Determining the relative maximum of a 2 variable function in less than 2.5 minuites.(GRE Exam 9768 Q.41)

Question

I know that I have to calculate the gradient of $f$, equate it to zero to find the critical points, calculate A,B,C,$\vartriangle$, then if $\vartriangle$ < 0 & A < 0 then f has relative(local) maximum at the given point.

But the critical points that I found are (0,0),($\sqrt{y/3}$,$\sqrt{x/3}$),($-\sqrt{y/3}$,$\sqrt{x/3}$),($\sqrt{y/3}$,$-\sqrt{x/3}$),($-\sqrt{y/3}$,$-\sqrt{x/3}$).

And this will take so long time to complete, and the correct answer is E which I do not have in my critical points.

Could anyone find my mistake or tell me a way for solving this question more rapid than this?

Answer 1

When you take partial derivatives: $$\begin{cases}f_x=y-3x^2=0 \\ f_y=x-3y^2=0\end{cases} \stackrel{-}\Rightarrow (y-x)(3y+3x+1)=0 \Rightarrow$$ $$1) \ y=x \Rightarrow x-3x^2=0 \Rightarrow x_1=0=y_1, x_2=\frac13=y_2.$$ $$2) \ y=-\frac{3x+1}{3} \Rightarrow 3x^2+\frac{3x+1}{3}=0 \Rightarrow \emptyset.$$ Second derivatives: $$f_{xx}=-6x,f_{yy}=-6y,f_{xy}=1,\Delta=f_{xx}f_{yy}-f_{xy}^2=36xy-1.$$ $$1)\ (0,0): \ \Delta<0 \ (saddle).$$ $$2) \ (\frac13,\frac13): \ \Delta=3>0,f_{xx}=-2<0,f_{yy}=-2<0 \ (max).$$

The answer is $E$.

Answer 2

Further my comment into a bigger hint:
Replacing the $y$ in $x=3y^2$ with $y=3x^2$ gives $x=3(3x^2)^2$.

You have the equation $27x^4-x=0$ to solve now. Do some fun factoring and win!

P.S. Don't forget to go back to $y=3x^2$ to find $y$