Question 1183538
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x(p)\ =\ 1200\  -\ 100p]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  R(p)\ =\ x(p)\cdot p\ = 1200p\ -\ 100p^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{dR}{dp}\ =\ 1200\ -\ 200p]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  1200\ -\ 200p\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  p\ =\ 6]  is an extreme point


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{d^2R}{dp^2}\ =\ -200\ <\ 0\ \forall\ p\ \in\ \mathbb{R}]


Therefore *[tex \LARGE p\ =\ 6] is a maximum.


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

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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