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First I solved $f_y=0$ then plugged in my variable into $f_x$ to get an output and then plugged that output back into $f_y$ to get a point and did this again for $f_x$ I think I screwed up on the $(-7,\sqrt{3})$

$f_y=-12x+24y^2$

$-12x+24y^2=0$

$-12x=24y^2$

$x=2y^2$

$f_x(2y^2,y)=3(2y^2)^2-12y$

$3(4y^4)-12y=0$

$12y^4 - 36y=0$

$12y(y^3-3=0) \to y=0$ or $y=\sqrt{3}$

$f_y(x,0)=-12x=0 \to (0,0)$

$f_y(x,\sqrt{3}) = -12x+24(3)$

$f_y(x.sqrt{3})= -12x-84=0$

$f_y(x,\sqrt{3})=-12x=84$

$x=-7 \to (-7,\sqrt{3})$

$f_y=-12x+24y^2=0$

$-12x=24y^2$

$x=2y^2$

$f_x(2y^2,y)=3(2y^2)^2-12y=0$

$12y^4-12y=0$

$12y(y^3-1)=0$

$y=0$ or $y=1$

$f_y(x,o)= -12x=0 \to x=0$

$f_y(x,1)=-12x+24=0$

$f_y(x,1)=x=2$

Critical points: $(2,-1),(0,0),(-7,\sqrt{3})$

$D=(6x)(48y)-(-12)^2$

$D(0,0)=-144 so (0,0) is a saddle point

$D(,2,-1)= Min

D(-7,\sqrt{3})= Saddle point

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  • $\begingroup$ From passage $6$ to passage $7$ your $-12y$ became $-36y$; thus you have $y=0$ or $y=1$ $\endgroup$ – Giulio Feb 23 '17 at 23:28
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$$f_x=3x^2-12y$$ $$f_y=24y^2-12x$$

Now we let $\nabla f(x,y)=(f_x,f_y)=0$ which gives us the following: $$ \left\{ \begin{array}{c} f_x=3x^2-12y=0 \\ f_y=24y^2-12x=0 \end{array} \right. $$

Solving it we have $$ \left\{ \begin{array}{c} x=2y^2 \\ 12y^4-12y=12y(y^3-1)=0 \end{array} \right. $$

Where $y=0\lor y=1$ and respectively $x=0\lor x=2$. Can you continue now?

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