Chapter 7, Problem 18CR

To determine

To find: The sell in a year to make the second offer is better.

Answer to Problem 18CR

The sale is more than $600000$ to make second offer is better.

Explanation of Solution

Given information: One company offer annual salary of $\$55000$ plus a year end bonus 0f $1.5\%$ of the sales and Other company offer annual salary of $\$52000$ plus a year end bonus 0f $2\%$ of the sales.

Calculation: One company offer annual salary of $\$55000$ plus a year end bonus 0f $1.5\%$ of the sales and Other company offer annual salary of $\$52000$ plus a year end bonus 0f $2\%$ of the sales.

Let the total sale $=x$ .

From the first offer income is:

  $\begin{array}{l}52000+\frac{2}{100}x>55000+\frac{15}{1000}x\\ \frac{2}{100}x-\frac{15}{1000}>55000-52000\\ \frac{20x-15x}{1000}>3000\\ \frac{5}{1000}x>3000\\ x>3000\times \frac{1000}{5}\\ x>600000\end{array}$

From the second offer income is:

  $=52000+\frac{2}{100}x$

According to problem second offer is best

So :

  $\begin{array}{l}52000+\frac{2}{100}x>55000+\frac{15}{1000}x\\ \frac{2}{100}x-\frac{15}{1000}>55000-52000\\ \frac{20x-15x}{1000}>3000\\ \frac{5}{1000}x>3000\\ x>3000\times \frac{1000}{5}\\ x>600000\end{array}$

Thus, the sale is more than $600000$ to make second offer is better.

Let the width of the rectangle $=\frac{8}{9}x$ .

Perimeter of the rectangle $=2(length+breadth)$ .

  $\begin{array}{l}2(x+\frac{8}{9}x)=68\\ 2(\frac{9x+8x}{9})=68\\ 2(\frac{17}{9})x=68\\ x=\frac{68\times 9}{2\times 17}=18\end{array}$

So The length of the rectangle $=18\mathrm{feet}$ .

The width of the rectangle $=18\times \frac{8}{9}=16\mathrm{feet}$ .

Thus, the length of the rectangle is $18\mathrm{feet}$ and the width of the rectangle is $16\mathrm{feet}$ .