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    Application of Derivatives Maximum Revenues

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    Find the number of units x that produce a maximum revinue R.

    R=800x-0.2x^2

    R=48x^2-0.2x^3

    Find the number of unites x that produce the minimum average cost per unit C.

    C=1.25x^2+25x+8000

    C=0.001x^3+5x+250

    find the amounts of advertising that maximizes the profit P. (s and p are measeured in thousands of dollars.) Find the point of diminishing returns.

    P= -2s^3+35s^2-100s+200

    P=-0.1s^3+6s^2+400

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    Solution Preview

    find the number of units x that produce a maximum revinue R.

    R=800x-0.2x^2
    dR/dx = 800 - 0.4x = 0 implies x = 800/0.4 = 2000

    R=48x^2-0.2x^3
    dR/dx = 96x - 0.6x^2 = 0 implies x = 96/0.6 = ...

    Solution Summary

    The applications of derivatives for maximum revenues are determined.

    $2.49

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