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Application of Integration II

  1. Mar 13, 2009 #1
    1. The problem statement, all variables and given/known data

    The monthly cost C(t) at time t of operating a certain machine in a factory can be modelled by C(t) = aebt-1 (0 < t <= 36),
    where t is in month and C(t) is in thousand dollars.
    The following table shows the values of C(t) when t = 1, 2, 3, 4.

    C(1) = 1.21; C(2) = 1.44; C(3) = 1.70; C(4) = 1.98.

    (a)(i) Express ln[C(t)+1] as a linear function of t.
    (a)(ii) Use the given table and the graph paper below to estimate graphically the values of a and b correct to 1 decimal place.
    (a)(iii) Using the values of a and b found in (a)(ii), estimate the monthly cost of operating this machine when t = 36.
    (b) The monthly income P(t) generated by this machine at time t can be modelled by P(t) = 439 - e0.2t (0 < t <= 36),
    where t is in month and P(t) is in thousand dollars.
    The factory will stop using this machine when the monthly cost of operation exceeds the monthly income.
    (i) Find the value of t when the factory stops using this machine. Give the answer correct to the nearest integer.
    (ii) What is the total profit generated by this machine? Give the answer correct to the nearest thousand dollars.

    (Answers
    (a)(i) ln a + bt
    (a)(ii) a = 2.0, b = 0.1
    (a)(iii) 72.1965 thousand dollars
    (b)(i) 30
    (b)(ii) 10806 thousand dollars)

    2. Relevant equations

    Definite Integration Formulae

    3. The attempt at a solution

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

    [tex]\int[/tex][tex]^{29}_{0}[/tex] (P(t) - C(t)) dt

    =[440t - [tex]\frac{1}{0.2}[/tex]e0.2t - [tex]\frac{2}{0.1}[/tex]e0.1t][tex]^{29}_{0}[/tex]

    =10770 (but not 10806)
     
    Last edited: Mar 13, 2009
  2. jcsd
  3. Mar 13, 2009 #2
    From the answer you got in (b)(i) I would say you have to integrate from 0 to 30, this gives the correct answer to (b)(ii).
     
  4. Mar 13, 2009 #3
    Yes, I got it!

    Thank you very much!
     
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