RE: #3 on the Quiz from 11/6/06
Since we got the quizzes back yesterday, I think I'm allowed to post a solution now...
For number 3, we are told that the number of "widgets" purchased is n(x)=200-(x^2/100), and that the price per widget is $x. Since revenue is equal to the number of products sold times the price per product, the revenue here can be given by the equation r(x)=x*(200-(x^2/100)). To find the price that results un the maximum possible revunue, you graph y1=x*(200-(x^2/100)) and find the maximum on the graph. The maximum is at (81.6497, 10886.6), so this means the maximum revenue is $10,886.60 and that the price per widget should be $81.65 (we only had to include the second part to answer the question).
I hope this helps.
For number 3, we are told that the number of "widgets" purchased is n(x)=200-(x^2/100), and that the price per widget is $x. Since revenue is equal to the number of products sold times the price per product, the revenue here can be given by the equation r(x)=x*(200-(x^2/100)). To find the price that results un the maximum possible revunue, you graph y1=x*(200-(x^2/100)) and find the maximum on the graph. The maximum is at (81.6497, 10886.6), so this means the maximum revenue is $10,886.60 and that the price per widget should be $81.65 (we only had to include the second part to answer the question).
I hope this helps.
posted by Casey Blue at 1:18 PM
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