# Local extrema w/2 var

1. Aug 24, 2005

### Guero

we're doing partial derivatives, and i thought i understood them until:

$$f(x,y)=-x^2-y^2-10xy+4y-4x+2$$

we are meant to find and classify all local extrema. i got:

$$f_x=-2x-10y-4$$ and $$f_y=-2y-10x+4$$

i've never had two variables at this stage, and i can't solve for any of the points. the text doesn't seem to say anything about this...

2. Aug 24, 2005

### Guero

yay! i'm now a level 30 thumb twiddler

3. Aug 24, 2005

### TD

You solve $$\nabla f = \vec0$$ (gradient of f) which basically gives you a system of 2 equations here. The solutions are the 'stationary points' which may be extrema, but you have to check that (e.g. with a test)

So you solve:

$$\left\{ \begin{gathered} \frac{{\delta f}} {{\delta x}} = 0 \hfill \\ \frac{{\delta f}} {{\delta y}} = 0 \hfill \\ \end{gathered} \right.$$

4. Aug 24, 2005

### Guero

mm, i get: $$10y=-2x-4$$ and $$10x=-2y+4$$ for each of which there are infinite solutions for x and y. i thought that if i combined the equations (no reason) it might do it, but i still got $$y=x-1$$

5. Aug 24, 2005

### TD

Yes, you have to combine them in a system.

$$\left\{ \begin{gathered} 2x + 10y + 4 = 0 \hfill \\ 10x + 2y - 4 = 0 \hfill \\ \end{gathered} \right.$$

Solving it should give $\left( {\frac{1} {2}, - \frac{1} {2}} \right)$

However, although it's a stationary point, it's not an extremum.

6. Aug 24, 2005

### Guero

aha! thanks, i get it now