What is the smallest perimeter possible for a rectangle of area 16 in^2?

1 Answer
Nov 8, 2015

The minimum perimeter is #16#in for equal sides of #4#in.

Explanation:

If we denote one side of the rectangle with #a#, and the other with #b# we can write, that:

#a*b=16#

so we can write, that #b=16/a#

Now we can write perimeter #P# as a function of #a#

#P=2*(a+16/a)#

We are looking for the smallest perimeter, so we have to calculate derivative:

#P(a)=2a+32/a#

#P'(a)=2+((-32)/a^2)#

#P'(a)=2-32/a^2=(2a^2-32)/a^2#

The extreme values can only be found in points where #P'(a)=0#

#P'(a)=0 iff 2a^2-32=0#

#2a^2-32=0#
#color(white)(x)a^2-16=0#
#color(white)(xxx..)a^2=16#
#color(white)(xxxxx)a=-4 or a=4#

Since, length is a scalar quantity, therefore, it cannot be negative,

When #a=4#,

#b=16/4#
#color(white)(b)=4#

You may be thinking, since both sides are of equal lengths, does it not become a square instead of a rectangle?

The answer is no because the properties of a rectangle are as follows:

  1. opposite sides are parallel
  2. opposite sides are congruent
  3. diagonals bisect each other
  4. diagonals are congruent
  5. each of the interior angles must be #90^@#

Since there is no rule that states a rectangle cannot have all sides of equal length, all squares are rectangles, but not rectangles are squares.

Hence, the minimum perimeter is #16#in with equal sides of #4#in.

P.S. What is a comedian's favourite square? a PUNnett square.