# Find and classify the stationary points for $f(x,y) = (4x_1^2 - x_2)^2$

So first I calculated the gradient which was $$\nabla f(x) = (64x_1^3 - 16x_1x_2, -8x_1^2 + 2x_2)^T.$$ Then setting the equations in this equal to $$0$$ I solved for $$x_2$$ and got $$x_2=4x_1^2$$. Then calculated the Hessian matrix which I calculated to be $$\begin{equation*} \nabla ^2 f(x)=\left( \begin{array}{rr} 192x_1^2 -16x_2 & -16x_1 \\ -16x_1 & 2 \end{array} \right). \qquad \end{equation*}$$ Then I plugged in $$4x_1^2$$ for $$x_2$$ and got $$\begin{equation*} \nabla ^2 f (x)= \left( \begin{array}{rr} 128x_1^2 & -64x_1^2 \\ -64x_1^2 & 2 \end{array} \right). \qquad \end{equation*}$$ After this I'm stuck and don't really know how to figure out what the stationary points are. I know that there are stationary points whenever $$x_2 = 4x_1^2$$, but I'm not sure how to classify these as saddle points, local minimums, maximums or anything. Any help?

Thanks

$$16x_1(4x_1^2−x_2) = 0\\ -2x_2(4x_1^2-x_2) = 0$$

We have this parabola $$4x_1^2 = x_2$$ where the partial derivatives are $$0.$$

$$(0,0)$$ is on the parabola, so no need for a special call out.

$$f(x_1,x_2) = 0$$ for all points on the parabola.

$$f(x_1,x_2) > 0$$ for all points not on the parabola.

The parabola forms a sort of trough, and they are all minima.

$$\begin{bmatrix} f_{xx} & f_{xy}\\ f_{xy}& f_{yy}\end{bmatrix}$$
(by the way, I think you mistyped something) If the determinant of this matrix is greater than $$0,$$ you have a max or a min. If it is less than 0 you have a saddle. In this case it is exactly $$0$$ and no help at all.
• For all the points on the parabola $f(x_1,x_2) = 0$ And $f(x_1,x_2) \ge 0$ for all $(x_1,x_2)$ This set must be at the minimum of the function. The function doesn't have the right kind of curvature for the second derivative test to tell us anything, so we need to look for other ways to think about it. – Doug M Sep 28 '18 at 2:39