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Math Help - Optimization

  1. #1
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    Red face Optimization

    According to postal regulations, the girth plus length of parcels sent by fourth class mail may not exceed 108 inches. what is the largest volume of a cylindrical parcel that can be sent by fourth-class mail?

    I got:

    2r+1=108→Length=108-2r
    V(d)= πr^2(108-2r)=108πr^2-2πr^3
    Ví(d)=216 πr-6 πr^2=0
    r(36-r)=0
    So , r=36, Length = 36


    Is this Right?
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by lisa1984wilson View Post
    According to postal regulations, the girth plus length of parcels sent by fourth class mail may not exceed 108 inches. what is the largest volume of a cylindrical parcel that can be sent by fourth-class mail?

    I got:

    2r+1=108→Length=108-2r
    V(d)= πr^2(108-2r)=108πr^2-2πr^3
    Ví(d)=216 πr-6 πr^2=0
    r(36-r)=0
    So , r=36, Length = 36


    Is this Right?
    Lisa: I could be wrong but I think 'girth' refers to the circumference 2Pi r.
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  3. #3
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    Lightbulb

    girth = 2PI r
    height=h
    h+2 PI r =108
    h=108-2 PI r
    Volume = PI r^2 h
    V = PI r^2 (108-2PI r)
    Differentiate with respect to r
    dV/dr = PI (2r) (108-2PI r) +PI r^2 (-2PI) =0
    216 PI r -4 PI^2 r^2 -2 PI^2 r^2=0
    216 PI r -6 PI ^2 r^2=0
    PI r [ 216-6 PI r] =0
    6 PI r = 216
    r = 36/PI --- dimensions
    h = (108-2 PI (36/PI)) = 108 -72=36 -- dimensions

    Largest volume = PI r^2 h = PI (36/PI)^2 (36)
    =1,296 / PI

    d^2V/dr^2 =216 PI -6 PI^2 (2r)
    when r=36/PI, d^2V/dr2 = 216 PI - 6PI^2 (2)(36)/PI
    =216 PI -432 PI < 0, indicates V has been maximized.
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  4. #4
    Senior Member apcalculus's Avatar
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    Lisa: Your work and justification look good!
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