These qucstions are designed to be more challenging than a typical lecture, quiz
ID: 2257425 • Letter: T
Question
These qucstions are designed to be more challenging than a typical lecture, quiz or test/exam question. You are strongly encouraged to try the practice qucstions in Chapter 10.3 and 11.1 before continuing. 1. Show that e cos(ry) r+1 when r and y are both small. Show all your work. 2. A model for the yicld Y of an agricultural crop as a function of the nitrogen level N and phosphorous level P (measured in appropriate units) in the soil is where k is a positive constant. What levels of nitrogen and phosphorous result in the best yield? Show all your work.Explanation / Answer
1.
ex = 1+x/1! + x2/2! + ......
cos(xy) = 1- (xy)2/2! + (xy)4/4! + ....
ex cos(xy)
= (1+x/1! + x2/2! + ......)*( 1- (xy)2/2! + (xy)4/4! + ....)
Since x and y are very small , we can ignore the terms of order 2 or above which are x2/2! , - (xy)2/2! and so on
Therefore
ex cos(xy)
= (1+x/1 + x2/2! + ......)*( 1- (xy)2/2! + (xy)4/4! + ....)
= (1+x + 0 )(1- 0) = 1+x
2.
Y(N,P) = kNPe-N-P) = kNPe-Ne-P) = k*(Ne-N)*(Pe-P)
For maximum yield
dY/DN = 0 and dY/DP = 0
dY/DN = 0
=>d/DN ( k*(Ne-N)*(Pe-P)) = 0
=> k*(Pe-P)*d/DN (Ne-N) = 0
=> d/DN (Ne-N) = 0
=> Nd(e-N)/dN+ e-N d(N)/dN = 0
=> -Ne-N+e-N = 0
=> N = 1
and
dY/DP = 0
=>d/DP ( k*(Ne-N)*(Pe-P)) = 0
=> k*(Ne-N)*d/DP (Pe-P) = 0
=> d/DP (Pe-P) = 0
=> Pd(e-P)/dP+ e-P d(P)/dP = 0
=> -Pe-P+e-P = 0
=> P = 1
hence N= 1 and P = 1 for maximum yield.
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