Find the local maxima and minima of $ f(x,y) = x^3+y^3 $ in $ g(x,y) = x^2+y^2-1 $

Question

I've been trying to solve this problem for a while but I seem to get to the wrong answer every time.
My steps are the following:

I have to find the solution to the system:
$$ \begin{cases} \nabla f = \lambda \nabla g \\ g(x,y)=0, \end{cases} $$ and that is $$ \begin{cases} 3x^2 = 2 \lambda x\\ 3y^2 = 2 \lambda y \\ x^2+y^2-1=0. \end{cases} $$ But now, when I try to solve for $ (x,y,\lambda) $, I keep getting stuck.

I have the feeling that I'm making a stupid mistake somewhere, or assuming something that isn't quite right.

If anyone could give me some ideas, I would greatly appreciate it :)

Thanks in advance!

Answer 1

If you start with what you properly wrote $$\begin{cases} 3x^2 = 2 \lambda x\\ 3y^2 = 2 \lambda y \end{cases}$$ you have four solutions, namely $$\left\{x= \frac{2 \lambda }{3},y= 0\right\},\{x= 0,y= 0\},\left\{x= 0,y= \frac{2 \lambda }{3}\right\},\left\{x= \frac{2 \lambda }{3},y= \frac{2 \lambda }{3}\right\}$$ Just use them in the last equation $x^2+y^2-1=0$. You can discard the second one.

The first and third solutions would give $1=\frac{4 \lambda ^2}{9}$ while the fourth would give $1=\frac{8 \lambda ^2}{9}$.

Just continue.

Answer 2

$$x(3x-2\lambda )=0$$ $$y(3y-2\lambda )=0$$

Case $1$: $x=0$, then the problem become optimizing $y^3$ subject to $y^2=1$.

Case $2$, $y=0$, then the problem become optimizing $x^3$ subject to $x^2=1$.

Case $3$: $x \neq 0, y \neq 0$, then we have $x=y$ and $x^2+y^2=1$.Hence $x=y = \frac1{\sqrt2}$.

Remark:

If we use polar coordinate, the problem become optimizes $\cos^3(\theta) + \sin^3(\theta)$.