# Help with Stationary Points of a surface sadle points thing ..

• May 5th 2006, 04:31 AM
AVVIT
Help with Stationary Points of a surface sadle points thing ..
i have an exam coming up and this is going to be in it i have a problem in my notes i cant seem to do can anyone help please

Find the stationary points of the following surface. and determine thier nature of the stationary points.

f(x,y) = x^3 + y^3 - 3x - 12y +20

i understand the steps but for some reason when a Partial diff it never seems to come out like anything els in my notes any help would be very much apreciated
• May 5th 2006, 08:59 AM
ThePerfectHacker
Quote:

Originally Posted by AVVIT
i have an exam coming up and this is going to be in it i have a problem in my notes i cant seem to do can anyone help please

Find the stationary points of the following surface. and determine thier nature of the stationary points.

f(x,y) = x^3 + y^3 - 3x - 12y +20

i understand the steps but for some reason when a Partial diff it never seems to come out like anything els in my notes any help would be very much apreciated

I might be totally wrong but it seems to me to find the critical point for this function.

Thus, you need that,
$\left\{ \begin{array}{c}\frac{\partial f}{\partial x}=0\\ \, \\\frac{\partial f}{\partial y}=0$
But,
$\frac{\partial f}{\partial x}=3x^2-3$
$\frac{\partial f}{\partial y}=3y^2-12$

Solving the equations we have,
$x=-1,1$
$y=-2,2$
Thus, the only possible points are,
$(-1,-2,38),(-1,2,-6),(1,-2,34),(1,2,2)$
• May 5th 2006, 09:03 AM
AVVIT
This is the only problem thats one of the answers i get but it cant be write as its suposed to be harder than that lol... or so i thought examples in my notes when partial differentiated always end up with an x and Y in the resultant making them harder and then transposing the forumula to calc maximas and minimas ..
• May 5th 2006, 09:22 AM
ThePerfectHacker
To determine whether these points are maximum or minimum you need to find the value of,
$D=\left| \begin{array}{cc} f_{xx}& f_{xy}\\ f_{xy}& f_{yy} \end{array} \right|$

We find that,
$f_{xx}=6x^2$
$f_{yy}=6y^2$
$f_{xy}=0$
Therefore,
$D=\left| \begin{array}{cc} 6x^2&0\\ 0&6y^2 \end{array} \right|=36x^2y^2>0$
Looking at,
$f_{xx}(x_0,y_0)>0$
Thus it is a relative minimuma.