Chapter 8.3, Problem 110E

a.

To determine

To write: a formula in series notation for the number of unites that will be operative after n years.

Answer to Problem 110E

  $N={\displaystyle \sum _{n=0}^{\infty }8000{(0.9)}^n}.$

Explanation of Solution

Given information: The manufacturer of a new food processor plans to produce and sell.

8000 units per year. Each year, 10% of all units sold becomes inoperative , so 8000 units will be in use after 1 year, [8000+0.9(8000)] units will be in use after 2 years and so on. Calculation:

The common ratio is 0.9 so the series notation will be:

  $N={\displaystyle \sum _{n=0}^{\infty }8000{(0.9)}^n}.$

b.

To determine

To find: the number of units that will be operative after 10 years, 15 years, and 20 years.

Answer to Problem 110E

  $\begin{array}{c}{S}_{10}\approx 52106units.\\ S_{15}\approx 63529units.\\ S_{20}\approx 70274units\end{array}$

Explanation of Solution

Given information: The manufacturer of a new food processor plans to produce and sell.

8000 units per year. Each year, 10% of all units sold becomes inoperative, so 8000 units will be in use after 1 year, [8000+0.9(8000)] units will be in use after 2 years and so on.

Calculation:

Use the partial sum for geometric series:

  $\begin{array}{c}{S}_n=a_1(\frac{1-r^n}{1-r})\\ a_1=8000,r=0.9\\ S_n=8000\left(\frac{1-{0.9}^n}{1-0.9}\right).\\ n=10\\ S_{10}=8000\left(\frac{1-{0.9}^{10}}{1-0.9}\right)\approx 52106units.\\ n=15\\ S_{15}=8000\left(\frac{1-{0.9}^{15}}{1-0.9}\right)\approx 63529units\\ n=20\\ S_{20}=8000\left(\frac{1-{0.9}^{20}}{1-0.9}\right)\approx 70274units\end{array}$

c.

To determine

To find: this trends continues indefinitely, will the number of units that will be operative be finite, if so, how many and if not explain it.

Answer to Problem 110E

  $S=80000units.$

Explanation of Solution

Given information: The manufacturer of a new food processor plans to produce and sell.

8000 units per year. Each year, 10% of all units sold becomes inoperative, so 8000 units will be in use after 1 year, [8000+0.9(8000)] units will be in use after 2 years and so on.

Calculation:

Yes, using the infinite sum formula,

  $\begin{array}{l}S=a_1\left(\frac{1}{1-r}\right)\\ S=8000\left(\frac{1}{1-0.9}\right)=80000units.\end{array}$