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Difficulty: Very hardInterpreting Linear Relationships in Context

An environmental cleanup crew is removing a contaminant from a soil site. The remaining mass of the contaminant, CC, in kilograms, after dd days of treatment is modeled by the equation C=400−pdC = 400 - p d, where pp is the daily removal rate, in kilograms per day, under the original protocol. Under a new treatment protocol, the daily removal rate is increased by 25%25\%, and the treatment time required to completely remove the contaminant is reduced by 88 days. What was the daily removal rate, in kilograms per day, under the original protocol?

Answer: 10 kilograms per day

Answer

The daily removal rate under the original protocol was 10 kilograms per day.
Under the original protocol, complete removal of the 400400 kg contaminant occurs when C=0C = 0, giving a duration of d=400pd = \frac{400}{p} days. Under the new protocol, the removal rate increases by 25%25\% to 1.25p1.25p, and the duration is reduced by 88 days to 400p−8\frac{400}{p} - 8. Because the total mass removed must still equal 400400 kg, we write the equation (1.25p)(400p−8)=400(1.25p)\left(\frac{400}{p} - 8\right) = 400. Distributing 1.25p1.25p yields 500−10p=400500 - 10p = 400. Solving for pp gives 10p=10010p = 100, which simplifies to p=10p = 10 kilograms per day.

Step-by-Step Solution

1
Set C=0C = 0 in the original equation to represent complete removal.
0=400−pd  ⟹  d=400p0 = 400 - p d \implies d = \frac{400}{p}
Complete removal of the contaminant means that the remaining mass CC is 00 kilograms.
2
Express the new daily removal rate and the new treatment duration using the given percentage increase and day reduction.
pnew=1.25pp_{\text{new}} = 1.25p and dnew=d−8=400p−8d_{\text{new}} = d - 8 = \frac{400}{p} - 8
The new protocol increases the daily removal rate by 25%25\% and reduces the total treatment time by 88 days.
3
Set up the equation for complete removal under the new protocol using the new rate and duration.
400−(1.25p)(400p−8)=0400 - (1.25p) \left(\frac{400}{p} - 8\right) = 0
The total initial contaminant mass of 400400 kilograms must be completely removed by the new daily rate over the new duration.
4
Solve the equation for pp.
1.25p(400p−8)=400  ⟹  500−10p=400  ⟹  10p=100  ⟹  p=101.25p \left(\frac{400}{p} - 8\right) = 400 \implies 500 - 10p = 400 \implies 10p = 100 \implies p = 10
Distribute 1.25p1.25p into the parentheses to eliminate the fraction, then isolate the variable pp.

Key Concept

Interpreting the rate (slope) and intercepts of a linear relationship in context, and modeling variations of those parameters.
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Interpreting Linear Relationships in Context

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