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 ?

Thanks for your help !