Originally Posted by
HallsofIvy I assume you are asking for the minimum value $\displaystyle ax+ by^2$ on a regions bounded by some lines buy $\displaystyle 0<= x>= 1$ and $\displaystyle 0<=y>=1$ make no sense. If $\displaystyle x>= 0$ then x must be "$\displaystyle >= 0$. Did you mean "$\displaystyle 0<= x< = 1$ and $\displaystyle 0<= y<= 1$?
If so then the minimum value must occur:
1) in the interior of the set where the gradient is 0 (or undefined) or
2) on the boundary of the set.
$\displaystyle \nabla ax+ by^2= a\vec{i}+ 2by\vec{j}$ which is never 0 because of the "$\displaystyle a\vec{i}$" term. On the line x= 0, the y axis, the function is $\displaystyle by^2$ which has its minimum at y= 0. On the line y= 0, the x-axis, the function is $\displaystyle ax$ which has its minimum at x= 0. On the line x+ y= 1, x= 1- y so the function is $\displaystyle a- ay+ by^2$. The derivative of that is $\displaystyle -a+ 2by$ and is 0 at $\displaystyle y=\frac{a}{2b}$. Then $\displaystyle x= 1- \frac{a}{2b}= \frac{2b- a}{2b}$.
Of course, we also need to consider the vertices (1, 0) and (0, 1). That is, the minimum value must occur at one of (0, 0), (1, 0), (0, 1), or $\displaystyle \left(\frac{2b-a}{2b}, \frac{a}{2b}\right)$. Evaluate $\displaystyle ax+ by^2$ at each of those points to decide where it is smallest.