Solve this problem? It's so hard for me

Question

Samantha Adkison · Accepted Answer

Nitrogen level of #1# gives best yield.

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.

William Abdalla · Answer

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

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#

Charles Arocha · Answer

undefined Of course, I'd be happy to help. Please provide the problem you'd like assistance with, and I'll do my best to guide you through solving it.