Phoenix86 wrote:

First of this is somewhat embarassing as I really shouldn't be strugling with this, however I'm having a hard time getting back into this after a break of about a year.

I'm trying to obtain the critical points of the function bellow and then determin their nature. I'm having no trouble with the differentiation however solving the first derivatives for x and y is driving me nuts. Once I have these values I know how to solve the rest of the problem.

A surface has the following equation:

z=x^3-3xy+8y^3+20

So:

dz/dx= 3x^2-3y = 0

dz/dy= -3x+24y^2 = 0

thus:

3x^2=3y so x^2=y

and

24y^2=3x so 8y^2=x

CP's:

x = (A)??? & (B)???

y = (A)??? & (B)???

Z= (A)x^3-3(A)x(A)y+8(A)y^3+20

&

Z= (B)x^3-3(B)x(B)y+8(B)y^3+20

Any help is greatly appreciated.

Craig

You have a critical point when the tangent plane is "vertical", so first you find the equation for your tangent plane:

Write

then the tangent plane is

now

and similarly,

, and the plane is vertical when both are equal to zero.

So all you need to do is decide when

Using the second equation and squaring we get:

and substituting into the first you get:

, so we get

which works in the original set of equations, otherwise

so

, this gives the other option as:

so

and

from the first equation, and checking with the second equation, we see that only the positive possibility works.