document.write( "Question 1104103: An ice cream company finds that at a price of $3.00 demand is 3500 units. For every $0.10 decrease in price, demand increases by 50 units. Find the price and quantity sold that maximizes revenue.\r
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Algebra.Com's Answer #718832 by josgarithmetic(39617) $\"\"$ $\"About$

You can put this solution on YOUR website!
You can make a data table to help find the price and demand, and form your revenue function; $\"price%2Aunits=revenue\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Let n be how many increments of $0.10.
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document.write( "PRICE per UNIT     Demand in UNITS\r\n" );
document.write( "   3                        3500\r\n" );
document.write( "   3-0.1                  3500+50\r\n" );
document.write( "   3-2*(0.1)              3500+2(50)\r\n" );
document.write( "   .\r\n" );
document.write( "   .\r\n" );
document.write( "   3-0.1n                3500+50n\r\n" );
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\n" ); document.write( "\n" ); document.write( " $\"highlight%28R%28n%29=%283500%2B50n%29%283-0.1n%29%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "You can leave R in its factored form. R is a parabolic function and has a maximum point. Find the ROOTS for R, and the maximum will happen exactly in the middle of the two values of n.\r
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\n" ); document.write( "\n" ); document.write( " $\"%283500%2B50n%29%283-0.1n%29=0\"$ -------find the two n values and identify the value exactly in the middle.\r
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\n" ); document.write( "\n" ); document.write( "Do that and then finish the rest. \n" ); document.write( "

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