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

1 Answer
Jan 3, 2016

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.