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Calculus

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The product of 2 positive numbers is 48. find the value of the numbers if the sum of one of the numbers and the cube of the other is a minimum.

  • Calculus -

    a = first number

    b = second number


    a * b = 48 Divide both sides by a

    b = 48 / a


    S = the sum of one of the numbers and the cube of the other number


    S = a + b ^ 3

    S = a + ( 48 / a ) ^ 3

    S = a + 110,592 / a ^ 3

    S = a + 110,592 * a ^ - 3


    First derivaton :

    d S / d a = 1 - 3 * 110,592 * a ^ - 4

    d S / d a = 1 - 331,776 / a ^ 4


    Second derivation :

    d ^ 2 S / d a ^ 2 = - 3 * 331,776 ( - 4 ) * a ^ - 5

    d ^ 2 S / d a ^ 2 = 1,327,104 / a ^ 5


    A function has minimum or maximum in poit where first derivation = 0

    If second derivaton < 0 function has maximum.

    If second derivaton > 0 function has minimum.


    In this case:

    d S / d a = 1 - 331,776 / a ^ 4 = 0

    1 = 331,776 / a ^ 4 Multiply both sides by a ^ 4

    a ^ 4 = 331,776

    a = fourth root of 331,776

    a = + OR - 24


    For a = - 24

    d ^ 2 S / d a ^ 2 = 1,327,104 / a ^ 5 =

    1,327,104 / - 7,962,624 = - -0.166667 < 0

    function has maxsimum.



    For a = 24

    d ^ 2 S / d a ^ 2 = 1,327,104 / a ^ 5 =

    1,327,104 / -7,962,624 = 0.166667 > 0

    function has minimum.


    So a = 24

    b = 48 / a = 48 / 24 = 2

    The mumbers are a = 24 and b = 2

    Local minimum = a + b ^ 3 = 24 + 2 ^ 3 = 24 + 8 = 32

  • Calculus -

    Thank you!!!

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