"Box With No Top" Optimization I am having some trouble with this problem,

A box with no top is to be constructed from a piece of cardboard of dimensions $A$ by $B$ by cutting out squares of length $h$ from the corners and folding up the sides as in the figure below. Suppose that the box height is $h = 1 \text{ inch}$ and that it is constructed using $186$ $\text{ inches}^2$ of cardboard (i.e., $AB = 186$). Which values $A$ and $B$ maximize the volume? 



My Attempt:
We have, 
$$SA=AB+1(A+B)=AB+2A+2B=186.$$
Solving for $A$,
$$A=(186-2B)/(B+2).$$
Now, use this to calculate the volume,
$$\begin{align} V &= ABH \\
       &= AB \cdot 1 \\
       &= AB \\
       &= \left( \frac{186-2B}{b+2} \right) B. \\
\end{align}$$
Differentiating with respect to $B$,
$$\frac{dV}{dB} = \dfrac{-2(B^2+4B-186)}{(b+2)^2}.$$
Setting $\frac{dV}{dB}$ equal to zero and solving for $B$ yields,
$$B= -2-\sqrt{190} \ \text{ or } \ B = \sqrt{190}-2.$$

However, neither of these solutions seem to be correct. What am I doing wrong? 

The volume is $(A-2)(B-2)$. We want to maximize this given $AB=186$.
Now $(A-2)(B-2)=AB-2(A+B)+4=190-2(A+B)$. We want to minimize $A+B$, which is $A+\frac{186}{B}$. The minimum exists, at a place in "the middle." So the derivative there is $0$. Now it's over: we get the expected $A=B=\sqrt{186}$. 
