assuming profit = revenue - cost
P(x) = 15*e^(0.08*t) - 12*e^(−0.04*t)
P'(x) = 1.2e^(.08t) + .48e^(-.04t)
= 0 for a max of P(x)
1.2 e^ .08t + .48e^-.04t = 0
divide by .48
2.5e^.08t + e^-.04t = 0
e^-.48t( e^.56t + 1) = 0
I get no real solution for this, neither does Wolfram
http://www.wolframalpha.com/input/?i=e%5E(-.48t)(+e%5E(.56t)+%2B+1)+%3D+0
The revenue, in millions of dollars, for a company in year t is given by the function:
R(t)=15*e^(0.08*t), 0≤t≤15
and the cost, in millions of dollars, to run the company in year t is approximated by:
C(t)=12*e^(−0.04*t), 0≤t≤15
where t is the number of years after January 1st of the year 2000. What was the net profit (in millions of dollars) for the company from January 1st in the year 2000 until January 1st in the year 2007? Round your answer to the nearest million dollars.
1 answer
1 answer