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    collegegirl08's Avatar
    collegegirl08 Posts: 20, Reputation: 1
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    #1

    Jun 13, 2010, 09:30 AM
    Derivative Help
    Hi, I am working on my homework and was wondering if someone could tell me if I did this correctly?

    Problem:
    The occupancy rate at a certain hotel is given by the function:
    r(t) = (10/81)t^3 - (10/3)t^2+ (200/9)t + 55

    where t is measured in months, and t=0 corresponds to the beginning of January.

    It has been estimated that the monthly revenue in thousands of dollars is approximated by the function:
    R(r)= (3/5000)r^3 + (9/50)r^2

    where r is the occupancy rate.


    Questions:

    a.) What is the occupancy rate at the beginning of January?
    =10/81*(0^3)-10/3*(0^2)+200/9*(0)+55 =55

    b.)What is the occupancy rate at the beginning of June?
    =10/81*(5^3)-10/3*(5^2)+200/9*(5)+55 = 98.20987654

    c.) What is the monthly revenue at the beginning of January?
    =3/5000*(55^3)+9/50*(55^2)=$644.325

    d.) What is the montly revenue at the beginning of June?
    =3/5000*(98.20987654^3)+9/50*(98.20987654^2)

    e.) Find an expression that gives the rate of change of the occupancy rate with respect to time.
    (10/27)tē-(20/3)t+(200/9)

    *I found this by taking the derivative of the occupancy rate.

    f.) Find an expression that gives the rate of change of the monthly revenue with respect to time.
    (9/5000)rē+(9/25)r

    *I found this by taking the derivative of the revenue function.

    g.) What is the rate of change of the monthly revenue with respect to time at the beginning of January?

    (Hint: Use the Chain Rule to find R'(r(0))r'(0) and R'(r(5))r'(5))

    =((9/5000)*(0*0)+(9/25)*0)*(B11)*((10/27)*(0*0)-(20/3)*(0)+(200/9)) = $0.00

    At the beginning of June?
    =((9/5000)*(5*5)+(9/25)*5)*(B14)*((10/27)*(5*5)-(20/3)*(5)+(200/9)) = -$7,873.65
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #2

    Jun 13, 2010, 10:39 AM

    I don't see any mistake :), but I didn't check the calculations. :o

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