# Finding Stationary Points

1. Jan 7, 2014

### MMCS

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

Attempt at solution

∂f/∂x = 3x2+12x
∂f/∂y = 3y2-12y

Factorise

x(3x+12)

x=0 or,
3x+12 = 0 → x= -4

y(3y-12)

y=0 or,
3y-12=0 → y=4

I find the next stage difficult, because there are no mixed terms in the polynomial any of the values i substitute into another gives me 0, because of this i need help finding the stationary points, I'm fine with classifying them

Thanks

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2. Jan 7, 2014

### haruspex

I don't see the difficulty. You have two values for x and two for y. How many combinations does that give for (x, y)?

3. Jan 7, 2014

### MMCS

This just gives me two stationary points doesn't it?

4. Jan 7, 2014

### MMCS

I see, it is 4, i thought there would be other values i would need, so my points would be

(0,0)
(0,4)
(-4,0)
(-4,4)

Is this correct?

5. Jan 7, 2014

### HallsofIvy

Staff Emeritus
If you had had "mixed" terms, xy, then you would have equations with both x and y leading to the result that a specific value of x leads to a specific value of y. Here, that does not happen. The derivative with respect to x will be 0 if x= 0 or -4 no matter what y is. The derivative with respect to y will be 0 if y= 0 or 4 no matter what x is. In particular, both derivatives will be 0 at (0, 0), (0, 4), (-4, 0), and (-4, 4) as you say.

6. Jan 7, 2014

### MMCS

Thanks, that explanation helps me understand that better

7. Jan 7, 2014

### Ray Vickson

Yes, those are the stationary points. As for classification: have you taken (multivariate) second-order tests yet?