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?

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.