# Classify stationary points of $f(x,y) =8x^3-3x^4+48xy-12y^2$

Find and classify the stationary points (min,max,saddle)$$f(x,y) =8x^3-3x^4+48xy-12y^2$$

For the most part, I can solve this problem... I am actually just stuck at identifying the critical points. (I'm used to easier, or differently styled problems).

What I know I need is the partial in terms of $x$ and $y$ and set them equal to $0$. But when I do I honestly don't understand how to solve them.

$$f_x = 24x^2 -12x^3+48y =0$$ $$f_y=48x-24y=0$$ for the 2nd equation id solve for y, which is 2x=y. But if i plug that in I get mumbo jumbo I can't figure out. Help please!

We have the function

$$f(x, y) = 8x^3-3x^4+48xy-12y^2$$

We have

$$f_x = -12 x^3+24 x^2+48 y, f_y = 48 x-24 y$$

If we set both equations equal to zero and simultaneously solve them, from the second equation, we get $y = 2x$. Substituting this into the first equation, we get $$-12 x^3+24 x^2+96 x = -12 x(x-4) (x+2) =0 \implies x = -2, 0, 4$$

Using these three $x-$values, we substitute them back in $y = 2x$ and this produces three critical points:

$$(x, y) = (-2, -4), (0,0), (4, 8)$$

• when you find the 3 x values, do you plug them back into the original function? or into the partial of y function? or... the same one you got them from? Dec 13, 2016 at 8:56
• That is correct, please see update.
– Moo
Dec 13, 2016 at 8:57
• awesome thanks for the clarification. Dec 13, 2016 at 8:58