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

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Answer 1 · 2015-11-08T18:10:28

The minimum perimeter is $16$ in for equal sides of $4$ in.

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:

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.