How to find the the minimum non-negative value of a function:
$$f(x,y)=ax^2+by^2+cx+dy+e$$
s.t. $x$ lies in $[0, A]$ and $y$ lies in $[A, \infty),$
where $A$ is a known constant.
or simply $0\leq x\leq A\leq y$
Example: $f(x,y)=-x^2 + 2y^2 + 3y +8$ has a minimum positive value of $12$ for $A=1$. I have found this graphically but I would like to find the solution analytically.
Any help would be beneficial.
first you can find the critical points of $f$ (under which the inner extrema occur) in $(0, A) \times (A, \infty)$ by solving $$ \frac{\partial f}{\partial x}(x,y) = \frac{\partial f}{\partial y}(x,y) = 0. $$ After that find the critical points of $f(\cdot, A)$, $f(0, \cdot)$ and $f(A, \cdot)$ by solving $$ \frac{\partial f}{\partial x}(x, A) = 0, \frac{\partial f}{\partial y}(0, y) = 0 \text{ resp. } \frac{\partial f}{\partial y}(A, y) $$ Then compute the values of $f$ at each critical point and $f(0,A)$ and $f(A, A)$. You have found your minimum.
Concernig your example: We have \begin{align*} \partial_x f(x,y) &= -2x\\ \partial_y f(x,y) &= 4y + 3 \end{align*} Hence there are no inner critical points, no critical points on $(0, A) \times \{A\}$ and no on $\{0, A\} \times (A, \infty)$. So we compute $f(0,1) = 2 + 3 + 8 = 13$ and $f(1,1) = -1 + 2 + 3 + 8 = 12$. The mimimal positive value is therefore 12 (as $f$ is increasing for $y \to \infty$).