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Thread: Cost , Maximum Revenue, Profit - Help needed with formula required

  1. #1
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    Question Cost , Maximum Revenue, Profit - Help needed with formula required

    -Fixed costs are given at $1000.
    -Variable cost per unit is given by (x+25)
    -Selling Price per unit is given by (250-x)

    a) determine break even points
    b) find maximum revenue
    c) Find maximum profit

    Total Cost I believe is Variable Cost(x) + Fixed Cost = (x+25)x + 1000 ; Where I am getting confused is the wording of the problem.
    So instead should Total Cost be x+25+1000 & linear. I suspect my functions would have to be quadratic as some point since that's what the
    current topic is in school.

    To find break even point do I have to say Profit which is (Selling price(x)Units )-(total cost)(x)Unit) = 0
    would this be ((250-x)x)-((250-x)+1000) = 0 ; then solve for x? - since asks for B.Even Points ie plural.

    Maximum revenue is the Vertex of the (250-x)x function?

    Thanks for any insights.
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  2. #2
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    Re: Cost , Maximum Revenue, Profit - Help needed with formula required

    Quote Originally Posted by Njords View Post
    -Fixed costs are given at $1000.
    -Variable cost per unit is given by (x+25)
    -Selling Price per unit is given by (250-x)

    a) determine break even points
    b) find maximum revenue
    c) Find maximum profit

    Total Cost I believe is Variable Cost(x) + Fixed Cost = (x+25)x + 1000 ; Where I am getting confused is the wording of the problem.
    So instead should Total Cost be x+25+1000 & linear. I suspect my functions would have to be quadratic as some point since that's what the
    current topic is in school.

    To find break even point do I have to say Profit which is (Selling price(x)Units )-(total cost)(x)Unit) = 0
    would this be ((250-x)x)-((250-x)+1000) = 0 ; then solve for x? - since asks for B.Even Points ie plural.

    Maximum revenue is the Vertex of the (250-x)x function?

    Thanks for any insights.
    For Total Cost, yes. $(x+25)x+1000$ does not equal $x+25+1000$. You dropped an $x$.

    $(x+25)x+1000 = x^2+25x+1000$

    Then, the break even point is when $(250-x)x-(x^2+25x+1000) = 0$ which becomes $225x-2x^2-1000 = 0$ or $2x^2-225x+1000 = 0$

    For (b), the maximum revenue is as you said.
    For (c), the maximum profit is the maximum of $225x-2x^2-1000$ (the formula you used for the break-even points, but not setting it equal to zero).

    Do you know how to complete the square?
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  3. #3
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    Re: Cost , Maximum Revenue, Profit - Help needed with formula required

    Hello Thanks for the reply... I'm all good with completing the square.

    Appreciate the help.
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