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AP Calculus: Find the number of units x that produce a maximum revenue R. R=600x^2 - .02x^3
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x=20000 to maximize set the first derivative equal zero.. then you can solve it easily
1200x-.06x^2=0
so how did u get 20000
x(1200-0.06x)=0 then u set both part equal zero so u have the minimum and maximum for x... x=0 and 1200-0.06x=0 ... 1200/0.06=x
ty can u help me with another problem?
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yes i can try
k ima open it in a new question
ok
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