Question 974830
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Since Revenue = *[tex \Large xp] and *[tex \Large p\ =\ -\frac{1}{10}x\ +\ 150], we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ R(x)\ =\ -\frac{1}{10}x^2\ +\ 150x]


The graph of this function is a parabola opening downward.  Find the vertex.  The *[tex \Large x]-coordinate of the vertex is the value of *[tex \Large x] that maximizes revenue.  The *[tex \Large y]-coordinate is the value of the maximum revenue.


Hint:


For a parabola with equation of the form 


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


The *[tex \Large x]-coordinate of the vertex is given by *[tex \Large x_v\ =\ -\frac{b}{2a}].  The *[tex \Large y]-coordinate of the vertex is simply the value of the function at the *[tex \Large x]-coordinate of the vertex, to wit:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_{v}\ =\ \rho\left(-\frac{b}{2a}\right)\ =\ a\left(-\frac{b}{2a}\right)^2\ +\ b\left(-\frac{b}{2a}\right)\ +\ c]


Just plug in the numbers and do the arithmetic.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \