Finding constrained optima of $f(x,y) = x^3 - y^3 -x$

Question

For a function $$f(x,y) = x^3 - y^3 -x$$, the minimum and maximum under the constraint $$x^2 + y^2 =1 $$

is searched.

So as usual, my approach is to set up the Lagrangian and the FOC:

$$ L(x,y, \lambda) = x^3 - y^3 -x + \lambda(x^2 + y^2 -1)$$ $$ \frac{\delta L}{\delta x} = 3 x^2-1+ 2\lambda x = 0 $$ $$ \frac{\delta L}{\delta y} = -3 y^2+ 2\lambda y = 0 $$ $$x^2 + y^2 - 1 = 0 $$

That, however, appears to be fairly complicated to solve, as both substituting $y=\sqrt{1-x^2}$ and $x=\sqrt{1-y^2}$ ends up in a pretty messy equation. Thus, I wonder, am I on the right path at all?

Given past assignments, I observed a tendency of 'nice' solutions, but the current path doesn't seem nice at all, which is why I would be grateful for some input. Thanks!

Answer 1

Idea: Substitute $x$ by $\cos \varphi$ and $y$ by $\sin \varphi$ for $\varphi \in [0,\pi/2)$ and see if the objective function can be put in a nice form using trigonometric identities. The constraint is satisfied.

Answer 2

it is not difficult to solve the $x,y$.

first :$y=0$ case, $x^2=1$, you get $2$ pair $(x,y)$

second: $y \not=0$, $x\not=0$, then $y=\dfrac{1}{3x}-2x$, you put in $x^2+y^2=1$, you will have something like $ax^4+bx^2+c=0$, so you have $4$ pair $(x,y)$, some may not real numbers.

last step you should know what to do with these $(x,y)$.