Use the Second Derivative Test to find all critical points of the function.

Question

The function is $$f(x,y)=x^3-6xy +8y^3$$ Finding the first partial derivative of $x$ and $y$: $$f_x(x,y)= 3x^2-6y$$ $$f_y(x,y)=-6x+24y^2$$ Now we set these to zero and solve. When I first did this problem I did it like so, $$3x^2=6y \to x^2=2y$$ $$24y^2=6x \to 4y^2=x$$ So now I have $x=4y^2$ so I plugged that into the partial of $x$ to get $y =0$ or $y= \frac{1}{2}$. $$3(4y^2)-6y=0 \to y(12y-6)=0$$ But I now realize that I forgot to square my $4y^2$ but still got the right values for $y$. I guess I got lucky but I would like to know how it's supposed to go. I asked my professor and he said to solve them simultaneously, I tried that but found in either case I have to multiply one of the partial derivative by the other variable (to get $y^2$ or $x^2$). So that's where I am stuck at right now.

Answer 1

Take $x=4y^2$ and plug it into $x^2=2y$. You get $$(4y^2)^2=2y$$or $$y(8y^3-1)=0$$ You have two real solutions, $y=0$ and $y=1/2$, and two complex solutions, which we can ignore. The corresponding $x$ values are $0$ and $1$.