After exploring the first Box Problem applet, some essential elements of moving a seemingly two-variable problem into a single variable problem have come to light. In particular, students are now prepared to explore the single-variable box problem. Specifically, the introduction provided by the first box problem applet has provided a means of reducing the complexity of the problem. Instead of having two variables with which to contend, we have found that without loss of generality, we can set the position of the cut to be twice the length of the cut.
What exactly does this provide? We can now express the volume of the box with respect to a single variable thereby opening this problem to the techniques from Calculus 1. In particular, we can turn our attention to maximizing the volume of a box based upon the cut length. Now, the second Box Problem applet, will focus its attention on looking at
\[ V(l) = \left( {B - 2l} \right)\left( {{1 \over 2}A - 2l} \right)\left( {2l} \right) \]
where A and B are the correspondent length and width of the rectangular piece of cardboard and l corresponds to the length of the cut.
We should note that this description is equivalent to the formula presented in Dundas (1984) of
\[ V(l) = T\left( {{1 \over 2} - T} \right)\left( {{A \over l} - T} \right) \]
as long as the following adjustments are made:
In addition, the l of Dundas (1984) needs to correspond to the position of the cut instead of the length of the cut.