maxima, minima, saddle point problem

**Dr Zoidburg** · September 4th 2008, 09:02 PM

Find and classify all local maxima, minima and saddle points for the surface:

$<br /> f(x,y)=x^3+y^3-3x-3y<br />$

last question for my assignment
thanks!

**Chris L T521** · September 4th 2008, 09:27 PM

Originally Posted by Dr Zoidburg

Find and classify all local maxima, minima and saddle points for the surface:

$<br /> f(x,y)=x^3+y^3-3x-3y<br />$

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

**Dr Zoidburg** · September 4th 2008, 09:56 PM

thanks for that!

**Chris L T521** · September 5th 2008, 02:27 PM

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

].

Here's the graph:

--Chris

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