Answer:
- A) 2.5 tons
- D) 2.4 tons/year
Step-by-step explanation:
Given the number of tons of recycled aluminum as a function of population is T(p) = √(1.5p² +0.25), and the estimated population is p(t) = 0.2(t -1)² -3, you want the tons of aluminum recycled 6 years from now, and its rate of change.
Quantity
The population in 6 years is estimated as ...
p(6) = 0.2(6 -1)² -3 = 2 . . . . . thousands
The amount of recycling that population does is ...
T(2) = √(1.5(2²) +0.25) = √6.25 = 2.5 . . . . tons
Six years from now, 2.5 tons of aluminum are expected to be recycled.
Rate of change
The rate of change of the quantity recycled is given by the derivative of the function:
T'(p) = 1.5(2p)/(2√(1.5p² +0.25))·p' = 1.5p·p'/T(p)
In 6 years, this is ...
T'(6) = 1.5·2·p'/2.5 = 1.2p'
The rate of change of population in 6 years is expected to be ...
p'(t) = 0.2(2)(t -1) = 0.4(t -1)
p'(6) = 0.4(6 -2) = 2
So, the rate of change of recycled amount is ...
T'(6) = 1.2(2) = 2.4 . . . . . tons/year
Six years from now the rate of change of recycled amount will be about 2.4 tons per year.
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Additional comment
The rate of change of recycled amount varies substantially. The problem statement does not give a time frame for the desired rate of change. We have assumed it is 6 years, since the value at that time matches one of the answer choices.
The derivative of T has units of tons per thousand population. Since we want tons per year, we need to include the effect of population on the derivative.
