Çözüldü Maximization of Profit - Differentiation

Konusu 'TOEFL - IELTS - SAT - ACT - GRE - GMAT Hazırlık' forumundadır ve Honore tarafından 28 Ocak 2021 başlatılmıştır.

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  1. Honore

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    Test adaptation of a problem from the University of New Brunswick:

    A factory makes TV sets and the selling price is $1,000 each. If the manufacturing cost of x sets is $10,000 + 2x + .01x^2, how many sets should they make to maximize the profit?
    A) 52100
    B) 49900
    C) 42500
    D) 39200
    E) 35750

    Profit = Price - Cost
    Profit = P(x) = 1000x - 10,000 - 2x - .01x^2 = -0.01x^2 + 998x - 10000
    dP(x) / dx = -0.02x + 998 = 0 ⇒ x = 49900

    http://www.math.unb.ca/exams/x1823-99/ (7th Question)
    http://www.math.unb.ca/exams/1823-99-ans/ (Answers)
    Honore, 28 Ocak 2021
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    2. Benzer Konular: Maximization Profit
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