The minimum value of f(x,y)=x^2+13y^2-6xy-4y-2 is?

Jul 23, 2018

$f \left(x , y\right) = {x}^{2} + 13 {y}^{2} - 6 x y - 4 y - 2$

$\implies f \left(x , y\right) = {x}^{2} - 2 \cdot x \cdot \left(3 y\right) + {\left(3 y\right)}^{2} + {\left(2 y\right)}^{2} - 2 \cdot \left(2 y\right) \cdot 1 + {1}^{2} - 3$
$\implies f \left(x , y\right) = {\left(x - 3 y\right)}^{2} + {\left(2 y - 1\right)}^{2} - 3$

Minimum value of each squared expression must be zero.

So ${\left[f \left(x , y\right)\right]}_{\text{min}} = - 3$

Jul 23, 2018

There is a relative minimum at $\left(\frac{3}{2} , \frac{1}{2}\right)$ and $f \left(\frac{3}{2} , \frac{1}{2}\right) = - 3$

Explanation:

I think that we must calculate the partial derivatives.

Here,

$f \left(x , y\right) = {x}^{2} + 13 {y}^{2} - 6 x y - 4 y - 2$

The first partial derivatives are

$\frac{\partial f}{\partial x} = 2 x - 6 y$

$\frac{\partial f}{\partial y} = 26 y - 6 x - 4$

The critical points are

$\left\{\begin{matrix}2 x - 6 y = 0 \\ 26 y - 6 x - 4 = 0\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}3 y = x \\ 26 y - 6 \cdot 3 y - 4 = 0\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}3 y = x \\ 8 y = 4\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}x = \frac{3}{2} \\ y = \frac{1}{2}\end{matrix}\right.$

The second partial derivatives are

$\frac{{\partial}^{2} f}{\partial {x}^{2}} = 2$

$\frac{{\partial}^{2} f}{\partial {y}^{2}} = 26$

$\frac{{\partial}^{2} f}{\partial x \partial y} = - 6$

$\frac{{\partial}^{2} f}{\partial y \partial x} = - 6$

The determinant of the Hessian matrix is

$D \left(x , y\right) = | \left(\frac{{\partial}^{2} f}{\partial {x}^{2}} , \frac{{\partial}^{2} f}{\partial x \partial y}\right) , \left(\frac{{\partial}^{2} f}{\partial {y}^{2}} , \frac{{\partial}^{2} f}{\partial y \partial x}\right) |$

$= | \left(2 , - 6\right) , \left(- 6 , 26\right) |$

$= 52 - 36$

$= 16 > 0$

As $D \left(x , y\right) > 0$

and

$\frac{{\partial}^{2} f}{\partial {x}^{2}} = 2 > 0$

There is a relative minimum at $\left(\frac{3}{2} , \frac{1}{2}\right)$

And

$f \left(\frac{3}{2} , \frac{1}{2}\right) = {1.5}^{2} + 13 \cdot {0.5}^{2} - 6 \cdot 1.5 \cdot 0.5 - 4 \cdot 0.5 - 2 = - 3$