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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 number of sales will increase by 10 units for each $0.10 decrease in price. The company has fixed costs of $250. The cost to make each unit is $4.20. Find the level of production that maximizes the company's profit.

The Should Produce _________
Units at a price of $__________
Which will yield a profit of $_______

Sagot :

sqdancefan

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

__

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

_____

Additional comment

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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