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Suppose the cost of producing x items is given by C(x)=1000-x^3, and the revenue made on the sale of x-items is R(x)=100x-10x^2. Find the number of items which serves as a break-even point.

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  1. Break even point is when cost equals revenue (i.e. zero profit).
    So for
    C(x)=R(x), we have
    1000-x^3=100x-10x^2
    Rearrange to give
    -x^3+10x^2-100x+10x^2+1000=0
    which solves easily to
    x=10

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