# What are the critical points, if any, of f(x,y) = 5x^2 + 4xy + 3y^2 - 52x - 56y + 13?

Jun 7, 2016

$f \left(x , y\right)$ has a local minimum point at $\left\{x = 2 , y = 8\right\}$

#### Explanation:

We will determine the stationary points instead. Those points observe the condition

$\nabla f \left(x , y\right) = \vec{0}$

In the present case we have

$\nabla f \left(x , y\right) = \left\{\frac{\partial f}{\partial x} , \frac{\partial f}{\partial y}\right\} = \left\{- 52 + 10 x + 4 y , - 56 + 4 x + 6 y\right\}$

and for stationary points determination

 { (-52 + 10 x + 4 y = 0), (-56 + 4 x + 6 y=0) :}

solving for $x , y$ we have

$\left\{x = 2 , y = 8\right\}$

Point qualification is obtained by computing

${\nabla}^{2} f \left(x , y\right) = \left(\begin{matrix}{f}_{x x} & {f}_{x y} \\ {f}_{y x} & {f}_{y y}\end{matrix}\right)$

In this point we have

${\nabla}^{2} f \left(2 , 8\right) = \left(\begin{matrix}10 & 4 \\ 4 & 6\end{matrix}\right)$

This matrix is positive definite because it have two positive eigenvalues $\left\{2 \left(4 + \sqrt{5}\right) , 2 \left(4 - \sqrt{5}\right)\right\}$ so the point is a local minimum point.