Question

Solve this problem? It's so hard for me

Answer 1

Nitrogen level of `1` gives best yield.

Explanation:

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

it is maximized when `(dY)/(dN)=0` and `(d^2Y)/(dN^2)<0`

As `Y=(kN)/(1+N^2)`, using quotient rule

`(dY)/(dN)=(k(1+N^2)-2NxxkN)/(1+N^2)^2`

= `(k-kN^2)/(1+N^2)^2=k(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 `(d^2Y)/(dN^2)=((1+N^2)^2xx(-2kN)-2(1-N^2)(1+N^2)xx2N)/(1+N^2)^4`

When `N=1`, it is `(-8k)/16=-k/2` and hence negative.

Hence nitrogen level of `1` gives best yield.

Answer 2

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=(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`