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Yesterday, 16:17

Rite cut riding lawnmower obeys the demand equation p=-1/20x+1060. The cost of producing x lawnmowers is given by the function c (x) = 120x+5000

a) Express the revenue R as a function of x

b) Express the profit P as a function of X

c) Fine the value of x that maximizes profit. What is the maximum profit?

d) What price should be charged to maximize profit? ... ?

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  1. Yesterday, 18:52
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    A.) Revenue = price * quantity = px = - 1/20x^2 + 1060x

    R (x) = - 1/20x^2 + 1060x.

    b.) Profit = Revenue - Cost = R (x) - C (x) = - 1/20x^2 + 1060x - 120x - 5000

    P (x) = - 1/20x^2 + 940x - 5000

    c.) For maximum profit, dP/dx = 0

    -1/10x + 940 = 0

    1/10x = 940

    x = 940 * 10 = 9,400

    x = 9,400

    Maximum profit = P (9400) = - 1/20 (9400) ^2 + 940 (9400) - 5000 = $4,413,000

    d.) The price to be charged for maximum profit = - 1/20 (9400) + 1060 = $590
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