Find the dimensions of the rectangle of area A that has the smallest perimeter. There are no given values.

neela | Student

Let the area of the rectangle be A. We are required to determine the smallest perimeter.

The perimeter of p the  rectanangle is given by:

p = 2(l+w) where l  is length and w  is width.

We assume that length  l = x , then  width w = A/x.

Therefore p(x) =2(l+w) = 2(x+A/x).

Therefore P(x) is minimum,  for some x=c for which p'(c) = 0 and p"(c)  > 0.

Therefore we differentiate p(x) and solve for x = c and then find  whether p''(c) < 0.

P'(x) = {2(x+A/x)}'

P'(x) = 2(1-A/x^2).

P'(x) = 0 gives 1-A/x^2 = 0 , or x^2 = A. So x = A^(1/2).

 Therefore x = c = A^(1/2).

We calculate P"(x) = {2(1-A/x^2)} =  -2A(-2/x^3) = +4A/x^3 .

It is obvious that  p''(c) = 4A/c^3 is > 0. This confirms that P(c) is the minimum value of P(x) at x = c.

Therefore P(c) = p(A^(1/2)) = 2{A^(1/2) + A/A(1/2)} = 4^A.(1/2).

Therefore , the rectangle with area A has the smallest perimeter 4A^(1/2)  when it is a square with each side  measuring ^A(1/2) .