Question 142358
{{{P=2L + 2W}}}
{{{A=LW}}}


Given P = 22, so {{{2L + 2W=22}}}


A little algebra gets us: {{{L=11-W}}}


So {{{A=LW=(11-W)W=-W^2+11W}}}


This equation graphs to a convex down parabola with vertex at {{{-11/(2(-1))=11/2}}}, telling us that the maximum area is achieved when the length of the side is {{{11/2}}}.  The difficulty is that the problem asks for the maximum area of a rectangle with <b><i>integer</i></b> side lengths.  The nearest integer smaller than {{{11/2}}} is 5 and the nearest integer larger than {{{11/2}}} is 6.


Substituting 5 for W in {{{2L + 2W=22}}} gives us {{{L=6}}}, so obviously substituting 6 for W in {{{2L + 2W=22}}} will give us {{{L=5}}}.  Since 5 and 6 are the nearest integers to the side length that provides the maximum area (notice that all 4 sides would be {{{11/2}}} making a square) these must be the required side lengths and the maximum area for integer side lengths is {{{5*6=30}}}