Chapter 6.5, Problem 80SR

To determine

To Find:

The width of the table that satisfies the given situation.

Answer to Problem 80SR

  $\text{The width of the table could be between 0 and 3 feet or between 10 and 13 feet}\text{.}$

Explanation of Solution

Given Information:

A model car builder is building a display table for model cars. He wants the perimeter of the table to be 26 feet, but he wants the area of the table to be no more than 30 square feet.

  $\begin{array}{l}\text{Let 'w' and 'l' be the width and length of the table respectively}\text{.}\\ \text{ So, by the given information , we have :}\\ \text{Perimeter = 26}\\ \Rightarrow \text{2(w+l)=26}\\ \text{Area }\le \text{ 30}\\ \Rightarrow \text{ wl }\le \text{ 30}\end{array}$

Formula Used:

• ZERO PRODUCT PROPERTY:

If ab=0

Than , either one of a or b is zero.

That is , a = 0 or b = 0.

• Area of a rectangle having length ‘l’ and width ‘w’ = lw .
• Perimeter of a rectangle having length ‘l’ and width ‘w’ = 2( l+w).
• To solve quadratic inequalities:
• Find the ‘=0’ points. That is first find the roots of the equation by removing the inequality sign and placing the ‘=’ sign.
• In between the roots , the intervals are where the function has either
• Greater than zero value, or
• Less than zero value.
• Then pick a test value to find out which interval is the correct fit for the given inequality. That is , take any value from the interval and check if it satisfies the inequality.

Calculation:

Given

  $\begin{array}{l}\text{Let 'w' and 'l' be the width and length of the table respectively}\text{.}\\ \Rightarrow \text{2(w+l)=26 and wl }\le \text{ 30}\end{array}$

  $\begin{array}{l}2(w+l)=26\\ \Rightarrow 2w+2l=26\\ \Rightarrow 2w+2l-2w=26-2w\mathrm{.......}[\text{Subtract 2w from each side]}\\ \Rightarrow 2l=26-2w\\ \Rightarrow \frac{2l}{2}=\frac{26-2w}{2}\mathrm{......................}[\text{Divide each side by 2]}\\ \Rightarrow l=13-w\\ Also,\\ wl\le 30\\ Substitute"l=13-w"\\ \Rightarrow w(13-w)\le 30\\ \Rightarrow 13w-w^2\le 30\\ \Rightarrow -w^2+13w-30\le 30-30=\mathrm{0..}[\text{subtract 30 from each side]}\\ \Rightarrow {\text{w}}^2-13w+30\ge \mathrm{0....................}[\text{Multiply -1 each side]}\\ {\text{Solve w}}^2-13w+30=0\\ \Rightarrow w^2-10w-3w+30=0\\ \Rightarrow w(w-10)-3(w-10)=0\\ \Rightarrow (w-3)(w-10)=0\\ \Rightarrow w=3\text{ or }w=10\\ \text{Now , we check that which region satisfies the given inequality:}\\ \text{A={w | 3}\le \text{w}\le \text{10}}\\ \text{B={w | w}\le 3\text{ or w}\ge \text{10}}\\ \text{Let us consider w=0 from region B}\\ \Rightarrow {\text{w}}^2-13w+30=0^2-13(0)+30=30\ge 0\\ \text{So, the solution region of the inequality }\\ {\text{w}}^2-13w+30\ge 0\text{ is {w | w}\le 3\text{ or w}\ge \text{10}}\mathrm{......}\text{(1)}\\ \text{Since, w is the width of the table , it will always be}\\ \text{non-negative}\text{.}\\ \text{So, w}\ge \text{0}\mathrm{........................}\text{(2)}\\ \text{Also , l = 13-w }\text{.}\\ \text{Since, length of the table is l, so it will also be non-negative}\text{.}\\ \text{So, l=13-w}\ge \text{0}\\ \Rightarrow w\le \mathrm{13.........................}(3)\\ \text{Combining (1),(2) and (3):}\\ \text{The width of the table can be :}\\ \text{{w | 0}\le \text{w}\le \text{3 or 10}\le \text{w}\le \text{13}}\\ \text{So, the width of the table could be between 0 and 3 feet}\\ \text{or between 10 and 13 feet}\text{.}\end{array}$