<PRE><B>Question 190488</B>
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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 &lt;i&gt;&lt;b&gt;R(a)&lt;/b&gt;&lt;/i&gt;, just substitute &lt;i&gt;&lt;b&gt;a&lt;/b&gt;&lt;/i&gt; for the independent variable, &lt;i&gt;&lt;b&gt;p&lt;/b&gt;&lt;/i&gt; in this case, and then do the arithmetic.


&lt;i&gt;&lt;b&gt;R(5)&lt;/b&gt;&lt;/i&gt;:


*[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&#39;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}]


For your function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  p = {-2000 \over 2(-100)} = 10]


Meaning &lt;i&gt;&lt;b&gt;R(10)&lt;/b&gt;&lt;/i&gt; will give the largest &lt;i&gt;&lt;b&gt;R&lt;/b&gt;&lt;/i&gt; for any possible value of &lt;i&gt;&lt;b&gt;p&lt;/b&gt;&lt;/i&gt;


&lt;i&gt;&lt;b&gt;Super-double-plus extra credit:&lt;/b&gt;&lt;/i&gt;


What is the &lt;i&gt;&lt;b&gt;domain&lt;/b&gt;&lt;/i&gt; of this function?  That is, what is the set of values that &lt;i&gt;&lt;b&gt;p&lt;/b&gt;&lt;/i&gt; could reasonably be?


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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