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    Can someone please answer the following problem thoroughly and step by step (show me how exactly i can answer a similar question)

    Given: TC= 100 +60Q-12Q^2+Q^3

    a) Find the equations of TVC, AVC, and MC functions.
    b) Find the level of output at which AVC and MC are minimum, and prove that AVC and MC curves are U-shaped.
    c) Find AVC and MC for the level of output at which the AVC curve is minimum.

    Thank you so much for any help
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    a)
    TC = 100 + 60Q - 12Q^2 + Q^3
    Fixed Cost =100 (unrelated to quantity produced(Q) )
    TVC = 60Q - 12Q^2 + Q^3
    AVC = 60 - 12Q + Q^2 (TVC divided by Q)
    MC = dTVC/dQ = 60 - 24Q + 3Q^2 (TVC differentiated by Q)

    >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >
    b)
    To find minimum differentiate and set to zero

    i)
    AVC = 60 - 12Q + Q^2
    dAVC/dQ = -12 + 2Q = 0
    Q*=6


    ii)
    MC = 60 - 24Q + 3Q^2
    dMC/dQ = -24 +6Q = 0
    Q*=4

    Both functions (AVC and MC) are concave up (ushaped) as their second derivatives (wrt Q) are both positive ( d2AVC/dQ2=2 d2MC/dQ2=6)

    >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >
    c)AVC curve minimum at Q*=6 (see b)i) above )
    put Q=6 into the AVC and MC equations to determine "the level of output at which the AVC curve is minimum."

    AVC = 60 - 12Q + Q^2
    = 60 - 12(6) + (6)^2
    = 60 - 72 + 36
    = 24

    MC = 60 - 24Q + 3Q^2
    = 60 - 24(6) + 3(6)^2
    = 60 - 144 + 3x(36)
    = 24

    Output Quantity = 24

    >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >
    ---

    Worth noting that when AVC is minimised AVC=MC
 
 
 
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