# Homework Help: Stationary Points

1. May 2, 2010

### Jenkz

1. The problem statement, all variables and given/known data

Find the four stationary points of the function:
u(x, y) = 4x^3 − 18(x^2)y + 24x(y^2) − 120y

Determine whether they are maxima, minima or saddle points.

2. Relevant equations

To find stationary points use:

E= (d^2u/dxy)^2 - [(d^2u/dx^2) * (d^2u/dy^2)]

E < 0 Maximum -> d^2u/dx^2 < 0
Minimum -> d^2u/dx^2 > 0

3. The attempt at a solution

du/dx = 12x^2 - 36xy +24y^2 (1)
du/dy = -18x^2 + 48xy-120 (2)

Stationary points mean both du/dx and du/dy are equal to 0. Here I should find simultaneous solution of equations (1) and (2). This is where I get stuck and I am not sure how to find them.

I have done the next part though:

E= (-36x + 48y)^2 - (1152x^2 - 1728xy)

But I need the stationary points to find the min/max/saddle points.

2. May 2, 2010

### D H

Staff Emeritus
Try factoring one of those equations.

3. May 2, 2010

### Jenkz

Ok i've tried:

du/dx = 12x^2 - 36xy +24y^2 (1)
du/dy = -18x^2 + 48xy-120 (2)

(1) x^2 - 3xy + 2y^2 = 0
(2)-3x^2 + 8xy -20 = 0

Eliminate x^2 ; xy +6y^2 - 20= 0 ; x = 20/y - 6y
Or
Eliminate xy ; x^2 - 16y^2 + 60 = 0 (not sure how to factorise)

Do I sub that back into the previous equtions?

I am still not sure where to go form here.

4. May 2, 2010

### Staff: Mentor

The first equation can be written in factored form as
(x - y)(x - 2y) = 0

BTW, you should post problems like this in the Calculus & Beyond section, not this section.

Last edited by a moderator: May 2, 2010
5. May 3, 2010

### Jenkz

Thank you! I think I know what I need to do now with the other equation now.

Noted, i'll post in the other section.