A certain business loses sales in a quantity proportional to

A certain business loses sales in a quantity proportional to its derivate. If at time zero sales was $300 (in thousands) a month, and at time 2 sales dropped to $280 then what is the sales value we would expect at time 5? Note: you will have first to find the \"c\" value, then the k value, then answer the final question.

Solution

Given that loss in sales is proportonal to its derivative.

-d[A]dt = k[A]

-d[A]/[A]= kt

On integrating,

Integral of -d[A]/[A] = Integral of kt

-ln[A]t = kt + C ------------- Eq (1)

At time t= 0,

C = 0+ -ln[A]o

C = -ln[A]o

Hence,

-log[A]t = kt + C ------------- Eq (1)

-log[A]t = kt - ln[A]o

kt = ln {[A]o / [A]t}

k =(1/t)  ln {[A]o / [A]t} --------Eq (2)

[At] = [Ao] e^-kt

Solving for k:

Given that  [Ao] = 300

[At] = 280

t =2

Then,

[At] = [Ao] e^-kt

280 = 300 x e^-k.2

e^-k.2= 280/300

-2k = In (280/300)

2k = In (300/280)

k = 0.0345

Sales at time 5:

[At] = [Ao] e^-kt

= 300 x  e^{-(0.0345 x 5)}

   = 252.5

[At] =  252.5

Therefore,

sales at time 5 =  252.5