# Max/min with partial derivatives

1. Jul 6, 2009

### compliant

1. The problem statement, all variables and given/known data
Show that f(x,y) = -(x^2 - 1)^2 - (yx^2-x-1)^2 has only two critical points, and both are maxima.

3. The attempt at a solution
Set partial derivatives (wrt x and y) to zero to find critical pts.

f_x = -2(x^2 - 1)(2x) - 2(yx^2 - x - 1)(2xy - 1) = 0
f_y = -2(yx^2 - x - 1)(x^2) = 0

-2(x^2 - 1)(2x) - 2(yx^2 - x - 1)(2xy - 1) = -2(yx^2 - x - 1)(x^2)
(x^2 - 1)(2x) + (yx^2 - x - 1)(2xy - 1) = (yx^2 - x - 1)(x^2)
2x^3 - 2x + 2(x^3)(y^2) - 2(x^2)y - 2xy - yx^2 + x + 1 = yx^4 - x^3 - x^2
3x^3 + x^2 - x + 1 = yx^4 -2(x^3)(y^2) + 2(x^2)y + 2xy
3x^3 + x^2 - x + 1 = y(x^4 -2(x^3)(y) + 2(x^2) + 2x)

And that's about as far as I got.

2. Jul 6, 2009

### queenofbabes

look at the f_y equation. What can you deduce about its solutions straightaway (it's already factorized!) ? Check if each solution works by comparing with f_x equation. I see only two workable ones. I suppose I'll leave it to you to check for maxima.

3. Jul 6, 2009

### compliant

well, from f_y, I can deduce that x=0 or yx^2 - x - 1 = 0.

If x=0, from f_x, y= -2.

If yx^2 - x - 1 = 0,
y = (x+1)/(x^2)

Subbing into f_x,

-2(x^2 - 1)(2x) - 2((x+1)-x-1)(2x(x+1)/(x^2) - 1) = 0
-2(x^2 - 1)(2x) = 0
x(x^2 - 1) = 0
x = 0 (as above), x = 1 or x = -1

If x = 1, y = 2
If x = -1, y = 0

Using the second derivatives test,
D = (f_xx)(f_yy) - (f_xy)^2
where

f_xx = -2[6x^2 - 2 + 6(x^2)(y^2) - 4xy - 2y]
f_yy = -2[x^4]
f_xy = -2[4yx^3 - 3x^2 - 2x]

At (0,-2), f_xx = -2(-2+4) = -4; D = (-4)(0) - 0 = 0
At (1,2), f_xx = -2(6-2+24-8-4) = -32; D = (-32)(-2) - (3)^2 > 0
At (-1,0), f_xx = -2(6-2) = -8, D = (8)(-2) - (-5)^2 < 0

So, I essentially have a saddle point at (-1,0), a maximum at (1,2) and an unknown at (0,-2). The problem with this is that:

1) there are three critical points, somehow, and not two.
2) only one is a maximum.

4. Jul 6, 2009

### queenofbabes

"If x=0, from f_x, y= -2."

Are you sure?

f_x = -2(x^2 - 1)(2x) - 2(yx^2 - x - 1)(2xy - 1)
= -2(0-1)(0) - 2(-1)(-1) = -2

its f_x, NOT y.

Thus, what does this mean?

5. Jul 6, 2009

### compliant

*backtracks*

ooooooooooookay. so.

when x=0, f_x = -2, but f_x = 0, so x=0 is not a solution.

which brings me back to the points (-1,0) and (1,2). looking at my calculations for (-1,0), turns out that it's horrendously wrong.

At (-1,0), f_xx = -2(6-2) = -8, D = (-8)(-2) - (0-3+2)^2 = 16 - 1 > 0

Which gives a maximum. Hooray. Thanks for the help.

6. Jul 6, 2009

np =)