Chapter 9.4, Problem 4LC

To determine

The dimensions of the table.

Answer to Problem 4LC

Length is 4 ft and width is $\frac{\text{5}}{2}ft.$

Explanation of Solution

Given:

The area of the rectangular table should be 10 square feet.

The length of the table should be 1 ft shorter than twice its width.

Calculation:

Let the width of the table be x and the length be 2x - 1.

  $\begin{array}{l}A=l\times w\\ x\left(2x-1\right)=10\\ 2x^2-x=10\\ 2x^2-x-10=0\\ 2x^2+4x-5x-10=0\\ \left(2x^2+4x\right)+\left(-5x-10\right)=0\\ 2x\left(x+2\right)-5\left(x+2\right)=0\\ \left(x+2\right)\left(2x-5\right)=0\\ x+2=0\text{ or 2x - 5 = 0}\\ x=-2\text{ or x = }\frac{\text{5}}{2}\end{array}$

As the dimension cannot be negative, neglect the value -2.

Therefore, the width of the rectangle is $\text{x = }\frac{\text{5}}{2}$ .

Length of the rectangle will be:

  $\begin{array}{l}2x-1=2\left(\frac{5}{2}\right)-1\\ =4ft\end{array}$

Hence, length is 4 ft and width is $\frac{\text{5}}{2}ft.$