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At a certain vineyard it is found that each grape vine produces about 10 pounds of grapes in a season when about 700 vines are planted per acre. For each additional vine that is planted, the production of each vine decreases by about 1 percent. So the number of pounds of grapes produced per acre is modeled by $$ A(n) = (700+n)(10-0.01n), $$ where $n$ is the number of additional vines planted.

Question: Shouldn't the $10-0.01n$ really be $0.1n$ since 1 percent of 10 is $0.1$ and not $0.01$?

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  • $\begingroup$ I agree with you, in principle but I would say $(700+n)(10)(1-0.01)^n$ when $n$ is small $(10)(1-0.01)^n \approx (10-0.1n)$ but as $n$ gets larger it diverges. By the book's model model, for a large enough value of $n$ production becomes negative. $\endgroup$ – Doug M Feb 25 '17 at 1:02
  • $\begingroup$ It's a model, so it will have a limited range of validity. If I were presented with such a model, I'd guess that the model wouldn't be reliable if the yield dropped below, say, 50% of optimum, because nobody was willing to do the experiment. $\endgroup$ – Joffan Feb 25 '17 at 1:10
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I agree with your correction but the question seems very ambiguous:

Let $P_{n}$ denote the pounds of graps produced by each vine after $n$ vines have been planted. The only relation which would make sense is that $P_{0} = 10, P_{n+1} = 0.99P_{n}$. Otherwise for $n>100$, we get a negative production.

$\Rightarrow P_{n} = 0.99^{n}P_{0} = 10*0.99^{n}$

Then $A(n) = 10*0.99^{n}(700+n)$

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  • $\begingroup$ There's a valid possibility that, within its working range, the model is talking about a percentage-point (of optimum) drop in production per extra vine. $\endgroup$ – Joffan Feb 25 '17 at 1:12
  • $\begingroup$ The model should then state a range of $n$ for which it is valid $\endgroup$ – mrnovice Feb 25 '17 at 1:21
  • $\begingroup$ Yes, of course, but it isn't likely to improve the model to use a different function in its range of validity, and who knows what to use anywhere else? $\endgroup$ – Joffan Feb 25 '17 at 1:27

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