are $(-1,1)$, $(-1,-1)$, $(1,-1)$, $(1,1)$.

Now apply the second partials test:

$D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-f_{xy}^2(x_0,y_0)$

Recall that if:

$D>0~\text{and}~f_{xx}(x_0,y_0)>0$, then $f$ has a relative minimum at $(x_0,y_0)$

$D>0~\text{and}~f_{xx}(x_0,y_0)><0$, then $f$ has a relative maximum at $(x_0,y_0)$

$D<0$, then $f$ has a saddle point at $(x_0,y_0)$

$D=0$, then no conclusion can be drawn.

Try to take it from here.

--Chris
• Sep 4th 2008, 09:56 PM
Dr Zoidburg
thanks for that!
• Sep 5th 2008, 02:27 PM
Chris L T521
I checked over my work and noticed a slight error...the critical points are really $(-1,-1),~(-1,1),~(1,-1),~(1,1)$. There shouldn't be any zeros in the critical points. My mistake [shouldn't be math when you're dead tired (Rofl)].

Here's the graph:

http://img.photobucket.com/albums/v4...3varmaxmin.jpg

--Chris