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14 November, 12:12

A model used for the yield y of an agricultural crop as a function of the nitrogen level n in the soil (measured in appropriate units) is y = kn 1 + n2 where k is a positive constant. what nitrogen level gives the best yield?

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  1. 14 November, 14:56
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    The maxima of an equation can be obtained by taking the 1st derivative of the equation then equate it to 0. The value of N that result in best yield is when dy/dn = 0.

    Taking the 1st derivative of the equation y = (kn) / (9+n^2):

    By using the quotient rule the form of the equation is:

    y = g (n) / h (n)

    where:

    g (n) = kn - - - > g' (n) = k

    h (n) = 9 + n^2 - - - > h' (n) = 2n

    dy/dn is defined as:

    dy/dn = [h (n) * g' (n) - h' (n) * g (n) ] / h (n) ^2

    dy/dn = [ (9 + n^2) (k) - (kn) (2n) ] / (9 + n^2) ^2

    dy/dn = (9k + kn^2 - 2kn^2) / (9 + n^2) ^2

    dy/dn = (9k - kn^2) / (9 + n^2) ^2

    dy/dn = k (9 - n^2) / (9 + n^2) ^2

    Equate dy/dn = 0, then solve for n

    k (9 - n^2) / (9 + n^2) ^2 = 0

    k (9 - n^2) = 0

    9 - n^2 = 0

    n^2 = 9

    n = sqrt (9)

    n = 3

    Answer: The nitrogen level that gives the best yield of agricultural crops is 3 units.
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