document.write( "Question 102423: Solve the following applications involving functions.\r
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document.write( "If the inventor in exercise 53 charges $4 per unit, then her
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document.write( "profit for producing and selling x units is given by the function\r
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document.write( "P(x)=2.25x-7000\r
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document.write( "(a) What is her profit if she sells 2000 units?
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document.write( "(b) What is her profit if she sells 5000 units?
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document.write( "(c) What is the break-even point for sales? \n" );
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Algebra.Com's Answer #74479 by JP(22)![]() ![]() You can put this solution on YOUR website! If the function is P(x)=2.25x-7000\r \n" ); document.write( "\n" ); document.write( "A.) For every instance of x, you must plug in 2000 units since x is equal to the units sold. Therefore, P(2000)=2.25(2000)-7000 \n" ); document.write( "2.25(2000)=4500-7000=-$2500 profit.\r \n" ); document.write( "\n" ); document.write( "B.)For every instance of x, you must plug in 5000 units since x is equal to the units sold. Therefore, p(5000)=2.25(5000)-7000 \n" ); document.write( "2.25(5000)=11250, then subtract 7000 to get a profit of $4250\r \n" ); document.write( "\n" ); document.write( "C.)To get the break-even point you must set the units to x. \n" ); document.write( "Your equation is P(x)=2.25(x)-7000=0 \n" ); document.write( "2.25(x)=2.25x-7000=0 Then add 7000 to the other side. \n" ); document.write( "2.25x=7000 divide now to get the x variable alone \n" ); document.write( "x=7000/2.25 \n" ); document.write( "x=$3111.11 \n" ); document.write( "Plug it into the original function to verify your work. \n" ); document.write( "P(3111.11)=2.25(3111.11)-7000 \n" ); document.write( "2.25(3111.11)=7000 then subtract 7000 \n" ); document.write( "7000-7000=0 or your break-even point\r \n" ); document.write( "\n" ); document.write( "I hope this helps...however, please verify that the function p(x)=2.25x-7000 is the correct function given by your book. If you have any further questions let me know. \n" ); document.write( " |