document.write( "Question 1157181: Total profit P is the difference between total revenue R and total cost C. Given the following total-revenue and total-cost functions, find the total profit, the maximum value of the total profit, and the value of x at which it occurs.\r
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document.write( "Upper R left parenthesis x right parenthesis equals 1200 x minus x squared
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document.write( "R(x)=1200x−x2,
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document.write( "Upper C left parenthesis x right parenthesis equals 3000 plus 20 x
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document.write( "C(x)=3000+20x \n" );
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Algebra.Com's Answer #780008 by Theo(13342)![]() ![]() You can put this solution on YOUR website! r(x) = 1200x - x^2 \n" ); document.write( "c(x) = 3000 + 20x \n" ); document.write( "p(x) = r(x) - c(x) \n" ); document.write( "that becomes: \n" ); document.write( "p(x) = 1200x - x^2 - (3000 + 20x) \n" ); document.write( "simplify to get: \n" ); document.write( "p(x) = 1200x - x^2 - 3000 - 20x \n" ); document.write( "combine like terms to get: \n" ); document.write( "p(x) = -3000 + 1180x - x^2 \n" ); document.write( "order the terms in descending order of degree to get: \n" ); document.write( "p(x) = -x^2 + 1180x - 3000 \n" ); document.write( "a = the coefficient of the x^2 term = -1 \n" ); document.write( "b = the coefficient of the x term = 1180 \n" ); document.write( "c = the constant term = -3000 \n" ); document.write( "the maxim profit is when x = -b/2a. \n" ); document.write( "that becomes x = -1180/-2 = 590. \n" ); document.write( "the maximum profit is the value of the equation at x = 590 \n" ); document.write( "that becomes p(x) = 345,100 \n" ); document.write( "that's your solution. \n" ); document.write( "here's the graph that confirms that. \n" ); document.write( " ![]() \n" ); document.write( " \n" ); document.write( " |