Given
Find
Solution
\(
\begin{array}{l}
\text{Extremum:} \\
\nabla f(x,y) = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}) = (2x, 2y) = 0 \\\\
\begin{eqnarray} \text{I:}\ & 2x = 0 \\ \text{II:}\ & 2y = 0 \end{eqnarray} \\\\
E_1 = (0,0) \\\\\\
\text{Types:} \\
H_f(x,y) = \begin{pmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \\\\
H_f(0,0) = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \\
H_1(0,0) = 2 \gt 0\\
H_2(0,0) = \begin{vmatrix} 2 & 0 \\ 0 & 2 \end{vmatrix} = 2 \cdot 2 – 0 \cdot 0 = 4 \gt 0 \\\\
H_f(0,0)\ \text{is positive-define and therefore}\ E_1\ \text{is a minima.}
\end{array}
\)
Answer