SOLUTION: a hospital revenue (in millions of dollars) is projected to be r(x)=9x^2+7x+81 and it costs (in millions of dollars) are projected to be c(x)=-2x^2-10x-14 where the x represent the
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Question 1137193: a hospital revenue (in millions of dollars) is projected to be r(x)=9x^2+7x+81 and it costs (in millions of dollars) are projected to be c(x)=-2x^2-10x-14 where the x represent the numbers of years into the future
a. create a profit function for the hospital
b. what is the average rate of change in the profit from year 3 to year 5 Answer by jim_thompson5910(35256) (Show Source):
x = number of years into the future
p(x) = profit, in millions of dollars
Example: x = 2 leads to p(x) = 173, telling us that 2 years into the future the profit will be estimated at 173 million dollars.
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Part B
We will use the formula
where,
R = average rate of change on the interval a < x < b
p(x) = the profit function
a,b = start and endpoint of the interval
In this case, a = 3 and b = 5. So we'll first need to compute p(3) and p(5)
plug in x = 3
p(x) = 11x^2+17x+95
p(3) = 11(3)^2+17(3)+95
p(3) = 11(9)+17(3)+95
p(3) = 99+51+95
p(3) = 245
then repeat for x = 5
p(x) = 11x^2+17x+95
p(5) = 11(5)^2+17(5)+95
p(5) = 11(25)+17(5)+95
p(5) = 275+85+95
p(5) = 455
Now we can use the average rate of change formula mentioned earlier to get...
Plug in a = 3 and b = 5
Replace p(5) with 455; replace p(3) with 245
The average rate of change in profit, from year 3 to year 5, is 105 million dollars per year
This represents the average speed in growth rate of the profit over this time span.
Recall that y = p(x) is measured in millions of dollars. When we subtracted p(b)-p(a), we found the change in millions of dollars. When we computed b-a in the denominator, we found the change in years. Dividing change in profit over change in time yields the units "millions of dollars per year"; or you can think of it as "dollars per year" in a sense.
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