Question

A recently-installed machine earns the company revenue at a continuous rate of 60,000t 45,000 dollars per year during the first six months of operation and at the continuous rate of 75,000 dollars per year after the first six months. The cost of the machine is $159,000, the interest rate is 7% per year, compounded continuously, and is time in years since the machine was installed. (a) Find the present value of the revenue earned by the machine during the first year of operation. Round your answer to the nearest integer. 130914.5 (b) Find how long it will take for the machine to pay for itself; that is, how long it will take for the present value of the revenue to equal the cost of the machine?

Answer 1

Answer:

(a) The first year of the machine total Rate (R) =112500 dollars

(b) time (t)=1.8 years

Step-by-step explanation:

First we collect the data given;

Rate = 60000*t +45000

For six month : t = 1/2 year = 0.5 then

The revised Rate = 75000

Cost of Machine (CP)= 159000 dollar

Interest rate (r) = 7% per year = 0.07

time = t

(a) for first 6 months (from t=0 to t=0.5)

Rate =R (t) = 60000*t + 45000

put t = 0.5 we get,

R(0.5) = 60000(0.5) +45000

= 30000 + 45000 = 75000

for next six month (from t=0.5 to t=1)

Rate = R(t) =R(t)*t

= 75000 (0.5) = 37500

So for the first year of the machine total Rate (R) = 75000 + 37500 = 112500 dollar

(b) The compounded cost of machine is P(t) = CP*[e^rt]

P(t) = 159000* [e^0.07t]

Again the revanue of the machine after one year t =1 is

R(t) = 112500 + 75000(t-1)

According to question, P(t) = R(t)

159000* [e^0.07t] = 112500 + 75000(t-1)

= 75000t + 37500

e^0.07t = (75000t + 37500)/159000

solving the equation we have to find the value of t

applying Newtons erritation method

on simplifying above question we get, t = 1.8 years