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