Question

Two models R1 and R2 are given for revenue (in millions of dollars) for a corporation. Both models are estimates of revenues from 2030 through 2035, with t = 0 corresponding to 2030.

R1 = 7.23 + 0.25t + 0.03t^2
R2 = 7.23 + 0.1t + 0.01t^2
How much more total revenue (in millions of dollars) does that model project over the six-year period ending at t = 5? (Round your answer to three decimal places.)

Answer 1

To find the difference in total revenue projected by the two models over the six-year period ending at t = 5, we need to calculate the difference between the values of R1 and R2 at t = 5.

For R1:
R1 = 7.23 + 0.25t + 0.03t^2

Substituting t = 5 into the equation:
R1 = 7.23 + 0.25(5) + 0.03(5^2)
R1 = 7.23 + 1.25 + 0.75
R1 = 9.23 + 0.75
R1 = 9.98 million dollars

For R2:
R2 = 7.23 + 0.1t + 0.01t^2

Substituting t = 5 into the equation:
R2 = 7.23 + 0.1(5) + 0.01(5^2)
R2 = 7.23 + 0.5 + 0.25
R2 = 7.73 + 0.25
R2 = 7.98 million dollars

To find the difference in total revenue, we subtract R2 from R1:
Difference = R1 - R2
Difference = 9.98 - 7.98
Difference = 2 million dollars

Therefore, the model R1 projects $2 million more in total revenue over the six-year period ending at t = 5 compared to model R2.

Answer 2

Explanation:

To find the difference in total revenue projected by the two models over the six-year period ending at t = 5, we need to calculate the revenue for each model from t = 0 to t = 5 and subtract the results.

For R1:

R1 = 7.23 + 0.25t + 0.03t^2

Substituting t = 5:

R1(5) = 7.23 + 0.25(5) + 0.03(5^2)

R1(5) = 7.23 + 1.25 + 0.75

R1(5) = 9.23 + 0.75

R1(5) = 9.98 million dollars

For R2:

R2 = 7.23 + 0.1t + 0.01t^2

Substituting t = 5:

R2(5) = 7.23 + 0.1(5) + 0.01(5^2)

R2(5) = 7.23 + 0.5 + 0.25

R2(5) = 7.73 + 0.25

R2(5) = 7.98 million dollars

To find the difference, we subtract R2(5) from R1(5):

Difference = R1(5) - R2(5)

Difference = 9.98 - 7.98

Difference = 2 million dollars

Therefore, the model R1 projects 2 million dollars more in total revenue than R2 over the six-year period ending at t = 5.