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C2- Applications of differentiation help! watch

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    Please can you help me with the following questions. I missed a few lessons on this and this is just confusing me.
    1) A man wishes to fence in a rectangular enclosure of are 128m^2. One side of the enlosure is formed by part of a brick wall already in position. What is the least possible length of fencing required for the other three sides?

    2) The stamp counter at a post office is open from 9am to 3 pm daily. The average queuing time q, measured in minutes, at time x hours after opening is modelled by the equation q=1/4(-x^3+6x^2+20)
    a) Show that using this model, the longest average queuing time occurs around 1 pm and find how long it is.
    b) Find when the average queuing time is rising at its maximum rate, and show that this maximum rate of increase is 3 minutes/hour.

    Thanks in advance
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    1.)
    Let x be the two sides and y the side parallel to the wall.
    area=128=xy y=128/x
    perimeter=P=2x+y
    P=2x+128x^-1
    dP/dx=2-128x^-2=0 (for a minimum)
    2=128/x^2
    2x^2=128
    x^2=64
    x=8 (must be positive as its a length)]
    y=128/8=16
    P=16+(2*8)=32
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    2) The stamp counter at a post office is open from 9am to 3 pm daily. The average queuing time q, measured in minutes, at time x hours after opening is modelled by the equation q=1/4(-x^3+6x^2+20)
    a) Show that using this model, the longest average queuing time occurs around 1 pm
    Q = (6x2-x3+20)/4
    dQ/dx = 3x - 3x2/4

    Let dQ/dx = 0, so
    => 3x - 3x2/4 = 0
    => 3x = 3x2/4
    => 12 = 3x
    :. x = 4
    and find how long it is.
    4 hours after opening, so 9 + 4 = 1pm.

    sub (x = 4) into Q => (6(4)2 - (4)3+20)/4 = 13 min.
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    2.)
    a.)
    q=1/4(-x^3+6x^2+20)
    dq/dx=1/4(-3x^2+12x)=0 (for a maximum)
    12x-3x^2=0
    3x(4-x)=0
    x=0 or 4
    d2q/dx2=1/4(-6x+12)
    If x=0 d2q/dx2 is +ve, so it is a minimum.
    If x=4 d2q/dx2 is -ve, so it is a maximum.
    9am+4 hours is 1pm.

    b.)
    d2q/dx2=1/4(-6x+12)=0 (for maximum)
    12=6x x=2
    Put this into dq/dx.
    dq/dx=1/4(-3x^2+12x)
    =1/4(-12+24)
    =3 min/hour.
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    2) The stamp counter at a post office is open from 9am to 3 pm daily. The average queuing time q, measured in minutes, at time x hours after opening is modelled by the equation q=1/4(-x^3+6x^2+20)
    b) Find when the average queuing time is rising at its maximum rate, and show that this maximum rate of increase is 3 minutes/hour.
    dQ/dx = 3x - 3x2/4
    d2Q/dx2 = 3 - 3x/2

    Let d2Q/dx2 = 0, so

    => 0 = 3 - 3x/2
    => 3 = 3x/2
    => x = 2

    sub (x = 2) into dQ/dx => 3(2) - 3(2)2/4 = 3min.
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    thanks guys:tsr2: :yy:
 
 
 
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Updated: February 19, 2006

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