Question

Find and classify the stationary points of f(x,y)=y2-x2+3xy .

Answer 1

Answer Question #41401, Integral Calculus

Find and classify the stationary points of $f(x,y) = y^{2} - x^{2} + 3xy$.

Solution:

$$\left\{ \begin{array}{l} \frac {\partial f}{\partial x} = - 2 x + 3 y = 0 \\ \frac {\partial f}{\partial y} = 2 y + 3 x = 0 \end{array} \right. \quad \Rightarrow \quad x = 0; y = 0$$

$(0,0)$ - is a critical (stationary) point

$$\left\{ \begin{array}{l} \frac {\partial^ {2} f}{\partial x ^ {2}} = - 2 \\ \frac {\partial^ {2} f}{\partial y ^ {2}} = 2 \quad \Rightarrow \quad D = \left( \begin{array}{cc} - 2 & 3 \\ 3 & 2 \end{array} \right) \\ \frac {\partial^ {2} f}{\partial x \partial y} = 3 \end{array} \right.$$

Determinant of matrix $D$ is:

$$\left| \begin{array}{cc} - 2 & 3 \\ 3 & 2 \end{array} \right| = - 2 * 2 - 3 * 3 = - 4 - 9 = - 13 < 0$$

And matrix $D$ don't depends on $x$ and $y$, so all critical points are saddle points. It means that $(0,0)$ is a saddle point.

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