# Finding critical points

1. Nov 13, 2015

### qq545282501

1. The problem statement, all variables and given/known data
find all critical points and identify the locations of local maximums, minimums and saddle points of the function $$f(x,y)=4xy-\frac{x^4}{2}-y^2$$

2. Relevant equations

3. The attempt at a solution
setting Partial derivative respect to x = 0 : $$4y-\frac{4x^3}{2}=0$$
setting partial derivative respect to y=0: $$4x-2y=0$$

from the second equation, Y=2X, plug it into the first equation. I get : $$2x(4-x^2)=0$$
now, x=0 or x=-2 or x=2. since y=2x, now I got 3 sets of critical points. (0,0), (-2,-4) and (2,4)

second partial derivative respect to x = $$-6x^2$$
second partial derivative respect to y=-2
and ƒxy=0

so $$D(a,b)= 12x^2$$

plug in these critical points into D(a,b) I found that local max value at (-2,-4) and (2,4) for which both give D value of 48, greater than 0, and ƒxx at both points are smaller than 0.
for D(0,0) , the test is inconclusive .

I did not find any local min, I am wondering if i missed something.
any help is appreciated

2. Nov 13, 2015

### ehild

fxy is not zero.

3. Nov 13, 2015

### Ray Vickson

$f_{xy} \neq 0$.

4. Nov 13, 2015

### qq545282501

opps, got it. thank you