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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