1. ## maxima, minima, saddle point problem

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

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

last question for my assignment
thanks!

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

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

last question for my assignment
thanks!
Find where $\displaystyle \frac{\partial f}{\partial x}=0$ and $\displaystyle \frac{\partial f}{\partial y}=0$

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

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

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

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

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

Now apply the second partials test:

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

Recall that if:

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

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

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

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

Try to take it from here.

--Chris

3. thanks for that!

4. I checked over my work and noticed a slight error...the critical points are really $\displaystyle (-1,-1),~(-1,1),~(1,-1),~(1,1)$. 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