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

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:

$a \cdot b = 16$

so we can write, that $b = \frac{16}{a}$

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

$P = 2 \cdot \left(a + \frac{16}{a}\right)$

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

$P \left(a\right) = 2 a + \frac{32}{a}$

$P ' \left(a\right) = 2 + \left(\frac{- 32}{a} ^ 2\right)$

$P ' \left(a\right) = 2 - \frac{32}{a} ^ 2 = \frac{2 {a}^{2} - 32}{a} ^ 2$

The extreme values can only be found in points where $P ' \left(a\right) = 0$

$P ' \left(a\right) = 0 \iff 2 {a}^{2} - 32 = 0$

$2 {a}^{2} - 32 = 0$
$\textcolor{w h i t e}{x} {a}^{2} - 16 = 0$
$\textcolor{w h i t e}{\times x . .} {a}^{2} = 16$
$\textcolor{w h i t e}{\times \times x} a = - 4 \mathmr{and} a = 4$

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

When $a = 4$,

$b = \frac{16}{4}$
$\textcolor{w h i t e}{b} = 4$

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 ${90}^{\circ}$

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.