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
<u>Price Function</u>
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
<u>Cost Function</u>
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
<u>Revenue Function</u>
Revenue is the product of price and quantity:
r(q) = qp
r(q) = (1/2)q(140 -q) . . . . revenue from sale of q items
<u>Profit Function</u>
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
<u>Break-Even Points</u>
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.
<u>Maximum Profit</u>
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
<u>Quantity for Maximum Profit</u>
This was found to be 30 in the previous section.
