Answer:
The nitrogen level that gives the best yield is 7.
Step-by-step explanation:
We have the yield Y in function of the nitrogen level N defined as:
[tex]Y=\frac{kN}{49+N^2}[/tex].
To know which nitrogen level N maximize the yield, we have to derive and equal to zero:
[tex]dY/dn=k\frac{1*(n^2+49)-n*(2n)}{(n^2+49)^2} =k\frac{n^2+49-2n^2}{(n^2+49)^2} \\\\dY/dn=k\frac{49-n^2}{(n^2+49)^2} =0\\\\49-n^2=0\\\\n=\sqrt{49} =7[/tex]
The nitrogen level that gives the best yield is 7.
In the graph, we model Y versus n, with k=100.
