Monday

December 22, 2014

December 22, 2014

Posted by **ssassie** on Monday, January 28, 2013 at 7:11pm.

- pre-calc -
**ssassie**, Tuesday, January 29, 2013 at 8:30pmSuppose the function p= -1/8 x + 100

(0 grtr&= x ~12) relates the selling

price p of an item to the quantity x

that is sold. Assume that p is in

dollars. What is the maximum

revenue possible in this situation?

- pre-calc -
**Anonymous**, Tuesday, November 18, 2014 at 10:52pmYou should use the formula for revenue R(x)=p*q.

In this case q is x. So you do R(x)= -1/8 x + 100(x). This results in R(x)= -1/8x^2 + 100x.

To find the max, set this to 0 and complete the square:

0=-1/8(x^2-800x)

0=-1/8(x-400)^2+20000

20,000 is your answer

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