document.write( "Question 933553: SOLVE FULLY. DEFINE VARIABLES, CREATE QUADRATIC EQUATION AND ANSWER PROBLEM.
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document.write( "Tickets to a dance cost $4 and the projected attendance is 300 people. For every $0.10 increase in ticket price, the school dance committee predicts attendance will decrease by 5 people. Determine price of ticket that produces greatest revenue. \n" );
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Algebra.Com's Answer #566982 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! Tickets to a dance cost $4 and the projected attendance is 300 people. \n" ); document.write( " For every $0.10 increase in ticket price, the school dance committee predicts attendance will decrease by 5 people. \n" ); document.write( " Determine price of ticket that produces greatest revenue. \n" ); document.write( ": \n" ); document.write( "Let x = the no. of 10 cent increases \n" ); document.write( "Also \n" ); document.write( "Let x = no. of 5 people reductions with each price increase \n" ); document.write( ": \n" ); document.write( "y = revenue for ticket sales \n" ); document.write( "y = (4+.10x)(300-5x) \n" ); document.write( "FOIL \n" ); document.write( "y = 1200 - 20x + 30x - .5x^2 \n" ); document.write( "A quadratic equation \n" ); document.write( "y = -.5x^2 + 10x + 1200 \n" ); document.write( "greatest y (revenue) occurs at the axis of symmetry, x = -b/(2a) \n" ); document.write( "x = \n" ); document.write( "x = +10 price increases for max revenue \n" ); document.write( "therefore the price will be: \n" ); document.write( "4 + 10(.10) = $5 is the price for max revenue \n" ); document.write( ": \n" ); document.write( "No. of tickets then (300 - 10(5)) = 250 people \n" ); document.write( ": \n" ); document.write( "Max revenue: 250 * 5 = $1250 \n" ); document.write( " \n" ); document.write( " |