Chapter 1.5, Problem 24PPSE

Solution

The dimensions of the frame.

The length of the frame should be less than $\frac{29}{2}$ in. and the width should be less than $\frac{23}{2}$ in.

Given:

The length of a picture frame is 3 in. greater than the width. The perimeter is less than 52 in.

Concept Used:

• If there is the strict inequality symbol $<\text{ or}>$, then use a parenthesis $\left(\right)$ on the interval
• If there is the inequality symbol $\le \text{ or}\ge$, then use a bracket $\left[\right]$ on the interval
• To graph the endpoint with strict inequality $<\text{ or}>$, use the open parenthesis or a hollow circle at that endpoint.
• To graph the endpoint with inequality symbol $\le \text{ or}\ge$, use the bracket or a solid circle at that endpoint.
• Multiplying or dividing an inequality by a negative number reverse the inequality sign.
• Perimeter of a rectangle: $P=2\left(\text{Length}+\text{Width}\right)$

Calculation:

Let w be the width of the frame.

Since the length of the frame is 3 in. greater than the width, the expression that represents the length of the frame is $w+3$ .

Since the perimeter of the frame is less than 52 in., it follows

  $\begin{array}{c}P<52\\ 2\left(\text{Length}+\text{Width}\right)<52\\ 2\left(w+3+w\right)<52\\ 2\left(2w+3\right)<52\\ 4w+6<52\\ 4w+6-6<52-6\\ 4w<46\\ w<\frac{46}{4}\\ w<\frac{23}{2}\end{array}$

And the length is

  $\begin{array}{c}L=w+3\\ L<\frac{23}{2}+3\\ L<\frac{23+6}{2}\\ L<\frac{29}{2}\end{array}$

Thus, the length of the frame should be less than $\frac{29}{2}$ in. and the width should be less than $\frac{23}{2}$ in.