# How do you solve this optimization 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 $y = \frac{k N}{1 + {N}^{2}}$ where k is a positive constant. What nitrogen level gives the best yield? N=?

Apr 22, 2018

$N = 1$

#### Explanation:

Take the first derivative with respect to $N :$

$y ' = \frac{\left(1 + {N}^{2}\right) k - k N \left(2 N\right)}{1 + {N}^{2}} ^ 2$

$y ' = \frac{k + k {N}^{2} - 2 k {N}^{2}}{1 + {N}^{2}} ^ 2$

$y ' = \frac{k - k {N}^{2}}{1 + {N}^{2}} ^ 2$

Equate to $0$ and solve for $N$:

$\frac{k - k {N}^{2}}{1 + {N}^{2}} ^ 2 = 0$

$k \left(1 - {N}^{2}\right) = 0$

$1 - {N}^{2} = 0$

${N}^{2} = 1$

$N = \pm 1 \to N = 1$ is the only possible answer as we cannot have a negative nitrogen level.

The "best yield" would entail $y$ being at its maximum. To ensure that $N = 1$ gives a maximum for $y$, evaluate $y '$ in the following intervals:

$\left[0 , 1\right) , \left(1 , \infty\right)$ to determine whether $y '$ is positive ($y$ is increasing) or $y '$ is negative ($y$ is decreasing) in each interval.

If $N = 1$ is a maximum, then $y '$ will be positive before we reach $N = 1$ and negative afterwards:

$\left[0 , 1\right) :$

$y ' \left(0\right) = \frac{k - k \left(0\right)}{1} ^ 2 = k > 0$ So, $y$ is increasing on $\left[0 , N\right)$

$\left(1 , \infty\right) :$

$y ' \left(2\right) = \left(\frac{k - 4 k}{1 + 4} ^ 2\right) = - \frac{3 k}{25} < 0$ So, $y$ is decreasing on $\left(1 , \infty\right)$ and the maximum possible crop yield happens with $N = 1$.