Chapter 8.3, Problem 110E

(a)

To determine

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

Answer to Problem 110E

The formula is

  $S_n={\displaystyle \sum _{i=1}^{n}8000{\left(0.9\right)}^{i-1}}$

Explanation of Solution

Given information:

The manufacturer of a new food processor plans to produce and sell 8000 units per year. Given year, 10% of all units sold become inoperative. So, 8000 units will be in use after 1 year, $\left[8000+0.9\left(8000\right)\right]$ units will be in use after 2 years.

Calculation:

The formula in series notation for the number of units that will be operative after $n$ years is obtained as shown below:

The corresponding geometric sequence is given by $8000,8000{\left(0.9\right)}^1,8000{\left(0.9\right)}^2,\mathrm{...}$

Now we have,

  $\begin{array}{c}{S}_n=8000+8000{\left(0.9\right)}^1+8000{\left(0.9\right)}^2+8000{\left(0.9\right)}^3+\mathrm{...}+8000{\left(0.9\right)}^{n-1}\\ ={\displaystyle \sum _{i=1}^{n}8000{\left(0.9\right)}^{i-1}}\end{array}$

(b)

To determine

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

Answer to Problem 110E

The number of units that will be operative after 10 years, 15 years, and 20 years are $52106\text{units,63529}\text{units,}\text{and}\text{70274 units}$ respectively.

Explanation of Solution

Given information:

The manufacturer of a new food processor plans to produce and sell 8000 units per year. Given year, 10% of all units sold become inoperative. So, 8000 units will be in use after 1 year, $\left[8000+0.9\left(8000\right)\right]$ units will be in use after 2 years.

Calculation:

Find the number of units that will be operative after 10 years, 15 years, and 20 years as shown below:

  $\begin{array}{c}{S}_n={\displaystyle \sum _{i=1}^{n}8000{\left(0.9\right)}^{i-1}}\\ S_{10}={\displaystyle \sum _{i=1}^{10}8000{\left(0.9\right)}^{i-1}}\\ =8000\left(\frac{1-{0.9}^{10}}{1-0.9}\right)\\ =52105.724792\\ =52106\text{units}\end{array}$

Continue to determine $S_{15}$ as shown in above method

  $\begin{array}{c}{S}_n={\displaystyle \sum _{i=1}^{n}8000{\left(0.9\right)}^{i-1}}\\ S_{15}={\displaystyle \sum _{i=1}^{15}8000{\left(0.9\right)}^{i-1}}\\ =8000\left(\frac{1-{0.9}^{15}}{1-0.9}\right)\\ =63528.709432428\\ =63529\text{units}\end{array}$

Continue to determine $S_{20}$ as shown in above method

  $\begin{array}{c}{S}_n={\displaystyle \sum _{i=1}^{n}8000{\left(0.9\right)}^{i-1}}\\ S_{20}={\displaystyle \sum _{i=1}^{20}8000{\left(0.9\right)}^{i-1}}\\ =8000\left(\frac{1-{0.9}^{20}}{1-0.9}\right)\\ =70273.8676327544\\ =70274\text{units}\end{array}$

Thus, the number of units that will be operative after 10 years, 15 years, and 20 years are $52106\text{units,63529}\text{units,}\text{and}\text{70274 units}$ respectively.

(c)

To determine

To find: whether the number of operative units will be finite for the continued trend and to calculate the finite number of units.

Answer to Problem 110E

This trend continuous indefinitely then the number of units that are operative will be finite and the number of units are 80000 units.

Explanation of Solution

Given information:

The manufacturer of a new food processor plans to produce and sell 8000 units per year. Given year, 10% of all units sold become inoperative. So, 8000 units will be in use after 1 year, $\left[8000+0.9\left(8000\right)\right]$ units will be in use after 2 years.

Calculation:

If this trend continuous indefinitely then the number of units that are operative will be finite because for the following reason.

  $\begin{array}{c}S={\displaystyle \sum _{n-0}^{\infty }8000{\left(0.9\right)}^n}\\ =\frac{8000}{1-0.9}\\ =80000\text{units}\end{array}$

Thus, if this trend continuous indefinitely then the number of units that are operative will be finite and the number of units are 80000 units.