# Solve this problem? It's so hard for me

Jan 4, 2018

Nitrogen level of $1$ gives best yield.

#### Explanation:

As yield $Y$ is given by $Y = \frac{k N}{1 + {N}^{2}}$, where $N$ is a nitrogen level in soil and $k$ isa positive constant>

it is maximized when $\frac{\mathrm{dY}}{\mathrm{dN}} = 0$ and $\frac{{d}^{2} Y}{{\mathrm{dN}}^{2}} < 0$

As $Y = \frac{k N}{1 + {N}^{2}}$, using quotient rule

$\frac{\mathrm{dY}}{\mathrm{dN}} = \frac{k \left(1 + {N}^{2}\right) - 2 N \times k N}{1 + {N}^{2}} ^ 2$

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

It is $0$ when $1 - {N}^{2} = 0$ i.e. $N = 1$ (we ignore $N = - 1$ as nitrogen level can take only real positive values).

Further $\frac{{d}^{2} Y}{{\mathrm{dN}}^{2}} = \frac{{\left(1 + {N}^{2}\right)}^{2} \times \left(- 2 k N\right) - 2 \left(1 - {N}^{2}\right) \left(1 + {N}^{2}\right) \times 2 N}{1 + {N}^{2}} ^ 4$

When $N = 1$, it is $\frac{- 8 k}{16} = - \frac{k}{2}$ and hence negative.

Hence nitrogen level of $1$ gives best yield.

Jan 4, 2018

The best yield will be when $N = 1$ and then function stands: F(k)=k/2; k>0

#### Explanation:

I have no idea how plants work but I'm assuming the more nitrogen a plant has the better. Therefore we are looking for maximum. We can find that by first derivative: (k is constant)

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

Y'=(k(1+N^2)-kN(2N))/(1+N^2)^2=k(1-N^2)/(1+N^2)^2;quadquadk>0

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

$\left(1 - {N}^{2}\right) = 0 \quad \implies \quad \left(1 - N\right) \left(1 + N\right) = 0$

$\implies {N}_{0} = \pm 1$

$N \in \left(- \infty , - 1\right) \Leftrightarrow Y ' < 0$ function is decreasing

min: $N = - 1 \implies$ minimum of function (f goes down and then up=>must be min.)

$N \in \left(- 1 , 1\right) \Leftrightarrow Y ' > 0$ function is increasing

max: $\textcolor{red}{N = 1 \implies F \left(1\right) = \frac{1 \cdot k}{1 + {1}^{2}} = \frac{k}{2}}$

$N \in \left(1 , \infty\right) \Leftrightarrow Y ' < 0$ function is decreasing

The best yield will be when $N = 1$ and then function stands: F(k)=k/2; k>0