document.write( "Question 167650: I really do not understand this problem at all... help, please! ---use the fact that revenue = price*quantity to solve the problem. the price (p) in dollars and the quantity (x) sold of a certain product are described by the following: p=-1/6x+100, 0<=x<=600.
\n" ); document.write( "a)express the revenue R as a function of x.
\n" ); document.write( "b)what quantity x maximizes the revenue?
\n" ); document.write( "c)what price should the company charge for the maximum revenue?
\n" ); document.write( "

Algebra.Com's Answer #123555 by stanbon(75887)\"\" \"About 
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use the fact that revenue = price*quantity to solve the problem. the price (p) in dollars and the quantity (x) sold of a certain product are described by the following: p=-1/6x+100, 0<=x<=600.
\n" ); document.write( "a)express the revenue R as a function of x.
\n" ); document.write( "R(x) = x[(-1/6)x + 100] = (-1/6)x^2 + 100x
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\n" ); document.write( "b)what quantity x maximizes the revenue?
\n" ); document.write( "max occurs when x = -b/2a = -100/(2(-1/6)) = 300
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\n" ); document.write( "c)what price should the company charge for the maximum revenue?
\n" ); document.write( "P(300) = (-1/6)(300) + 100
\n" ); document.write( "p(300) = -50 + 100 = $50
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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