document.write( "Question 871669: A bus company has 4000 passengers daily, each paying a fare of $2. For each $0.15 increase, the company estimates that it will lose 40 passengers per day.
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document.write( "- What fare will maximize the revenue? How many passengers will ride at this price?\r
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document.write( "I'm not sure if it's correct but I calculated 43.3 to be the price increases would maximize the revenue. What I did was plugged the 43.3 into the equation (2+0.15x) which is the cost of each item and (4000-40x)but i'm not sure if 8.50 is right for the fare and 2264 people will ride that day. Can you just check if I've done it right? \n" );
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Algebra.Com's Answer #525774 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! A bus company has 4000 passengers daily, each paying a fare of $2. \n" ); document.write( " For each $0.15 increase, the company estimates that it will lose 40 passengers per day. \n" ); document.write( "- What fare will maximize the revenue? \n" ); document.write( " How many passengers will ride at this price? \n" ); document.write( ": \n" ); document.write( "I think you have this right, using (4000-40x)(2-.15x), equation I got an axis \n" ); document.write( "of symmetry of 43.3 also, A fare of $8.50, but I had 2268 passengers \n" ); document.write( "A revenue of 8.5 * 2268 = $1978 \n" ); document.write( ": \n" ); document.write( "But you can prove this to yourself, find the revenue when the fare is .15 less with 40 more passengers. \n" ); document.write( " 8.35 * 2308 = $1971.80, slightly less \n" ); document.write( "and \n" ); document.write( "With a fare .15 more and 40 passengers less: 8.65 * 2228 = $1972.20, less also \n" ); document.write( "You obviously have the maximum \n" ); document.write( " |