document.write( "Question 1146565: The monthly revenue R achieved by selling x wristwatches is figured to be R(x)=71x-0.2x^2. The monthly cost C of selling wristwatches is C(x)=26x+1600.
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document.write( "a) How many wristwatches must the firm sell to maximize revenue? What is the maximum revenue?
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document.write( "b. Profit is given as P(x)=R(x)-C(x). What is a profit function? \n" );
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Algebra.Com's Answer #767865 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! Maximum revenue maximizes -0.2x^2+71x, where a=-0.2 and b=71 \n" ); document.write( "the x-value for this quadratic's maximum if -b/2a=-71/-0.4 or 177.5 watches \n" ); document.write( "177 revenue is \n" ); document.write( "$18832.80 \n" ); document.write( "178 revenue is $6301.20 \n" ); document.write( "177 revenue is $6301.20 \n" ); document.write( "either of those maximizes revenue \n" ); document.write( "profit function is their difference or -0.2x^2+45x-1600=P(x)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |