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 $a=b=4$

We have the Perimeter as

$p=2(a+b)=2(a+16/a)$
So we get by $AM-GM$

$a+16/a>=2sqrt(a*16/a)=8$
Multiplying by $2$

$2(a+16/a)>=4sqrt(16)=16$
the equal sign holds if
$a=b=4$

Answer 2 · 2018-06-11T09:44:05

$P=16$

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

$b=16/x$

measured also in inches.

The perimeter is then:

$P= 2(x+16/x)$

Evaluate the derivative:

$(dP)/dx = 2-32/x^2$

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

$(d^2P)/dx^2 = 64/x^3 > 0$

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.