Question

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

Answer 1

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.