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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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    You can check for critical points inside the domain, and also check for minimum values along the boundary.2012-03-27
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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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    Thanks. Also the partial derivatives w.r.t x and y are single variable only, therefore critical points of f(.,A), f(0,.) and f(A,.) are same as that of f(x,y). The result is that minimum would lie either at the boundary values f(0,A) , f(A,A) or at the critical points if they fall inside the region defined by the constraint.? Is this right? Also does it guarantee that the minimum would be non-negative?2012-03-27
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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