What is the greatest area of a rectangle with a perimeter of 50?

P = 2(L + W)

50 = 2(L + W)

25 = L + W

Let L = x, so W = 25 - x

A = LW

A = x(25 - x)

A = -x2 + 25x

Since the parabola that represents the above equation opens downward, we have a maximum point (the y-coordinate value of the vertex of the parabola, gives us the maximum value of the area).

vertex x-coordinate value = - b/2a = - 25/-2 = 25/2

vertex y-coordinate value = -(25/2)2 + 25(25/2) = - 252/4 + 252/2 = - 252/4 + 2(252)/4 = 252/4 = 156.25

Thu, the maximum area will be 156.25 unit2.