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

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Answer 1 · 2018-06-11T09:27:16

We get $a = b = 4$ $a=b=4$

We have the Perimeter as

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

$a + \frac{16}{a} \geq 2 \sqrt{a \cdot \frac{16}{a}} = 8$ $a+16/a>=2sqrt(a*16/a)=8$
Multiplying by $2$ $2$

$2 (a + \frac{16}{a}) \geq 4 \sqrt{16} = 16$ $2(a+16/a)>=4sqrt(16)=16$
the equal sign holds if
$a = b = 4$ $a=b=4$

Answer 2 · 2018-06-11T09:44:05

$P = 16$ $P=16$

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

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

measured also in inches.

The perimeter is then:

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

Evaluate the derivative:

$\frac{d P}{d x} = 2 - \frac{32}{x^{2}}$ $(dP)/dx = 2-32/x^2$

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

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

thus $x = 4$ $x=4$ is a minimum.

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