Question

A retangler has a length at least four more than five times the width. If the perimeter is greater than or equal to 32 units, find the least possible with and length.

Answer 1

The perimeter of a rectangle is given by the formula

P=2(L+W)

In this problem

$L\ge5W+4\text{ ----> inequality 1}$$P\ge32$

substitute given values

$2(L+W)\ge32$

Solve for L

$\begin{gathered} L+W\ge\frac{32}{2} \\ \\ L+W\ge16 \\ L\ge16-W\text{ ---->inequality 2} \end{gathered}$

Equate both inequalities

$\begin{gathered} 5W+4=16-W \\ 5W+W=16-4 \\ 6W=12 \\ W=2 \end{gathered}$

Substitute the value of W in the inequality 1 or inequality 2

$\begin{gathered} L\ge5W+4 \\ L\geqslant5(2)+4 \\ L\geqslant14 \end{gathered}$

therefore

the least possible width and length are

L=14 units and W=2 units