Question

What are the values and types of the critical points, if any, of `f(x,y)=4y(1 - x^2)`?

Answer 1

There are two critical points, with coordinates `(x,y)=(pm 1, 0)`, which are both saddle points.

Explanation:

Let `z=f(x,y)=4y(1-x^2)=4y-4x^2y`, then `(partial z)/(partial x)=-8xy` and `(partial z)/(partial y)=4-4x^2`.

After setting both of these partial derivatives equal to zero, we get that `4=4x^2` so that `x^2=1` and `x= pm 1` and we also have `-8xy=0`, so that `y=0`. The two critical points are therefore `(x,y)=(pm 1, 0)`.

The second-order partial derivatives are `(partial^{2} z)/(partial x^{2})=-8y`, `(partial^{2} z)/(partial y^{2})=0`, and `(partial^{2} z)/(partial x partial y)=-8x`, giving a discriminant of

`(partial^{2} z)/(partial x^{2}) * (partial^{2} z)/(partial y^{2})-((partial^{2} z)/(partial x partial y))^{2}=0-(-8x)^{2}=-64x^{2}`

This discriminant is negative at both critical points, making them both saddle points.