Calculate stationary points and local maxima and minima

maxima-minimapartial derivativestationary point

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.
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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?

Best Answer

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.
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