Question 1113247
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The way you posed this question was confusing.  *[tex \LARGE c(x)\ =\ x^2\ +\ x\ +\ ] implies that the coefficients on the high order and first-degree terms must be one, which, of course, is impossible.  Had you said "find the coefficients for *[tex \LARGE c(x)\ =\ ax^2\ +\ bx\ +\ d]" then it would make sense.  And that is the only way that I can make sense of your question so that's what I'm going to show here.


We know that he gets 14 students @ $12 each, so if *[tex \Large x\ =\ 0] then his revenue is $168.  If *[tex \Large x\ =\ 1] he gets 13 students @ $14 each, so his revenue is $182.  And if *[tex \Large x\ =\ 2] we have 12 @ $16 for $192.  Symbolically:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c(0)\ =\ a(0)^2\ +\ b(0)\ +\ d\ =\ 168]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c(1)\ =\ a(1)^2\ +\ b(1)\ +\ d\ =\ 182]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c(2)\ =\ a(2)^2\ +\ b(2)\ +\ d\ =\ 192]


Since the first equation reduces to *[tex \LARGE d\ =\ 168] we can reduce the other two equations to:


(1)  *[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ +\ b\ =\ 14]


(2)  *[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4a\ +\ 2b\ =\ 24]


From (1):

 
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b\ =\ 14\ -\ a]


Substituting into (2)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4a\ +\ 2(14\ -\ a)\ =\ 24]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2a\ =\ 24\ -\ 28]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ -2]


Substituting back into (1)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -2\ +\ b\ =\ 14]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b\ =\ 16]


And the desired function becomes:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c(x)\ =\ -2x^2\ +\ 16x\ +\ 168]


Extra credit:  What is the price point to maximize revenue?  Hint:  What is the *[tex \Large x]-coordinate of the vertex of this concave-down parabola?


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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