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Solve this problem? It's so hard for me

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Answer 1 · 2018-01-04T15:28:55

Nitrogen level of $1$ $1$ gives best yield.

Explanation:

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

it is maximized when $\frac{d Y}{d N} = 0$ $(dY)/(dN)=0$ and $\frac{d^{2} Y}{d N^{2}} < 0$ $(d^2Y)/(dN^2)<0$

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

$\frac{d Y}{d N} = \frac{k (1 + N^{2}) - 2 N \times k N}{{(1 + N^{2})}^{2}}$ $(dY)/(dN)=(k(1+N^2)-2NxxkN)/(1+N^2)^2$

= $\frac{k - k N^{2}}{{(1 + N^{2})}^{2}} = k \frac{1 - N^{2}}{{(1 + N^{2})}^{2}}$ $(k-kN^2)/(1+N^2)^2=k(1-N^2)/(1+N^2)^2$

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

Further $\frac{d^{2} Y}{d N^{2}} = \frac{{(1 + N^{2})}^{2} \times (- 2 k N) - 2 (1 - N^{2}) (1 + N^{2}) \times 2 N}{{(1 + N^{2})}^{4}}$ $(d^2Y)/(dN^2)=((1+N^2)^2xx(-2kN)-2(1-N^2)(1+N^2)xx2N)/(1+N^2)^4$

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

Hence nitrogen level of $1$ $1$ gives best yield.

Answer 2 · 2018-01-04T15:33:30

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

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)

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

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

$k(1-N^2)/(1+N^2)^2=0hArr(1-N^2)=0$

$(1-N^2)=0quad=>quad(1-N)(1+N)=0$

$=>N_0=+-1$

$N in (-oo, -1) hArr Y'<0$ function is decreasing

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

$N in (-1, 1) hArr Y'>0$ function is increasing

max: $color(red)(N=1=>F(1)=(1*k)/(1+1^2)=k/2)$

$N in (1,oo) hArr Y'<0$ function is decreasing

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

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