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Math Help - Econometrics: MR, Revenue, profit.

  1. #1
    Member princess_anna57's Avatar
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    Cool Econometrics: MR, Revenue, profit.

    The cost (in 000's) of producing x thousands of loudspeaker systems is TC = x^3 + 3x^2 - 10x + 2. Price (p) and quantity demanded (x) which are both required to be non-negative are related by p = 130 - 2x. Marginal Cost (MC) = 3x^2 + 6x - 10.

    a) Find Marginal Revenue (MR) as a function of output
    b) What quantity maximises revenue, and what is the corresponding revenue?
    c) What quantity maximises profit, and what is the corresponding profit?
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  2. #2
    Member
    Joined
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    Hello!
    a) MR = \frac{\mbox{d}TR}{\mbox{d}x} = \frac{\mbox{d}}{\mbox{d}x}(P \cdot x) = \frac{\mbox{d}P}{\mbox{d}x} \cdot x +P \cdot \frac{\mbox{d}x}{\mbox{d}x} = \frac{\mbox{d}P}{\mbox{d}x} \cdot x +P

    Since P = 130 - 2x,  \frac{\mbox{d}P}{\mbox{d}x}=-2 so

    MR = -2x + 130-2x = 130 -4x.

    b) TR = P \cdot x = (130-2x)x = -2x^2+130x
    To find the quantity that maximizes revenue TR, set \frac{\mbox{d}TR}{\mbox{d}x} = 0. This gives -4x+130 = 0, calculate x and use it to calculate the corresponding revenue TR. Check that it really is a maximum.

    c) Profit is equal to total revenue TR minus total cost TC. To find the maximum profit with respect to quantity sold x, set

    \frac{\mbox{d}}{\mbox{d}x}(TR-TC)=0. You get

    \frac{\mbox{d}}{\mbox{d}x}\left((-2x^2+130x)-(x^3 + 3x^2 - 10x + 2)\right) =

    = \frac{\mbox{d}}{\mbox{d}x}(-x^3-5x^2+140x-2) = -3x^2-10x+140 =0, find x (discard negative solution)

    and use it to calculate corresponding profit TR-TC. Check that it really is a maximum.

    Maybe I forgot to multiply by 1000 somewhere but I hope you get the idea.
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