document.write( "Question 54922: I have this problem that I desperately need help with. I am supposed tro use factoring on this but I'm having trouble with it. It goes:\r
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document.write( "Tropical Pools sells an aboveground model for \"p\" dollars each. The monthly revenue from the sale of this product is given by\r
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document.write( "R = -0.08p^2 + 300p\r
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document.write( "Revenue is the product of \"p\" and the demand (quantity sold).\r
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document.write( "a.) Factor out the price on the right-hand side of the formula\r
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document.write( "b.) What is an expression of the monthly demand?\r
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document.write( "c.) What is the monthly demand for this pool when the price is $3000?\r
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document.write( "d.) What is the approximate maimum revenue?\r
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document.write( "Can anyone help me out with this?\r
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document.write( "Thanks \n" );
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Algebra.Com's Answer #37143 by stanbon(75887) ![]() You can put this solution on YOUR website! The monthly revenue from the sale of this product is given by \n" ); document.write( "R = -0.08p^2 + 300p \n" ); document.write( "Revenue is the product of \"p\" and the demand (quantity sold). \n" ); document.write( "a.) Factor out the price on the right-hand side of the formula \n" ); document.write( "R=p(-0.08p+300) \n" ); document.write( "b.) What is an expression of the monthly demand? \n" ); document.write( "D=-0.08p+300 \n" ); document.write( "c.) What is the monthly demand for this pool when the price is $3000? \n" ); document.write( "D(3000)=-0.08(3000)+300 \n" ); document.write( "D(3000)=240+300=540 \n" ); document.write( "d.) What is the approximate maximum revenue? \n" ); document.write( "Maximum at -b/2a where a=-0.08, b=300 \n" ); document.write( "Max at -300/(-0.16)= 1875 \n" ); document.write( "R(1875)= -0.08(1875)^2+300(1875)=$281,250 (Maximum Revenue) \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " \n" ); document.write( " |