document.write( "Question 453695: an art museum charges $10 for an adult admission. The museum estimates they will lose 15 adult visitors per day for each one dollar increase in price. If the museum averages 300 adult visitors a day, which is the most profitable adult admission price? \n" ); document.write( "
Algebra.Com's Answer #311695 by htmentor(1343)\"\" \"About 
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an art museum charges $10 for an adult admission. The museum estimates they will lose 15 adult visitors per day for each one dollar increase in price. If the museum averages 300 adult visitors a day, which is the most profitable adult admission price?\r
\n" ); document.write( "\n" ); document.write( "Let x = the admission price
\n" ); document.write( "Then the number of visitors will be 300 - 15(x-10) [300 for x=10, 285 for x=11, etc.]
\n" ); document.write( "The total revenue will be:
\n" ); document.write( "R = x(300 - 15(x-10))
\n" ); document.write( "R = 300x - 15x^2 + 150x
\n" ); document.write( "R = 450x - 15x^2
\n" ); document.write( "To maximize the profit, we set dR/dx = 0 and solve for x:
\n" ); document.write( "dR/dx = 0 = 450 - 30x
\n" ); document.write( "This gives x = 15
\n" ); document.write( "So the most profitable price is $15
\n" ); document.write( "The graph of the revenue function is below:
\n" ); document.write( "\"graph%28300%2C300%2C-20%2C20%2C-200%2C3600%2C450x-15x%5E2%29\"
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