document.write( "Question 1008705: The total profit fore Timez series VI watches is given by R(x)=1500x-0.02x^2 where c is the number of watches sold. Determine maximum possible revenue and how many watches must be sold to achieve that profit \n" ); document.write( "

Algebra.Com's Answer #624291 by addingup(3677) $\"\"$ $\"About$

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R(x)=1500x-0.02x^2: In this formula x is the quantity, there is no c. So if your quantity is 1,000:
\n" ); document.write( "Revenue for 1,000 watches: R(1,000)
\n" ); document.write( "R(1,000)= 1500(1,000)- 0.02(1,000)^2
\n" ); document.write( "R(1,000)= 1,500,000- 400= = 1,499,600
\n" ); document.write( "With this formula you can answer two questions:
\n" ); document.write( "A)find the marginal revenue at a production level of x.
\n" ); document.write( "B)find the production levels where the revenue is $n.
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\n" ); document.write( "Profit is maximized where price and demand intersect. You set the formula like this:
\n" ); document.write( "Quantity of demand (Qd)= Quantity of supply (Qs)
\n" ); document.write( "For example:
\n" ); document.write( "If the demand function is Qd = 100-20P
\n" ); document.write( "and the supply function is Qs = -60 + 50p
\n" ); document.write( "set 100-20P= -60+50P
\n" ); document.write( "and solve for P
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