document.write( "Question 1171901: The parent council plans to run an annual student talent show to raise money for new drama equipment. Last year, tickets sold for $11 each and 400 people attended. In order to raise more money, the council determined that for every dollar increase in price, attendance would drop by 20 people. What ticket price would maximize revenue and what is the maximum revenue?
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Algebra.Com's Answer #796802 by Boreal(15235)\"\" \"About 
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x=each $1 and each 20 people
\n" ); document.write( "(11+x)(400-20x) product
\n" ); document.write( "find the maximum
\n" ); document.write( "-20x^2+180x+4400
\n" ); document.write( "maximum is where x=-b/2a, the vertex. That is -180/-40 or 4.5\r
\n" ); document.write( "\n" ); document.write( "$15.50 price and have 310 people (4.5*-20)
\n" ); document.write( "=$4805
\n" ); document.write( "always check on both sides
\n" ); document.write( "$16 and 300 people $4800
\n" ); document.write( "$15 and 320 people $4800
\n" ); document.write( "\"graph%28300%2C300%2C-20%2C30%2C-250%2C5000%2C4800%2C%2811%2Bx%29%28400-20x%29%29\"
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