1.  A----------------------------------------------------------------------B

                                                  <-L->

    A rope AB with length L is given. With which one can construct many different rectangles ,but all with same perimeter P. These different rectangles have different area A.

                                  

                                                                           

                                                                           

 

Which rectangle will have maximum area ?

These rectangles have same perimeter P = 2(l+b)

Area of a rectangle is the product of its length and breadth A = l x b

This geometrical figures predict an arithmetical fact – If the  sum of  any two numbers equals with sum of two different numbers  then their product will not be equal and vice-versa.

If a + b = x +y , then ab ≠ xy

If ab = xy then a+b ≠ x + y

          a + b = x+y = 24     ab              ab =  xy =24    a+b                                                                                     .        ………………………………………………

             1 +23                  23                   1x24               25

             2+22                   44                    2x12               14

             3+21                   63                    3 x 8               11

             4 +20                  80                    4 x 6               10

              5 + 19                 95   

             6 + 18               108 

             7 + 17               119

            8 +  16               128

            9 + 15                135

          10 + 14               140

          11 + 13               143

          12 + 12               144  

  ………………………………………………………………

From the table we find that when the difference between the pair of two numbers giving fixed sum is smaller greater will be its product and when the difference between the pair of two numbers giving fixed product is smaller will be its sum. 

Any increase in the length will have equal decrease in the breadth of the rectangle to keep up its perimeter same.To have maximum area, the length of the rectangle must be equal to its breadth, l =  b, where the rectangle becomes a square.  If P = L, then l = b = L/4 for a square and l =(L/4- x) , b = (L/4+x) for any rectangle .The area of the rectangle  A = (L/4 – x)( L/4+x) =( L² /16)  - x²   . To be maximum x = 0 .For a given perimeter, the area of a rectangle will be maximum when it becomes a square, 

Logical answer: The area of the rectangle depends upon two parameters  length and breadth. If length is greater ,breadth will be smaller ,if length is shorter, the breadth will be greater. Which make the area smaller than the maximum. To have maximum are, both the parameter must have optimum value that is they must be equal.