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.
