Understanding Local Extrema with Two Variables and Partial Derivatives

[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...

 

[Guero]

yay! I'm now a level 30 thumb twiddler

 

[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:

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

 

[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 $$y=x-1$$

 

[TD]

Yes, you have to combine them in a system.

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

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

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

 

[Guero]

aha! thanks, i get it now