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Math Help - Math profit question

  1. #1
    needshelp
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    Math profit question

    My question is:

    A 135-kg steer gains 3.5kg/day and costs 80cents/day to keep. The market price for beef cattle is $1.65/kg, but the price falls by 1cent/day. When should the steer be sold to maximize profit.

    help is greatly appreciated

    the answer is susposed to be 52 days, but i can't figure out how to get that.
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  2. #2
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Profit = Income minus Expenses

    Income = (Total weight of steer, in kg) times (unit price, in $/kg)
    Let x = number of days from now when the steer is sold -------------------***
    The total weight after x days = (135 +3.5*x) kg
    The unit price per kg, after x days = $1.65 -$0.01*x = (1.65 -0.01x) dollars

    The expenses after x days = $0.80*x = 0.80x dollars

    So,
    Profit = Income minus expenses
    P = (135 +3.5x)(1.65 -0.01x) -0.80x
    P = (135*1.65 -135*0.01x +3.5x*1.65 -3.5x*0.01x) -0.80x
    P = 222.75 -1.35x +5.775x -0.035x^2 -0.80x
    P = 222.75 +3.625x -0.035x^2 ---------------(i)

    Now here, we can solve for maximum P by
    a) using the properties of a parabola (because Eq.(i) is a vertical parabola that opens downward and so its vertex is its highest point which gives the maximum P.)
    or,b) using Calculus. P is maximum or minimum when dP/dx = 0.

    Let me assume you know Calculus, so,
    Differentiate both sides of (i) with respect to x,
    dP/dx = 3.625 -(0.035)[2x]
    dP/dx = 3.625 -0.070x
    Set that to zero,
    0 = 3.3625 -0.07x
    0.07x = 3.625
    x = 3.625/0.07
    x = 51.7875
    or, x = 52 days ---------answer.
    Last edited by ticbol; October 12th 2005 at 01:30 AM.
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