Question 190488
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Revenue is the number of items sold times the price for each, so:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \  R(p) = \left(2000-100p\right)p=2000p - 100p^2]

To find the value of any <i><b>R(a)</b></i>, just substitute <i><b>a</b></i> for the independent variable, <i><b>p</b></i> in this case, and then do the arithmetic.

<i><b>R(5)</b></i>:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \  R(5)\ =\ 2000(5) - 100(5)^2]

You can do your own arithmetic.  And the other two are done exactly the same way.

I can't do a bar graph on this site, at least not without a great deal of work.  You could put your data into Excel and have Excel create a chart for you.

There is a better way to find the price that gives the maximum revenue.  Since the coefficient on the high order term is negative and this is a quadratic function, you know the graph of the function is a parabola opening downward.  Hence, the vertex of the parabola is a maximum point.  The value of the independent variable coordinate of the vertex of a parabola, and hence value of the independent variable that causes the function to be extreme, for a parabola of the form:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \  f(x) = ax^2 + bx + c]

is given by:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x = {-b \over 2a}]

*[tex \LARGE \ \ \ \ \ \ \ \ \ \  p = {-2000 \over 2(-100)} = 10]

Meaning <i><b>R(10)</b></i> will give the largest <i><b>R</b></i> for any possible value of <i><b>p</b></i>

<i><b>Super-double-plus extra credit:</b></i>

What is the <i><b>domain</b></i> of this function?  That is, what is the set of values that <i><b>p</b></i> could reasonably be?

John
*[tex \LARGE e^{i\pi} + 1 = 0]
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