Initial/Fixed Value Problems

Question

An automobile company has found that their rate of sales of a new automobile, in millions of dollars, can be modeled by  where t is number of years since the new automobile was introduced. Since this is a new automobile, there are no sales initially.

 A. Find the function that models the total sales of the new car after t years.

B. What were the total sales 3 years after the new automobile was introduced?

Answer 1

An automobile company has found that their rate of sales of a new automobile, in millions of dollars.

Simplifying function modeling the sales rate of the new car above into mathematical symbols and numbers:

S'(t) = 10-10*e^(-0.1*t), where t is number of years since the new automobile was introduced. Since this is a new automobile, there are no sales initially.

A) Our function S'(t) measures our rate of sales during a particular moment in time. In order to find the total amount of sales 3 years after the model was released, we will need to integrate our function. This will add up the area under the curve which represents all the sales that the new car model has made until that particular moment in time. This new function S(t) will then be able to gives total sales of the new car model, in millions, at any given point after the new car model's release.

We will integrate S'(t) to get S(t):

∫S'(t) → S(t)

∫(10-10*e^(-0.1*t)) → 10t+100*e^(-0.1*t)+C

S(t): 10t+100e^(-0.1*t)+C

Now we know that S(0)=0. This is because the total sales are zero at this point because the car is just released. We can use this information to solve for C:

0 = S(0) = 10*(0)+100e^(-0.1*(0))+C = 0+100+C → C = -100

S(t) = 10t+100e^(-0.1*t)-100

B) For this part we only have to use the function S(t) we integrated for in part (A). Let t=3:

S(3) = 10*(3)+100e^(-0.1*(3))-100 = 30+100*e^(-0.3)-100 = 30+100*0.7408-100 = 4.08

In 3 years, the new car model should produce 4.08 million in total sales from the day of it's release.

Answer 2

[I translate the given expression as S'(t) = 10 - 10e^-0.1t]

A) Integrate S', using u = -0.1t, du = -0.1dt, and -10du = dt: S(t) = 10[t - (-10e^-0.1t)] + C. S(0) = 100e⁰ + C = 0, so C = -100, and S(t) = 10t + 100e^-0.1t - 100.

B) S(3) ≅ 30 + 74.082 - 100 which yields sales of approximately $4,082,000 after 3 years.