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rodriguez.gaddiel

Joined: Sep 20, 2010
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HELP DUE TODAY!! 85x((199-.03x)^(1/2))

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P=85x((199-.03x)^(1/2)) where x is the number of items demanded. What should the number of items be to maximize producers total revenue.

Number of items demanded to maximize revenue = ???

I've been tryingto solve this for hours and no luck, please help

[Sep 23, 2011 8:19:08 PM]

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pskinner

Joined: Apr 2, 2005
Posts: 797
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Re: HELP DUE TODAY!! 85x((199-.03x)^(1/2))

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If you plot 85x((199-.03x)^(1/2)) using our function graphing calculator, you will find that x=4422.22 yields a maximum. This occurs by taking the derivative of P with respect to x, setting it equal to 0, and solving for x. Substituting this value of x back into P, you will find that the maximum revenue is 3.06144x(10^6). This latter value is also given in the solution steps provided with the function graph.
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Principal Skinner

[Sep 24, 2011 10:39:34 AM]

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webhis

Joined: Oct 19, 2011
Posts: 1
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Re: HELP DUE TODAY!! 85x((199-.03x)^(1/2))

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Long time no calculations, and really a bit difficult ah.
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So can we purchase wow gold from them with cheaper wow po prices?

[Oct 20, 2011 3:10:53 AM]

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