Maximum Area of a Rectangle

We know that a rectangle is a form of a quadrilateral whose opposites sides are equal. The longer sides are called its length and their shorter sides are called width. The area of the rectangle is given by length * width.

To obtain a maximum area then length = width. So if the perimeter of the rectangle is given then we need to take the value of both length and width to be equal.

Now let us see few problems of this kind.

Example Problem on Maximum Area of a Rectangle:

Ex 1: Find the maximum area of a rectangle whose perimeter is 144cm.

Soln: Given: Perimeter = 144

`implies ` 2(L + W) = 144

`implies` L + W = 144/2 = 72

`implies` L + W = 72

`implies` L = W =36

Therefore, the maximum area = 36 `xx` 36 = 1296cm2


 
Ex 2: If the maximum area of a rectangle is 2916cm2. Find the length and width of the rectangle.

Soln: Given: Area = 2916

Since the maximum area of rectangle is obtained by having length equal to the width,

Area = l * w = 2916

` implies` l * l = 2916    [since l = w]

`implies`         l2 = 2916

` implies`        l = `sqrt[2916]` = 54

Therefore the dimension is 54 * 54 to get the Maximum area.

More Example Problem on Maximum Area of a Rectangle:

Ex 3: Perimeter of a rectangular fence is 252cm. Find the maximum area the fence can cover.

Soln: Given: Perimeter = 252

2( L + W) = 252

`implies` L + W = `252/2` = 126

`implies` L = W = `126/2` = 63

The maximum Area = 63 * 63 = 3969 m2

Ex 4: Let the length of the rectangle be 91cm. Its area is 8281 cm2. Can this length form a maximum area?

Soln: Given: Area = 8281 cm2 and L = 91

`implies` L `xx` W = 8281

`implies` W =` 8281/91` = 91

Here since w = 91 = L, 8281cm2 is the maximum area that can be formed by the length of 91cm.