Questions: The cost C (in dollars) of supplying recycling bins to p% of the population in a rural township is modeled by C = 27,000 p / (100-p), 0 ≤ p < 100. (a) Find the costs (in dollars) of supplying bins to 20%, 50%, and 95% of the population. cost of supplying bins to 20% cost of supplying bins to 50% cost of supplying bins to 95% (b) Find the limit of C as p → 100^(-). lim (p → 100^(-)) 27,000 p / (100-p) = Interpret the limit in the context of the problem. Use a graphing utility to verify your result. As the percent of the rural population being supplied with recycling bins approaches 100%, the cost -Select-

The cost C (in dollars) of supplying recycling bins to p% of the population in a rural township is modeled by
C = 27,000 p / (100-p), 0 ≤ p < 100.

(a) Find the costs (in dollars) of supplying bins to 20%, 50%, and 95% of the population.
cost of supplying bins to 20%
cost of supplying bins to 50%
cost of supplying bins to 95%

(b) Find the limit of C as p → 100^(-).
lim (p → 100^(-)) 27,000 p / (100-p) =

Interpret the limit in the context of the problem. Use a graphing utility to verify your result.
As the percent of the rural population being supplied with recycling bins approaches 100%, the cost -Select-

Transcript text: The cost C (in dollars) of supplying recycling bins to $p \%$ of the population in a rural township is modeled by \[ C=\frac{27,000 p}{100-p}, 0 \leq p<100 . \] (a) Find the costs (in dollars) of supplying bins to 20\%, 50\%, and 95\% of the population. cost of supplying bins to $20 \% \quad \$$ $\square$ cost of supplying bins to $50 \% \quad \$$ $\square$ cost of supplying bins to $95 \% \quad \$$ $\square$ (b) Find the limit of $C$ as $p \rightarrow 100^{-}$. \[ \lim _{p \rightarrow 100^{-}} \frac{27,000 p}{100-p}= \] $\square$ Interpret the limit in the context of the problem. Use a graphing utility to verify your result. As the percent of the rural population being supplied with recycling bins approaches $100 \%$, the cost -Select- $\square$

Solution

Solution Steps

To solve part (a), substitute the given percentages (20%, 50%, and 95%) into the cost function $ C = \frac{27,000p}{100-p} $ and calculate the cost for each percentage. For part (b), find the limit of the function as $ p $ approaches 100 from the left, which involves analyzing the behavior of the function as the denominator approaches zero.

Step 1: Calculate Costs for Given Percentages

To find the costs of supplying recycling bins to 20%, 50%, and 95% of the population, we use the cost function:

\[ C = \frac{27,000p}{100 - p} \]

Substituting the values:

For $ p = 20 $: \[ C(20) = \frac{27,000 \times 20}{100 - 20} = 6750 \]
For $ p = 50 $: \[ C(50) = \frac{27,000 \times 50}{100 - 50} = 27000 \]
For $ p = 95 $: \[ C(95) = \frac{27,000 \times 95}{100 - 95} = 513000 \]

Step 2: Find the Limit as $ p $ Approaches 100

Next, we find the limit of the cost function as $ p $ approaches 100 from the left:

\[ \lim_{p \to 100^{-}} C = \lim_{p \to 100^{-}} \frac{27,000p}{100 - p} \]

As $ p $ approaches 100, the denominator $ (100 - p) $ approaches 0, causing $ C $ to approach infinity:

\[ \lim_{p \to 100^{-}} C = \infty \]

Step 3: Interpret the Limit

In the context of the problem, as the percentage of the rural population being supplied with recycling bins approaches 100%, the cost becomes infinitely large. This indicates that supplying bins to the entire population would require an unbounded amount of resources.