Respuesta :
Answer:
a
The model is [tex]R(t) = \frac{13.897t^2 }{2} + 284.653 * ln(t) -311.42[/tex]
b
The revenue for 2007 is [tex]R(7) = \$ 582.97[/tex]
Step-by-step explanation:
From the question we are told that
The rate of change in revenue is [tex]\frac{dR}{dt} = 13.897t + \frac{284.653}{t}[/tex]
The revenue at 2008 is [tex]R_8 = \$ 725.2 \ millon[/tex]
Generally we are told from the question that the revenue is denoted as R so to obtain R we integrate the both sides of rate of change in revenue, so
[tex]\int\limits dR = \int\limits ( 13.897t + \frac{284.653}{t})dt[/tex]
=> [tex]R(t) = \frac{13.897t^2 }{2} + 284.653 * ln(t) + c[/tex]
From the question at t = 8 , [tex]R_8 = \$ 725.2 \ millon[/tex]
So
[tex]725.2 = \frac{13.897* 8^2 }{2} + 284.653 * ln(8) + c[/tex]
=> [tex]725.2 = 1036.623 +c[/tex]
=> [tex]c = -311.42[/tex]
So
[tex]R(t) = \frac{13.897t^2 }{2} + 284.653 * ln(t) -311.42[/tex]
At t = 7 the Under Armour's revenue is mathematically represented as
[tex]R(7) = \frac{13.897 * 7^2 }{2} + 284.653 * ln(7) -311.42[/tex]
=> [tex]R(7) = \frac{13.897 * 7^2 }{2} + 284.653 * ln(7) -311.42[/tex]
=> [tex]R(7) = \$ 582.97[/tex]
From the differential equation given, we have that:
a) The model is:
[tex]R(t) = 6.9485t^2 + 284.653\ln{t} + 219.41[/tex]
b) The revenue in 2007 was of $594.67 million.
The differential equation for the rate of change of revenue is:
[tex]\frac{dR}{dt} = 13.897t + \frac{284.653}{t}[/tex]
It can be solved using separation of variables, hence:
[tex]\int dR = \int \left(13.897t + \frac{284.653}{t}\right) dt[/tex]
[tex]R(t) = 6.9485t^2 + 284.653\ln{t} + K[/tex]
In which K is the constant of integration.
Item a:
In 2008, the revenue was of $725.2 million, hence R(4) = 725.2, and this is used to find K.
[tex]R(t) = 6.9485t^2 + 284.653\ln{t} + K[/tex]
[tex]725.2 = 6.9485(4)^2 + 284.653\ln{4} + K[/tex]
[tex]K = 725.2 - 6.9485(4)^2 - 284.653\ln{4}[/tex]
[tex]K = 219.41[/tex]
Hence, the model is:
[tex]R(t) = 6.9485t^2 + 284.653\ln{t} + 219.41[/tex]
Item b:
This is R(3), as 2007 - 2004 = 4, hence:
[tex]R(3) = 6.9485(3)^2 + 284.653\ln{3} + 219.41 = 594.67[/tex]
Thus, the revenue in 2007 was of $594.67 million.
A similar problem is given at https://brainly.com/question/14338155
