Question

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?

Answer 1

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.