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

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.

question is incomplete.

an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.