document.write( "Question 1140883: 12. A tool rental company determines that it will rent 500 jackhammers per day at a daily rental fee of $30 per jackhammer. For each $1 increase in rental price, 10 fewer jackhammers will be rented. What rental price maximizes revenue? \n" ); document.write( "
Algebra.Com's Answer #761395 by Boreal(15235)\"\" \"About 
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the revenue function is (500-10x)(30+x)=15000-300x+500x-10x^2
\n" ); document.write( "=-10x^2+200x-15000
\n" ); document.write( "this has a maximum at x=-b/2a=-200/-20 or 10
\n" ); document.write( "when the price is $40, the number is 400, and the revenue is $16,000
\n" ); document.write( "check with $41 and 390 ($15,990) and 39 and 410 (same)\r
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