jmc

Let one pair of parallel sides have length $x$ and the other pair of parallel sides have length $12-x$. This means that the perimeter of the rectangle is $x+x+12-x+12-x=24$ as the problem states. The area of this rectangle is $12x-x^2$. Completing the square results in $-(x-6{)}^2+36\le 36$ since $(x-6{)}^2\ge 0$, so the maximum area of $\boxed{36}$ is obtained when the rectangle is a square of side length 6 inches.