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)Â :
By using the quotient
rule the form of the equation is:
y = g(n) / h(n)Â
where:
g(n) = kn   --->   g'(n) = kÂ
h(n) = 9 + n^2Â Â Â Â --->Â Â Â h'(n) = 2nÂ
dy/dn is defined as:
dy/dn = [h(n) * g'(n) - h'(n) * g(n)] / h(n)^2Â
dy/dn = [(9 + n^2)(k) - (kn)(2n)] / (9 +
n^2)^2Â
dy/dn = (9k + kn^2 - 2kn^2) / (9 + n^2)^2Â
dy/dn = (9k - kn^2) / (9 + n^2)^2Â
dy/dn = k(9 - n^2) / (9 + n^2)^2Â
Equate dy/dn = 0, then solve for nÂ
k(9 - n^2) / (9 + n^2)^2 = 0Â
k(9 - n^2) = 0Â
9 - n^2 = 0Â
n^2 = 9Â
n = sqrt(9)Â
n = 3Â
Answer: The nitrogen
level that gives the best yield of agricultural crops is 3 units.