(adsbygoogle = window.adsbygoogle || []).push({}); 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!

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# Application of Integration

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