Expert: Scotto - 12/3/2009

Question
A herpetologist is designing a rectangular experimental iguana pen to fit in the 8' by 20' corner of a Komodo dragon pen. If only 42 feet of fencing is available for the iguana pen, what dimensions should be used to obtain the largest area for the new pen?

Answer
If it were only fenced on two sides and put in the corner, it could easily take up all the room,
so I don't think that's it.

If it were fenced on all sides, it would have length L and width W.
The area would be A = LW and the perimeter would be 2L+2W.
We know that 2L+2W=42, or, dividing by 2, L+W=21, so L=W-21.

Since A = LW, we can then change that to A = (W-21)W = W² - 21W.
Setting the derivative equal to zero tells us 2W - 21 = 0.
This would give W = 10.5 and L = 10.5, but this is too big since the Komodo pen is only 8' wide.

By the way, Komodo dragons are found in Indonesia,
and we knew an exchange student from Indonesia.
She told us that Komodo is the Indnesian word for dragon,
so when we say Komodo dragon, to any Indonesian that speaks American, that is, "Dragon Dragon".
I still call them Komodo dragons, though.

Back to what's at hand.  Since it can only be 8' wide, that would mean the L could be 13'.
This would mean an iguana pen could be built that was 8' wide and 13' long.

This would be silly, since Iguanas are much smaller than Komono Dragons,
and yet the Iguana's would get the bigger part of the pen.

Now to get real about the problem.
I would put the fencing across the pen and fence off one end.
That would only use 8' of the fencing, leaving you 34' to do something else with.