Question 617719
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Not sure how you got to


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -0.01(n\ +\ 100)^2\ +\ 8100]


But it is wrong.


Start with


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(n)\ =\ (800\ +\ n)(10\ -\ 0.01n)]


And FOIL it:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(n)\ =\ 8000\ -\ 2n\ +\ 10n\ -\ 0.01n^2]


Collect terms then go to standard quadratic form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(n)\ =\ -0.01n^2\ +\ 2n\ +\ 8000]


Note:  If you multiply yours out and collect terms you are only off by the sign on the first degree term, but it is nevertheless incorrect.


Note that this is a polynomial function of the form *[tex \LARGE \rho(x)\ =\ ax^2\ +\ bx\ +\ c] the graph of which is a parabola.  Since it has a negative lead coefficient, the parabola opens downward.  Therefore, the vertex of the parabola is the maximum of the function.  Recall that the *[tex \Large x]-coordinate of the vertex of *[tex \LARGE \rho(x)\ =\ ax^2\ +\ bx\ +\ c] is given by *[tex \LARGE x_v\ =\ \frac{-b}{2a}]


So calculate *[tex \LARGE \frac{-2}{2(-0.01)}] to find the number of vines <i><b>in excess of 800</b></i> that should be planted to maximize the yield.  Don't forget to add the answer to 800 to get the answer to the question posed by the problem.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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