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An electronics store sells an average of 60 entertainment systems per month at an average of $800 more than the cost price. For every $20 increase in the selling price, the store sells one fewer system. What amount over the cost price will maximize revenue?

Mathematics
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c = cost price s = selling price n = number of systems sn = total revenue sn - c = income/profit/loss s = (c+800) n = 60 c? if s goes up 20 n = n -1 sorry this is all i got so far, hope it helps in any way
Q = 60 + (800-P)/20 R= QP = (100 -P/20)P so find the max of R = -P^2 /20 +100P and find the Price at that max... (set the derivative equal to zero and solve for P)
First thing you want is number sold in terms of price sold at lets call number sold n, and price sold p \[n = 60-\frac{p-800}{20}\] \[n = 60+\frac{800-p}{20}\] revenue (r) = number sold x price sold \[r = p(60+\frac{800-p}{20})\] \[r = 60p+40p-\frac{p^2}{20}\] \[r = 100p - \frac{p^2}{20}\] To find the max, we find the stationary point (we know since this is a quadratic, and that the coefficient of p^2 is negative the max is the stationary point) \[\frac{dr}{dp} = 100 -\frac{2p}{20}\] \[\frac{dr}{dp} = 100 -\frac{p}{10}\] \[0 = 100 - \frac{p}{10}\] \[0 = 1000 - p] \[p = 1000]

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Thank you very much! I finally understand it.
that's great :)

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