Calculate stationary points and local maxima and minima

Question

Let $a, b∈ \mathbb{Z}$ \ {$0$} and let $f: \mathbb{R^2}→\mathbb{R}$ be defined by $$f(x_1, x_2)=ax_1^2+bx_2^2-4ab^2x_1-2a^2bx_2.$$

Find all stationary points of $f$ and, if possible, determine which of the stationary points are strict local maxima and strict local minima. $$$$ Now, I know that, in order to calculate the stationary points, first I have to calculate the gradient of $f$. Afterwards, we calculate $\nabla{f}=0$, and then we find the stationary points. Since a and b are constants, we have $\frac{\partial{f}}{\partial{x_1}}=2ax_1-4ab^2$ and $\frac{\partial{f}}{\partial{x_2}}=2bx_2-2a^2b$, therefore $\nabla{f}=(2ax_1-4ab^2, 2bx_2-2a^2b)$ and $\nabla{f}=0$. The only way for $\nabla{f}=0$ is $x_1=2b^2$ and $x_2=a^2$. Are these stationary points correct?

Answer 1

The stationary points are solutions of the system $$ 2a x_1-4ab^2=0, \quad 2b x_2 -2a^2b =0 $$

which leads to $x_1 = 2b^2$ and $x_2=a^2$. This is the only stationary point.

• If both $a$ and $b$ are positive you get a (global) minimum.
• If $a<0$ and $b>0$ you get a (global) maximum.
• In other cases, you get a saddle point.