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
Found 2 solutions by stanbon, addingup:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The total profit for Timez series VI watches is given by R(x)=1500x-0.02x^2 where x is the number of watches sold. Determine maximum possible revenue and how many watches must be sold to achieve that profit
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R(x)=1500x-0.02x^2
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max occurs when x = -b/(2a) = -1500/(2*-0.02) = 1500/0.04 = 150000/4 = 37500
max revenue = f(37500) = $2.81x10^7
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Cheers,
Stan H.
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Answer by addingup(3677)  (Show Source):
You can put this solution on YOUR website! R(x)=1500x-0.02x^2: In this formula x is the quantity, there is no c. So if your quantity is 1,000:
Revenue for 1,000 watches: R(1,000)
R(1,000)= 1500(1,000)- 0.02(1,000)^2
R(1,000)= 1,500,000- 400= = 1,499,600
With this formula you can answer two questions:
A)find the marginal revenue at a production level of x.
B)find the production levels where the revenue is $n.
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Profit is maximized where price and demand intersect. You set the formula like this:
Quantity of demand (Qd)= Quantity of supply (Qs)
For example:
If the demand function is Qd = 100-20P
and the supply function is Qs = -60 + 50p
set 100-20P= -60+50P
and solve for P
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