jmc

Since the perimeter is 12, the sides of the rectangle add up to $12/2=6.$ Let $x$ be one side length of the rectangle. Then the other side length is $6-x,$ so the area is $$x(6-x)=6x-x^2.$$Completing the square, we get $$-x^2+6x=-x^2+6x-9+9=9-(x-3{)}^2.$$Thus, the maximum area of the rectangle is $\boxed{9}$ square inches, which occurs for a $3\times 3$ square.