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March 26, 2017

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Find the radius and hight of cylinder with voume 64π and radius r between 1 and 5 that has smallest possible surface area. A cylinder of radius r and hight aah 2π^2+2π r h and π r ^(2) h.

  • calculus - ,

    V = r ^ 2 ð h

    64 ð = r ^ 2 ð h Divide both sides with ð

    64 = r ^ 2 h Divide both sides with r ^ 2

    64 / r ^ 2 = h

    h = 64 / r ^ 2


    A = 2 r ^ 2 ð + 2 r ð h

    A = 2 r ^ 2 ð + 2 r ð 64 / r ^ 2

    A = 2 r ^ 2 ð + 128 r ð / r ^ 2

    A = 2 r ^ 2 ð + 128 ð / r

    d A / d r = ( 4 ð ( r ^ 3 - 32 ) ) / r ^ 2

    Function has extreme value where first derivative = 0
    dA / dr = 0

    Solutions:

    r = -1.5874 + 2.7495 i

    r = -1.5874 2.7495 i

    and

    r = 2 2 ^( 2 / 3 ) = 3,1748

    So :

    r = 2 2 ^( 2 / 3 ) = 3,1748

    Second derivative :

    4 ð ( 64 / r ^ 3 + 1 ) =

    4 ð ( 64 / 32 + 1 ) =

    4 ð ( 2 + 1 ) =

    4 ð 3 = 12 ð > 0

    Remark : r ^ 3 = [ 2 2 ^( 2 / 3 ) ] ^ 3 = 32

    If second derivative > 0

    that is minimum value of function.

    So :

    r = 2 64 / [ 2 2 ^( 2 / 3 ) ] ^ 2

    h = 64 / r ^ 2

    h = h = 64 / [ 2 2 ^( 2 / 3 ) ] ^ 2

    h = 4 2 ^( 2 / 3 )

  • calculus - ,

    ð = pi number

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