document.write( "Question 1151586: Determine the maximum sales and optimal price.
\n" ); document.write( "a) Bob can sell 200 cars at $20 000 each but finds that every $1000 increase in price causes a 5-car drop in sales
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Algebra.Com's Answer #773389 by ikleyn(52781) $\"\"$ $\"About$

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document.write( "According to the condition, the number of sold cars \"n\" is THIS function of the price \"p\" per single car\r\n" );
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document.write( "    n(p) =  = 200 - 0.005*(p-20000) = -0.005p + 300.     (1)\r\n" );
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document.write( "The revenue ( = the sales) is then the product  R(p) = p*n(p)\r\n" );
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document.write( "    R(p) = p*n(p) = p*(-0.005p + 300) = - 0.005p^2 + 300p.    (2)\r\n" );
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document.write( "Thus the revenue is this quadratic function (2) of the price.\r\n" );
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document.write( "It is well known fact that the general form quadratic function  y = ax^2 + bx + c\r\n" );
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document.write( "with the negative leading coefficient \"a\" has the maximum at  x = .\r\n" );
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document.write( "In our case,  a= -0.005,  b= 300.  Therefore, the quadratic function (2) has maximal value at  \r\n" );
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document.write( "    q =  =  = 30000.\r\n" );
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document.write( "So, the optimal price is $30000.\r\n" );
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document.write( "Then the number of the sold cars will be only  n(30000) = -0.005*30000 + 300 = 150,\r\n" );
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document.write( "but the revenue will be  R = 150*30000 = 4,500,000  against  200*20000 = 4,000,000  at the price of $20000.\r\n" );
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