Question

When a company produces 10 items, it charges a price of $65 per product. When it produces 20 items, it charges a price of $60 per product. The cost to produce 10 items is $650, and the cost to produce 20 items is $1050. Based on this information, find the following:
price function

cost function

revenue function

profit function

break even points ,

maximum profit

number of products to produce maximum profit

Answer 1

Answer:

• price function: p = 70-q/2 . . . . price per item
• cost function: c = 40q +250 . . . . cost of q items
• revenue function: r = q/2(140 -q) . . . . revenue from q items
• profit function: P = 1/2(50-q)(q-10) . . . . profit from q items
• break even points: (p, q) = (65, 10), (45, 50)
• maximum profit: $200
• quantity for maximum profit: 30

Step-by-step explanation:

For these we have to assume the price and cost functions are linear.

Let p, c, r, P, q represent price, cost, revenue, Profit, and quantity (of items), respectively. The 2-point form of the equation for a line is ...

  y = (y2 -y1)/(x2 -x1)(x -x1) +y1

Price Function

Using the two-point form for price, we get ...

  p = (60 -65)/(20 -10)(q -10) +65 = -5/10(q -10) +65

  p = (-1/2)q +70 . . . . price per item

Cost Function

Using the two-point form for cost, we get ...

  c = (1050 -650)/(20 -10)(q -10) +650 = 40(q -10) +650

  c = 40q +250 . . . . cost for q items

Revenue Function

Revenue is the product of price and quantity:

  r(q) = qp

  r(q) = (1/2)q(140 -q) . . . . revenue from sale of q items

Profit Function

Profit is the difference between revenue and cost.

  P(q) = r(q) -c = 1/2q(140 -q) -(40q +250)

  P(q) = -1/2q^2 +30q -250

  P(q) = (-1/2)(q -10)(q -50) . . . . factored form

Break-Even Points

The profit function will be zero when its factors are zero, at q=10 and q=50. The price function tells us the corresponding prices are $65 and $45 per item, respectively.

Maximum Profit

The profit function is a maximum at the quantity halfway between the break-even points. There, q = (10+50)/2 = 30, and P(30) is ...

  P(30) = -1/2(30-10)(30-50) = 1/2(20^2) = 200 . . . . dollars

Quantity for Maximum Profit

This was found to be 30 in the previous section.