Hi, For all rectangle having a perimeter constant, the square have the maximum area. So the rectangle is a square of 10 cm of side. On other method: Let assume x le length of the rectangle,(40-2*x)/2=20-x is the width area=y=x*(20-x) y=20x-x² a) y'=20-2x=0==>x=10 and y=10 b) y=-(x²-2*10*x+100)+100y=100-(x-10)² is maximum if x=10.
