Question 1491
 Let L& W be the length &  width of the rectangle.
 The perimeter = 2(L+W) = 200, so L+W =100 or W = 100 -L 
 Its area = LW = L(100 -L) = 100 L - L^2
          = - (L^2 - 100 L + (100/2)^2)  + (100/2)^2 [Complete square]
          = 2500 - (L -50)^2

 We see that when L= 50,the area has the maximum value 2500.
 Also, when L = 50,W = 100 -50 = 50.

 Answer: when L= 50,and W = 50, we attain the  maximum area 2500.