# Global extremas of multivariable function on circle : are my results correct?

I wanted to know if my method and my results are correct. There isn’t a solution to that question, so I don’t know if it’s correct.

Given $$f(x,y)=3x^2-y^3$$ and I need to find its global extremas on the circle $$x^2+y^2 \leq 25$$.

So inside the domain, I have $$\nabla f =(6x, -3y^2)= (0,0)$$ which yields the trivial case $$x=y=0$$
Now on the border, we have $$x^2+y^2=25$$. I use lagrange multiplier : $$\nabla f = \lambda \nabla g$$ where $$g(x,y) =x^2+y^2-25$$
This gives $$(6x, -3y^2)= \lambda (2x, 2y)$$.
If $$x \neq 0$$ and $$y \neq 0$$, then $$\lambda = 3$$, $$y=-2$$ and $$x= \pm \sqrt{21}$$. If $$x=0, y = \pm 5$$. If $$y=0, x = \pm 5$$. So if we plug our values into our function, we get $$0, 71, -125, 125, 75$$ So there is a global maximum of 125 at $$(0,-5)$$ and a global minimum of $$-125$$ at $$(0, 5)$$.
Are my solutions correct ?

Also, I guess I could have used polar coordinates on the border given that it’s a circle. But if we do that ($$r=\sqrt{25}=5$$) and try to take the derivative of our function in polar coordinates, we get $$5^2\frac{d}{d \theta}(3cos^2(\theta)-5sin^3(\theta))$$ which isn’t nice to derivate. Do you agree that it’s much simpler to use Lagrange multiplier in this case ?