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Suppose that a company needs new equipment, and that the machinery in question earns the company revenue at a continuous rate of 66000t 38000 dollars per year during the first six months of operation, and at the continuous rate of $71000 per year after the first six months. The cost of the machine is $160000. The interest rate is 5% per year, compounded continuously. a) Find the present value of the revenue earned by the machine during the first year of operation. Round your answer to the nearest cent. Value: $ equation editorEquation Editor b) Determine how long it will take for the machine to pay for itself; that is, how long until the present value of the revenue is equal to the cost of the machine. Round your answer to the nearest hundredth. Years: equation editorEquation Editor

Sagot :

danialamin

Answer:

a-The present value of revenue in the first year is $61,085.92.

b-The total time it would take to pay for its price is 2.44 years of 29.33 months.

Explanation:

a-

Let the function of the revenue earned is given as

[tex]S(t)=\left \{ {{66000t+38000} {\ \ 0

The present value is given as

[tex]PV=\int\limits^a_b {S(t)e^{-rt}} \, dt[/tex]

Here

  • a and b are the limits of integral which are 0 and 1 respectively
  • r is the rate of interest which is 5% or 0.05
  • S(t) is the function of value which is [tex]S(t)=\left \{ {{66000t+38000} {\ \ 0

So the equation becomes

[tex]PV=\int\limits^0_1 {S(t)e^{-0.05t}} \, dt\\PV=\int\limits^{0.5}_0 {(66000t+38000)e^{-0.05t}} \, dt+\int\limits^{1}_{0.5}{(71000)e^{-0.05t}} \, dt\\PV=\int\limits^{0.5}_0 {(66000t)e^{-0.05t}} \, dt+\int\limits^{0.5}_0 {(38000)e^{-0.05t}} \, dt+\int\limits^{1}_{0.5}{(71000)e^{-0.05t}} \, dt\\PV=8113.7805+18764.4669+34207.6751\\PV=61085.9225[/tex]

So the present value of revenue in the first year is $61,085.92.

b-

The time in which the machine pays for itself is given as

[tex]PV=\int\limits^0_1 {S(t)e^{-0.05t}} \, dt+\int\limits^t_1 {S(t)e^{-0.05t}} \, dt\\PV=61085.9225+\int\limits^{t}_{1}{(71000)e^{-0.05t}} \, dt[/tex]

The present value is set equal to the value of machine which is given as

$160,000 so the equation becomes:

[tex]PV=61085.9225+\int\limits^{t}_{0}{(71000)e^{-0.05t}} \, dt\\160000=61085.9225+\int\limits^{t}_{0}{(71000)e^{-0.05t}} \, dt\\\int\limits^{t}_{0}{(71000)e^{-0.05t}} \, dt=160000-61085.9225\\\int\limits^{t}_{1}{(71000)e^{-0.05t}} \, dt=98914.07\\\\t=-\dfrac{\ln \left(0.93034\right)}{0.05}\\t=1.44496[/tex]

So the total time it would take to pay for its price is 2.44 years of 29.33 months.

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