Question 1188855
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The revenue is found by multiplying the number of customers by the price per customer. Both the number of customers and the price per customer can be expressed as functions of the number of $50 price decreases.


Let *[tex \Large x] represent the number of incremental $50 price decreases.


Then *[tex \Large P(x)\ =\ 650\ -\ 50x] represents the price per customer as a function of $50 price decreases.


And *[tex \Large C(x)\ = 50\ +\ 10x] represents the number of customers as a function of $50 price decreases.


Note that the domain of these functions is a subset of the integers where the lower bound is zero because you are not given information about the effect of a price increase, and an upper bound of 12 because it would be ludicrous to consider the company either giving away their product or services for nothing or even worse, paying people to take their product.


Then the revenue as a function of incremental price decreases is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ R(x)\ =\ P(x)C(x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ R(x)\ =\ (650\,-\,50x)(50\,+\,10x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ R(x)\ =\ 32500\ +\ 4000x\ -\ 500x^2]


The vertex of this convex-down parabola that represents the maximum value of the function and therefore the maximum revenue is located at *[tex \Large x_v\ = \frac{-4000}{2(-500)}\ =\ 4].  Hence the maximum revenue is obtained when the number of price reductions is 4.


Evaluate *[tex \Large P(4)]

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