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ivano_logy

  • 2 years ago

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?

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  1. jayz657
    • 2 years ago
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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

  2. Algebraic!
    • 2 years ago
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    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)

  3. apple_pi
    • 2 years ago
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    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]

  4. ivano_logy
    • 2 years ago
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    Thank you very much! I finally understand it.

  5. apple_pi
    • 2 years ago
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    that's great :)

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