Question

Find the dimensions of a rectangle with area 1,728 m2 whose perimeter is as small as possible

Answer 1

The dimensions of a rectangle with area 1728 square meter whose perimeter is as small(minimum) as possible are 41.57m and 41.56m.

Let the dimension of the rectangle be x and y m.

According to the given question.

The area of the rectangle is 1728 square meter.

⇒ x × y = 1728

⇒ x = 1728/y

Since, the perimeter of the rectangle is the sum of the length of its boundary.

Therefore,

Perimeter of the recatngle with dimensions x and y is given by

Perimeter of the rectangle, P = 2(x + y)

⇒ P = 2(1728/y + y)

⇒ P = 3456/y + 2y

Differentiate the above equation with respect to y

⇒ $P^{'} = -\frac{3456}{y^{2} } + 2$

For the minimum(small) perimeter equate the above equation to 0.

⇒ $-\frac{3456}{y^{2} } + 2 = 0$

⇒ $2y^{2} =3456$

⇒ $y^{2} = \frac{3456}{2}$

⇒ y^2 = 1728

⇒ y = √1728

⇒ y = 41.56 m

Therefore,

x = 1728/41.56

⇒ x = 41.57 m

Hence, the dimensions of a rectangle with area 1728 square meter whose perimeter is as small(minimum) as possible are 41.57m and 41.56m.

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