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 levels gives the best yield?

Answer 1

Answer:

A nitrogen level of 1 gives the best yield.

Step-by-step explanation:

We are given the following:

Yield Y of an agricultural crop as a function of the nitrogen level N in the soil

$Y(N) = \displaystyle\frac{kN}{1+N^2}$

First, we differentiate Y(N) with respect to N, to get,

$\displaystyle\frac{d(Y(N))}{dN} = \displaystyle\frac{d}{dN}\Bigg(\displaystyle\frac{kN}{1+N^2}\Bigg)\\\\= \frac{(1+N^2)k-kN(2N)}{(1+N^2)^2}\\\\=\frac{k-kN^2}{(1+N^2)^2}$

Equating the first derivative to zero, we get,

$\displaystyle\frac{d(Y(N))}{dN} = 0\\\\\Rightarrow \frac{k-kN^2}{(1+N^2)^2} = 0$

Solving, we get,

$\displaystyle\frac{k-kN^2}{(1+N^2)^2} = 0\\\\k-kN^2 = 0\\k(1-N^2) = 0\\k \neq 0\\1-N^2 = 0\\N = \pm 1\\N \neq -1\\N = 1$

Again differentiation Y(N), with respect to N, we get,

$\displaystyle\frac{d^2(Y(N))}{dN^2} = \displaystyle\frac{2kN(N^2-3)}{(1+N^2)^3}$

At N = 1

$\frac{d^2(Y(N))}{dN^2} < 0$

Thus, by double derivative test, the maximum value of Y(N) occurs at N = 1.

Thus, largest yield of crop is given by:

Y(1) =

$Y(N) = \displaystyle\frac{k(1)}{1+(1)^2} = \frac{k}{2}$