# Thread: Calculus 3 - Optimization without Constraints

1. ## Calculus 3 - Optimization without Constraints

The behavior of a function can be complicated near a critical point where D=0. Suppose that f(x,y)= x^3 - 3xy^2.
a.) Show that there is one critical point at (0,0) and that D=0 at that point.
b.) Show that the contour for f(x,y)=0 consists of three lines intersecting at the origin where f alternates from positive to negative. Sketch a contour diagram for f near 0.

I already did part a, but I have no idea how to even start part b.

2. Originally Posted by alyssalynnx38
The behavior of a function can be complicated near a critical point where D=0. Suppose that f(x,y)= x^3 - 3xy^2.
a.) Show that there is one critical point at (0,0) and that D=0 at that point.
firstly solve for $\nabla f = 0$

3. Like I said, I already did part a.

$fx = 3x^2 - 3y^2$
$fy = -6xy$
$fxx = 6x - 3y^2$
$fyy = -6x$
$fxy = fyx= -6y$

If you set fx and fy equal to 0, both x and y are 0 because xy = 0 and x^2 + y^2 = 0

$D = fxx*fyy - (fxy)^2$
$D = (6x - 3y^2)*(-6x) - (-6y)^2$
$D(0,0) = (6*0 - 3*0^2)*(-6*0) - (-6*0)^2 = 0$

So, there is a critical point at (0,0) and D = 0. And that's as far as I was able to get.