The nitrogen level would give the best yield at N = 1
Given y = (kN)/(1+N^2)
first step is to take the derivative
the derivative is [tex]y' = \frac{k - kN^2}{( 1 + N^2)^2}[/tex]
ON equating this value of y' to 0
we get
y' = k - k[tex]N^2[/tex] = 0
k ( 1 - N^2) = 0
After dividing the above equation with k
and then subtracting we get ;
N = ± 1
However N = 1 is the only possible answer since the nitrogen level can't be negative
thus the nitrogen level gives the best yield at N = 1
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