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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!

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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:

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

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  • $\begingroup$ 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? $\endgroup$
    – Ryanori
    Dec 13, 2016 at 8:56
  • $\begingroup$ That is correct, please see update. $\endgroup$
    – Moo
    Dec 13, 2016 at 8:57
  • $\begingroup$ awesome thanks for the clarification. $\endgroup$
    – Ryanori
    Dec 13, 2016 at 8:58

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