document.write( "Question 187637This question is from textbook mathematical analysis
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document.write( ": The demand function for an office supply company's line of plasic rulers is p= 0.45- 0.00045q, where p is the price (in dollars) per unit when q units are demanded(per day) by consumers. Find the level of production that will minimize the mufacture's total revenue, and determine this revenue.
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document.write( "Please with full details, thanks \n" );
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Algebra.Com's Answer #140751 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! The demand function for an office supply company's line of plastic rulers is \n" ); document.write( " p = 0.45 - 0.00045q, where p is the price (in dollars) per unit when q units \n" ); document.write( " are demanded(per day) by consumers. \n" ); document.write( "I think it should be: \n" ); document.write( " Find the level of production that will maximize the manufacturer's total \n" ); document.write( " revenue, and determine this revenue. \n" ); document.write( ": \n" ); document.write( "p = 0.45 - 0.00045q \n" ); document.write( ": \n" ); document.write( "Level of production = q \n" ); document.write( ": \n" ); document.write( "Revenue = quantity * price \n" ); document.write( "r = q * p \n" ); document.write( "; \n" ); document.write( "Replace p with (.45-.00045q) in the above equation and you have \n" ); document.write( "r = q * (.45-.00045q) \n" ); document.write( "r = .45q - .00045q^2 \n" ); document.write( "Arrange as a quadratic equation: \n" ); document.write( "r = -.00045q^2 + .45q \n" ); document.write( ": \n" ); document.write( "We can find the value of q which gives max amt by using the eq: x = -b/(2a) \n" ); document.write( "in this equation that would be \n" ); document.write( "q = \n" ); document.write( "q = \n" ); document.write( "q = +500 units need to be produced for max revenue \n" ); document.write( ": \n" ); document.write( "Then it says,\"and determine this revenue\" \n" ); document.write( "Substitute 500 for q in the revenue equation (r = -.00045q^2 + .45q) \n" ); document.write( "r = -.00045(500^2) + .45(500) \n" ); document.write( "r = -.00045(250000) + 225 \n" ); document.write( "r = -112.5 + 225 \n" ); document.write( "r = $112.5 is the max revenue and occurs when you produce 500 units \n" ); document.write( ": \n" ); document.write( "Did this make sense to you, any questions? \n" ); document.write( " \n" ); document.write( " |