Absolute Minimum of a Quadratic Function

Taking the gradient of $f(x,y)$ equal to zero gives us:

$f_{x} = 10x + 3y + 4 = 0;$

$f_{y} = 3x + 20y - 20 =0$

which this 2x2 linear system yields the unique point $P(x,y) = (-\frac{140}{191}, \frac{212}{191})$ . The Hessian Matrix, $F(x,y)$ , computes to:

$F(x,y) = \begin{bmatrix} f_{xx} && f_{xy} \\ f_{yx} && f_{yy} \end{bmatrix} = \begin{bmatrix} 10 && 3 \\ 3 && 20 \end{bmatrix}$

which is positive-definite for all $x,y \in \mathbb{R} \Rightarrow$ $P$ is a global minimum point. Hence, the global minimum of $f$ computes to $f(-\frac{140}{191}, \frac{212}{191}) = \boxed{\frac{7150}{191}}.$

$(ax + by + c)^2 + (dy + e)^2 + f$ has a global minimum of $f$ when $ax + by + c = 0$ and $dy + e = 0$ , and the former can be expanded to:

$a^2x^2 + 2abxy + (b^2 + d^2)y^2 + 2acx + 2(bc + de)y + c^2 + e^2 + f$

Comparing this to the given equation:

$f(x, y) = 5x^2 + 3xy + 10y^2 + 4x - 20y + 50$

we can set $a^2 = 5$ , $2ab = 3$ , $b^2 + d^2 = 10$ , $2ac = 4$ , $2(bc + de) = -20$ , and $c^2 + e^2 + f = 50$ .

Solving (mostly) from left to right, $a = \sqrt{5}$ , $b = \cfrac{3\sqrt{5}}{10}$ , $c = \cfrac{2\sqrt{5}}{5}$ , $d = \cfrac{\sqrt{955}}{10}$ , $e = -\cfrac{106\sqrt{955}}{955}$ , and $f = \cfrac{7150}{191}$ , so that $f(x, y)$ can be rearranged to:

$f(x, y) = (\sqrt{5}x + \cfrac{3\sqrt{5}}{10}y + \cfrac{2\sqrt{5}}{5})^2 + (\cfrac{\sqrt{955}}{10}y - \cfrac{106\sqrt{955}}{955})^2 + \cfrac{7150}{191}$

which has an absolute minimum of $f = \cfrac{7150}{191} \approx \boxed{37.435}$ .