# The length of a rectangle is 5 cm more than twice its width. What is the largest possible width if the perimeter is at most 64 cm?

##### 1 Answer
Aug 13, 2016

$w i \mathrm{dt} h = 9 c m$

#### Explanation:

Perimeter of a rectangle$= 2 \left(a + b\right)$ where $a$ is the length and $b$ is the width
As given we can write
$a = 2 b + 5$ by plugging this value

Perimeter of a rectangle$64 = 2 \left(a + b\right)$
or
$a + b = \frac{64}{2}$
or
$a + b = 32$
or
$2 b + 5 + b = 32$
or
$3 b = 32 - 5$
or
$3 b = 27$
or
$b = \frac{27}{3}$
or
$b = 9$