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Question 49131: I have a word problem that I have been working on that I am not getting.
A manufacturer of compact stereos believes that the revenue the company receives is related to the price (p) of a stereo by the function R(p)= 180p-1/3p^2. What price will give the maximum revenue, and what is the maximum revenue? So what I am wondering is..... Do I just plug in different prices? I have done this, but I just kepp getting higher numbers. When do I know I have the max number? I was hoping it would just finally go into the negative, but it does not. so..... Any ideas?
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! I have a word problem that I have been working on that I am not getting.
A manufacturer of compact stereos believes that the revenue the company receives is related to the price (p) of a stereo by the function R(p)= 180p-1/3p^2. What price will give the maximum revenue, and what is the maximum revenue? So what I am wondering is..... Do I just plug in different prices? I have done this, but I just kepp getting higher numbers. When do I know I have the max number? I was hoping it would just finally go into the negative, but it does not. so..... Any ideas
THERE ARE 2 METHODS..ONE IS BASED ON CALCULUS..BUT I THINK YOU ARE IN SCHOOL.SO LET US DO BY OTHER METHOD USING ALGEBRA.
R=180P-P^2/3...I SUPPOSE THIS IS WHAT YOU MEAN....PLEASE TYPE PROPERLY USING BRACKETS.WHAT YOU TYPED MEANS 1/(3*P^2)...AND NOT (1/3)*P^2...OR P^2/3
R=(-1/3)[P^2-540P]=(-1/3)[(P)^2-2(P)(270)+270^2-270^2]
=(-1/3)[(P-270)^2-270^2]
=(-1/3)(P-270)^2+(1/3)*270*270
R=(-1/3)(P-270)^2+24300=24300-(1/3)(P-270)^2
NOW WE HAVE A PERFECT SQUARE WHICH IS SUBTRACTED FROM 24300.PERFECT SQUARE IS ALWAYS POSITIVE.ITS MINIMUM VALUE COULD BE ZERO.SO WE GET MAXIMUM VALUE FOR
R=24300 WHEN THAT SQUARE IS ZERO...OR...P-270=0...OR...P=270
SO MAXIMUM REVENUE IS
R=24300 WHEN P=270
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