Take the first partial derivatives of f(x, y):
∂f/∂x = -2 (x - 5) = 10 - 2x
∂f/∂y = -2y
The critical points occur where both derivatives are zero:
10 - 2x = 0 ===> x = 5
-2y = 0 ===> y = 0
So there is only critical point at (5, 0), which takes on a value of f(5, 0) = 25.
Compute the Hessian matrix for f :
[tex]H(x, y) = \begin{bmatrix}\frac{\partial^2f}{\partial x^2} & \frac{\partial^2f}{\partial x\partial y} \\ \frac{\partial^2f}{\partial y\partial x} & \frac{\partial^2f}{\partial y^2}\end{bmatrix} = \begin{bmatrix}-2&0\\0&-2\end{bmatrix}[/tex]
Since det(H(x, y)) = 4 > 0 for all x, y, the critical point (5, 0) is a relative minimum.
f has no saddle points.
