Chapter 9, Problem 9.1P

To determine

(a)

The $10$ year average disposal cost for the town.

Answer to Problem 9.1P

The $10$ year average disposal cost is $\$144,375$ per year.

Explanation of Solution

Given:

Town has $30,000$ citizens.

Each home generates $1.1\text{tons}$ of solid waste per year.

Space available in the local landfill is $100,000\text{tons}$ .

Time period is $10$ year.

Calculation:

Write the expression to calculate the annual waste tonnage.

  $w=\frac{\text{people}}{\text{people}/\text{house}}\times \frac{\text{ton}/\text{year}}{\text{house}}$   ...... (I)

Here, annual waste tonnage is w.

Substitute $30,000$ for $\text{people}$ , $2.4$ for $\text{people}/\text{house}$ and $1.1$ for $\text{ton}/\text{year}$ in Equation (I).

  $\begin{array}{c}w=\frac{\text{30,000}}{\text{2}\text{.4}}\times \frac{\text{1}\text{.1}}{\text{1}}\\ =13,750\text{tons}/\text{year}\end{array}$

Calculate the duration when the landfill will get filled.

  $\begin{array}{c}t=\frac{100,000\text{tons}}{13,750\text{tons}/\text{year}}\\ =7.27\text{years}\end{array}$

Here, the duration is t.

Write the expression to calculate the cost incurred up until the landfill reaches its capacity.

  $c=\frac{\text{people}}{\text{people}/\text{house}}\times \frac{\text{ton}/\text{year}}{\text{house}}\times \frac{\$10}{\text{ton}}\times 7.27\text{years}$   ...... (II)

Here, the cost incurred on landfill is c.

Substitute $30,000$ for people, $2.4$ for $\text{people}/\text{house}$ and $1.1$ for $\text{ton}/\text{year}$ in Equation (II).

  $\begin{array}{c}c=\frac{\text{30,000}}{\text{2}\text{.4}}\times \frac{\text{1}\text{.1}}{\text{1}}\times \frac{\$10}{\text{1}}\times 7.27\\ =\$999,625\end{array}$

Write the expression to calculate the cost for sending the waste to a remote landfill.

  $x=\frac{\text{people}}{\text{people}/\text{house}}\times \frac{\text{ton}/\text{year}}{\text{house}}\times \frac{\$90}{\text{ton}}\times 2.73\text{years}$   ...... (III)

Here, the cost for sending the waste to remote landfill is x.

Substitute $30,000$ for people, $2.4$ for $\text{people}/\text{house}$ and $1.1$ for $\text{ton}/\text{year}$ in Equation (III).

  $\begin{array}{c}x=\frac{\text{30,000}}{\text{2}\text{.4}}\times \frac{\text{1}\text{.1}}{\text{1}}\times \frac{\$90}{\text{1}}\times 2.73\\ =\$375,375\end{array}$

Calculate the total cost with interest.

  $y=\left(x+c\right)\times 1.05$   ...... (IV)

Here, the total cost is y.

Substitute $\$999,625$ for c and $\$375,375$ for x in Equation (IV).

  $\begin{array}{c}y=\left(\$999,625+\$375,375\right)\times \left(1+0.05\right)\\ =1375000\times 1.05\\ =\$1,443,750\end{array}$

Write the expression to calculate the average cost for $10$ years.

  $a=\frac{y}{10\text{years}}$   ...... (V)

Here, the average cost is a.

Substitute $\$1,443,750$ for y in Equation (V).

  $\begin{array}{c}a=\frac{\$1,443,750}{10\text{years}}\\ =\$144,375\text{per}\text{year}\end{array}$

Conclusion:

Thus, the average annual cost is $\$144,375$ per year.

To determine

(b)

The total cost per home.

Answer to Problem 9.1P

The cost per home is $\$115.5$ .

Explanation of Solution

Write the expression to calculate the total cost per home.

  $q=\frac{y}{\text{number}\text{of}\text{homes}}$   ...... (VI)

Here, total cost per home is q.

Calculate the number of homes.

  $\begin{array}{c}\text{number}\text{of}\text{homes}=\frac{30,000\text{people}}{2.4\text{people}/\text{house}}\\ =12,500\end{array}$

Substitute $\$1,443,750$ for y and $12,500$ for number of homes in Equation (VI).

  $\begin{array}{c}q=\frac{\$1,443,750}{12,500}\\ =\$115.5\end{array}$

Conclusion:

Thus, the cost per home is $\$115.5$ .