CALC REVENUE PROBLEM

Question

It is estimated that the annual advertising revenue received by a certain newspaper will be
R(x) = 0.5x2 + 3x + 160
thousand dollars when its circulation is x thousand. The circulation of the paper is currently 12,000 and is increasing at a rate of 3,000 per year. At what rate will the annual advertising revenue be increasing with respect to time 3 years from now?

Answer 1

Let x(t) = circulation (in thousands) t years from now = 3t + 12

 

Revenue in t years = R(x(t))

 

                            = R(3t+12) = 0.5(3t+12)² + 3(3t+12) + 160

 

                                              = 0.5(9t² + 72t + 144) + 9t + 36 + 160

 

                                              = 4.5t² + 36t + 72 + 9t + 36 + 160

 

                                              = 4.5t² + 45t + 268

 

[ R(x(t) ]' = 9t + 45 = rate at which revenue is changing t years from now

 

So, when t = 3, 9t + 45 = 72

 

In 3 years, advertising revenue is increasing at the rate of $72,000 per year.  

 

Answer 2

Our circulation is currently 12000 and increases by 3000 per year.

The equation for circulation as a function of time in years is:

 

c(x) = 3000x + 12000

c(3) = 3000(3) + 12000

c(3) = 9000 + 12000

c(3) = 21000

 

In 3 years the circulation will be 21000 customers

 

We need to find the revenue associated with 21,000 circulation.

Since the function R(x) is for x in thousands:

 

R(21) = 0.5(21)² + 3(21) + 160

R(21) = 0.5(441) + 63 + 160

R(21) = 443.5

 

The advertising revenue will be $4,435,000

 

Rate of change is the slope... we have two points:

At beginning R(12) = 268000 ... (12, 268000)

3 years later R(21) = 4435000 ... (21, 4435000)

slope m = (4435000-268000)/(21-12) = 463,000 dollars per thousand