Shaping the largest pento-bridge

Contribution of each pentomino to shaping the rectangular outer border of
the bridge.

We
can see that the best choice leads to
6+6+5+4+3=24 units.

With this our enclosed rectangle should have a maximal area.

x + 2h = 24 => h = (24 - x) : 2

Area rectangle = f(x) = x.h

f(x)=x(12 -0,5x)

f(x)=-0,5x²+12x

Find out where this function has its maximum.

You can use your
TI84 Plus for this (if you don’t have
one, you can win one).

Take e.g. as window-settings:

Xmin=-30; Xmax=30;Xscl=5; Ymin=-100;Ymax=100; Yscl=10

You will find for the value of x at the top
12=>h=6

So we think that a maximal area will be reached if
x=12 en h=6.

We make also an exelfile.