# Thread: Finding and classifying all relative extrema

1. ## Finding and classifying all relative extrema

The function I have to find the relative extreme for is f(x,y)= x^3 + y^3 + 3x^2 - 3y^2 the question didn't say it was subject to any constraints and the only ones we,be doing have been subject to the constraint x^2+y^2=1 so I'm not sure what to do for this one, thanks in advance

2. ## Re: Finding and classifying all relative extrema

Try setting $\displaystyle \bigtriangledown f = 0$

That is, $\displaystyle \frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0$

And solve for $\displaystyle x$ and $\displaystyle y$.

This should give you four points which can be classified using the Hessian matrix.

3. ## Re: Finding and classifying all relative extrema

Originally Posted by TwoPlusTwo
Try setting $\displaystyle \bigtriangledown f = 0$

That is, $\displaystyle \frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0$

And solve for $\displaystyle x$ and $\displaystyle y$.

This should give you four points which can be classified using the Hessian matrix.
Thank you, so this would lead to $\displaystyle \frac{\partial f}{\partial x} =$ 3x^2+6x and $\displaystyle \frac{\partial f}{\partial y} =$ 3y^2-6y correct? And since they both equal zero does this bean that the realitive extrema would just be x=0 x=2 y=0 and y=-2? And I'm still unsure what my prfessor meant when he said to classify the extrema

4. ## Re: Finding and classifying all relative extrema

Originally Posted by grandmarquis84
Thank you, so this would lead to $\displaystyle \frac{\partial f}{\partial x} =$ 3x^2+6x and $\displaystyle \frac{\partial f}{\partial y} =$ 3y^2-6y correct? And since they both equal zero does this bean that the realitive extrema would just be x=0 x=2 y=0 and y=-2? And I'm still unsure what my prfessor meant when he said to classify the extrema
The partial derivatives are correct. The four critical points are:

$\displaystyle (0,0), (0,2), (-2,0), (-2,2)$

Classifying a critical point means to determine whether it's a minimum, maximum or a saddle point.

For this you can use the Hessian matrix

$\displaystyle H = \left(\begin{array}{cc}f_{xx} &f_{yx} \\f_{xy} &f_{yy}\end{array}\right)$.

Take it's determinant:

$\displaystyle |H| = f_{xx}f_{yy} - f_{xy}f_{yx}$

Then any given critical point is

1. a local minimum if $\displaystyle |H|>0$ and $\displaystyle f_{xx}>0$ in the point,
2. a local maximum if $\displaystyle |H|>0$ and $\displaystyle f_{xx}<0$ in the point,
3. a saddle point if $\displaystyle |H|<0$ in the point.

If $\displaystyle |H|=0$ the test is inconclusive.

Using this I got:

$\displaystyle (0,0)$ and $\displaystyle (-2,2)$ are saddle points.

$\displaystyle (0,2)$ is a local minimum, with $\displaystyle f(0,2)=-4$

$\displaystyle (-2,0)$ is a local maximum, with $\displaystyle f(-2,0)=4$

I suggest you try it yourself and see if you get the same results.