Problem+7+Danny+G

__This is excellent! __ (ms sweeney) __

Problem 7__ Model used for agricultural crop yield (Y) is function of nitrogen level (N) in soil. (k) is a constant. Find nitrogen level with the best yield.

Y = kN / (1 + N^2) - only one equation, this is objective equation Y' = [(1 + N^2)(k) - (kN)(2N)] / (1 + N^2)^2 Y' = [k + kN^2 - 2kN^2] / (1 + N^2)^2 Y' = k(1 - N^2) / (1 + N^2)^2 0 = k(1 - N^2) / (1 + N^2)^2 Critical Values = -1, 1

Elie P.:

I agree with Danny's work so far. To answer some of the questions, we are trying to maximize this function since the problem asks for what nitrogen level gives the __best__ yield. Also, k can never equal zero, as the problem describes k as a positive constant. This function is defined over the interval of (0, infinity).

To summarize Danny's work, to determine the numbers that will maximize the function, you need to find the critical values as well as the behavior at the endpoints. To find the critical values you take the derivative of objective function and set it equal to zero.

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

Y(1)=k/2 Y(-1)=-k/2 - Not possible to have a negative nitrogen level. Therefore, -1 is not a feasible answer. Elie already stated interval of (0, infinity).

The other options are the end points. lim Y= 0 n->infinity lim Y= 0 n-->0

Therefore, n=1 provides the largest yield.