# Application of Integration

1. Mar 10, 2009

### chrisyuen

1. The problem statement, all variables and given/known data

A textile factory plans to install a weaving machine on 1st January 1995 to increase its production of cloth. The monthly output x (in km) of the machine, after t months, can be modelled by the function x = 100e^(-0.01t) - 65e^(-0.02t) - 35.

(a)(i)
In which month and year will the machine cease producing any more cloth?

(a)(ii)
Estimate the total amount of cloth, to the nearest km, produced during the lifespan of the machine.

(b)
Suppose the cost of producing 1km of cloth is US$300; the monthly maintenance fee of the machine is US$300 and the selling price of 1km of cloth is US$800. In which month and year will the greatest monthly profit be obtained? Find also the profit, to the nearest US$, in that month.

(c)
The machine is regarded as "inefficient" when the monthly profit falls below US$500 and it should then be discarded. Find the month and year when the machine should be discarded. Explain your answer briefly. Answers (a)(i) February 2000 (a)(ii) 141km (b) March 1997; US$1430
(c) April 1999

2. Relevant equations

Quadratic Equation and Definite Integral Formulae

3. The attempt at a solution

I don't know how to solve the part (b).

I tried to establish a formula (800-300)x - 300t = 500x - 300t with no hope.

Can anyone tell me how to solve it?

Thank you very much!

Last edited: Mar 11, 2009
2. Mar 11, 2009

### tiny-tim

Hi chrisyuen!

(try using the X2 tag just above the reply box )

Just put x = 100e-0.01t - 65e-0.02t - 35 into 500x - 300t, and find the maximum

3. Mar 11, 2009

### chrisyuen

I followed your suggestion and then solve for t below:
650e-0.02t - 500e-0.01t - 300 = 0

Finally, I got e-0.01t = 1.17 OR e-0.01t = -0.4.

Both values should be rejected.

Did my work correct or not?

Thank you very much!

4. Mar 11, 2009

### tiny-tim

ah … on reading the question more carefully , your -300t should be left out … it's the same every month!

5. Mar 12, 2009

### chrisyuen

I can solve the part (b) using your method.

Thanks a lot!

I have another question for part (c).

I found that t = 6.2 OR t = 51.

How can I reject the first one?

Thank you very much!

6. Mar 12, 2009

### tiny-tim

Well, it says "Find the month and year when the machine should be discarded" … so surely it should be discarded the first month when it shows a loss, not the last one?

If you think the first one is wrong, you'd better show us your full calculation.

7. Mar 12, 2009

### chrisyuen

Monthly Profit y(t)
= 800x - 300x -300
= 500x - 300
= 500(100e-0.01t - 65e-0.02t -35) - 300
= 50000e-0.01t - 32500e-0.02t - 17800

y(t) = 500
50000e-0.01t - 32500e-0.02t - 18300 = 0
e-0.01t = 0.6 OR e-0.01t = 0.94
t = 51 OR t = 6.2

y(t) < 500
(e-0.01t - 0.6)(e-0.01t -0.94) < 0

Case 1: (e-0.01t - 0.6) < 0 and (e-0.01t -0.94) > 0

t > 51 and t < 6.2

Case 2: (e-0.01t - 0.6) > 0 and (e-0.01t -0.94) < 0

t < 51 and t > 6.2 ==> 6.2 < t < 51 (rejected)

But the answer provided by the textbook only gave me 51 months (April 1999).

Should I use inequality to solve this question?

Thank you very much!

Last edited: Mar 13, 2009
8. Mar 13, 2009

### tiny-tim

Hi chrisyuen

Yes, your calculations look completely correct

hmm … when t = 0, the output is 0 …

so the output starts very small, builds up, and then falls …

the question isn't clear on this , but it looks as if the machine has to operate for 6 months before it makes a profit … and it would be silly to close it down just then!

9. Mar 13, 2009

### chrisyuen

I got it!

Thank you very much!