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28 April, 13:20

An electronics store sells 70 digital cameras per month at a price of $320 each. For each $20 decrease in price, about 5 more cameras per month are sold. Use the verbal model and quadratic function to determine how much the store should charge per camera to maximize monthly revenue.

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  1. 28 April, 13:33
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    the price to maximize revenue monthly = $300

    Step-by-step explanation:

    From the model revenue equation, we have;

    Revenue = Price x Sales

    Let R (x) represent revenue

    Thus;

    R (x) = (320 - 20x) * (70 + 5x)

    R (x) = - 100x² + 200x + 22400

    Since the coefficient of x² is negative, it will be a downward opening parabola and thus the vertex will be at a maximum. Hence, we can use axis of symmetry formula to find x. Which is;

    x = - b/2a

    x = - 200 / (-2 x 100)

    x = - 200/-200

    x = 1

    Thus the price to maximize revenue monthly will be gotten by plugging in 1 for x into (320 - 20x)

    So, Price = (320 - 20 (1))

    Price = $300
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