# What is the smallest parameter possible for a rectangle whose area is 16 square inches and what are it’s dimensions?

Jun 11, 2018

We get $a = b = 4$

#### Explanation:

We have the Perimeter as

$p = 2 \left(a + b\right) = 2 \left(a + \frac{16}{a}\right)$
So we get by $A M - G M$

$a + \frac{16}{a} \ge 2 \sqrt{a \cdot \frac{16}{a}} = 8$
Multiplying by $2$

$2 \left(a + \frac{16}{a}\right) \ge 4 \sqrt{16} = 16$
the equal sign holds if
$a = b = 4$

Jun 11, 2018

$P = 16$

#### Explanation:

Let $a = x$ inches be one side of the rectangle, then the other side is:

$b = \frac{16}{x}$

measured also in inches.

The perimeter is then:

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

Evaluate the derivative:

$\frac{\mathrm{dP}}{\mathrm{dx}} = 2 - \frac{32}{x} ^ 2$

thus the derivative is null for $x = 4$, and in this point the second derivative:

$\frac{{d}^{2} P}{\mathrm{dx}} ^ 2 = \frac{64}{x} ^ 3 > 0$

thus $x = 4$ is a minimum.

We can conclude that the smallest perimeter is obtained when $a = b = 4$, that is when the rectangle is a square.