Thread: Help with Stationary Points of a surface sadle points thing ..

1. 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

2. 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,
$(-1,-2,38),(-1,2,-6),(1,-2,34),(1,2,2)$

3. 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 ..

4. 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.