Question

Find the dimensions of a rectangle (in m) with area 1,728 m2 whose perimeter is as small as possible. (Enter the dimensions as a comma separated list.)

Answer 1

Given :

Area of rectangle.

To Find :

The dimensions of a rectangle (in m) with area 1,728 m2 whose perimeter is as small as possible.

Solution :

Let, the dimensions of rectangle is x and y.

Area, A = xy.

x = A/y.              ....1)

Perimeter, P = 2( x + y )

Putting value of x in above equation, we get :

$P = 2( y + \dfrac{A}{y})$

For minimum P,

$\dfrac{dP}{dx}=2( 1 - \dfrac{A}{x^2})=0\\\\A = x^2$

SO, it is a square.

$x=\sqrt{A}\\\\x=\sqrt{1728\ m^2}\\\\x=24\sqrt{3}\ m$

Therefore, the dimensions are $(24\sqrt{3},24\sqrt{3})$.24\sqrt{3}

Hence, this is the required solution.