Chapter 8.8, Problem 36IP

To determine

Find the minimum dimensions of the rectangle.

Answer to Problem 36IP

The dimensions of the rectangle are $\text{length}=\text{19}\text{cm}\text{and}\text{Width}\text{=}\text{6}\text{cm}$

Explanation of Solution

Given:

The perimeter of a rectangle is at least 50 centimetres. The length of the rectangle is one more than 3 times the width of the rectangle. What are the minimum dimensions of the rectangle?

Concept Used:

Translate the conditions into an inequality.

The perimeter of a rectangle is at least 50 centimetres.

At least means the inequality sign is $\ge$

So perimeter $\ge$ 50 cm

The length of the rectangle is one more than 3 times the width of the rectangle.

Let the width be x cm, then length is (3x +1) cm.

  $\text{Perimeter = 2 }\left[\text{Length + width}\right]$

The required inequality is $2[3x+1+x]\ge 50$

Calculation:

The required inequality is $2[3x+1+x]\ge 50$

Now solve for x :

  $\begin{array}{l}2[3x+1+x]\ge 50\\ \frac{2[3x+1+x]}{2}\ge \frac{50}{2}[\text{Divide}\text{both}\text{sides}\text{by}7]\\ 3x+1+x\ge 25\\ 4x+1\ge 25\\ 4x+1-1\ge 25-1\text{[}\text{Subtract}1\text{from}\text{both}\text{sides}\text{]}\\ 4x\ge 24\\ \frac{4x}{4}\ge \frac{24}{4}[\text{Divide}\text{both}\text{sides}\text{by}4]\\ x\ge 6\end{array}$

  $\text{Width}\text{:}\text{6}\text{cm}\text{and}\text{length}=3·6+1=\text{19}\text{cm}$

Thus, the dimensions of the rectangle are $\text{length}=\text{19}\text{cm}\text{and}\text{Width}\text{=}\text{6}\text{cm}$