# 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 Summary

The applications of derivatives for maximum revenues are determined.

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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 = ...

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