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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.

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    yes, but its possible that critical points do not fall inside the constraint2012-03-27

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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 $y \to \infty$).

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    @Abhinav: The point is the following: For $(x,A)$ to be a critical point of $f(\cdot, A)$ we only want to have $\partial_x f(x,A) = 0$, but for an inner extreme point we need $\partial_x f(x,y) = \partial_yf(x,y) = 0$. And: No it doesn't garantee that the miminum would be non-negative, but: If you have $f(x,y) \to \infty$ for $y \in \infty$ as in your example and $f$ attains negative values, $f$ attains $0$ as its minimal non-negative value by continuity of $f$ and connectedness of $[0, A] \times [A, \infty)$.2012-03-27