Archived May, 2026.

Find and classify CPs

mark

Find and classify the critical points of

f(x,y) = \frac{1}{3}x^3-x y+y^2.
User 017

I found two critical points for this function: a saddle point at (0,0) and a minimum at (\frac12,\frac14).

Partial derivative set up:

f_{x}=x^{2}-y
f_{xx}=2x
f_{xy}=-1
f_{y}=2y-x
f_{yy}=2

By setting f_{x} and f_{y} equal to zero I solved the system as follows:

x^{2}=y
2x^{2}-x=0
x(2x-1)=0

So, x=0 and x=\frac12.

Then y=0 and y=\frac14.

The critical points found are (0,0) and (\frac12,\frac14).

Using the Discriminant,

D_{(0,0)}=0-1=-1 = Saddle point
D_{(\frac12,\frac14)}=2-1=1 = Extreme point. f_{xx}=2 so this extreme point is a minimum.