document.write( "Question 1171990: A bakery has created a new type of donut that is driving the competition out of the market. They are currently
\n" ); document.write( "selling the donuts for $3 each and sell 1400 donuts each month. Market research has shown that for every
\n" ); document.write( "$0.25 increase in price, they sell 70 fewer donuts. Determine the selling price of the donut that will maximize
\n" ); document.write( "revenue AND determine the maximum revenue. \n" ); document.write( "

Algebra.Com's Answer #796886 by ikleyn(52810) $\"\"$ $\"About$

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document.write( "Let n be the number of increments by $0.25.\r\n" );
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document.write( "Then the price for each donut is p(n) = 3 + 0.25n and the number of the sold donutes is  1400-70n.\r\n" );
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document.write( "Hence, the revenue is\r\n" );
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document.write( "    R(n) = p(n)*(1400 - 70n) = (3 + 0.25n)*(1400 - 70n).\r\n" );
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document.write( "It is a quadratic function , presented as the product of the two linear binomial.\r\n" );
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document.write( "The quadratic function has the roots there, where the linear binomials are zer0, i.e. at\r\n" );
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document.write( "      n = -12  and  n= 20.\r\n" );
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document.write( "The maximum of the quadratic function is achieved at the midpoint between its roots, i.e. at   = 4.\r\n" );
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document.write( "So, 4 (four) increases at $0.25 each are required to get the maximum revenue.\r\n" );
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document.write( "Thus the optimum price per donut is  3 + 4*0.25 = 4 dollars.\r\n" );
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document.write( "The optimum sale is then  1400 - 4*70 = 1120 donuts, and the maximum revenue is\r\n" );
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document.write( "    4*1120 = 4480 dollars.\r\n" );
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document.write( "Compare it with the starting revenue of  3*1400 = 4200 dollars.\r\n" );
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