A company estimates that it can sell 3,000 units each month of its product if it prices each unit at $80. However, its monthly n
1 answer:
9514 1404 393
Answer:
- 5290 units
- $57.10 per unit
- $279,591 profit
Step-by-step explanation:
For price p, the quantity q of units sold is said to be ...
q = 3000 +10(80-p)/0.10 . . . . for price in dollars
Then the revenue will be ...
r(p) = p·q = p(3000 +100(80-p)) = 100p(110 -p)
The cost is said to be ...
c(q) = 250 +4.20q
c(p) = 250 +4.20(100(110 -p)) = 250 +420(110 -p)
Then the profit function is ...
P(p) = r(p) -c(p)
P(p) = 100p(110 -p) -(250 +420(110 -p)) = (110 -p)(100p -420) -250
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Ignoring the vertical offset of -250, the quadratic has zeros at 110 and 4.2. So, its line of symmetry (and vertex) will be found at ...
p = (110 +4.2)/2 = 57.1 . . . . dollars per unit
The quantity that should be produced at that price is ...
q(57.1) = 100(110 -57.1) = 5290 . . . . units produced
The profit at that price is ...
P(57.1) = (110 -57.1)(100·57.1 -420) -250 = 52.9(5290) -250
= $279,591 . . . . best profit
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<em>Additional comment</em>
After you have seen a few of these, you can write down the revenue function based on the rate of change of units sold.
If p is the price of a unit, then 80-p is the number of dollars of decrease in price for a price of p. Then (80 -p)/0.10 is the number of 10¢ decreases in price. The increase in units sold is 10 times this, or 10(80 -p)/0.10 = 100(80 -p). When that is added to 3000 units sold, the total is ...
3000 +100(80 -p) = 3000 +8000 -100p
= 11000 -100p = 100(110 -p) . . . . . quantity sold at price p
Revenue is the product of price times the number of units sold at that price.
Profit is the difference between revenue and cost.
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