The Apex Widget Company has determined that its marginal cost function is C,(z)

Question

The Apex Widget Company has determined that its marginal cost function is C,(z) 0.5 50, where x represents the number of boxes of widgets which can be produced each week and C is in dollars. The cost of producing 200 boxes of widgets is $22, 500 Also, if they charge 850 per box, they will sell 350 boxes; and if they charge S75 per box they will sell 300 boxes. Assume this demand equation is linear 1. Find: (Assume that a box of widgets can be opened and divided before being shipped to a customer and that there are 120 widgets in a box.) (a) the demand equation. (b) the revenue function. (c) the number of boxes which they should sell to maximize revenue. (d) the maximum revenue. (e) the (total) cost function. (f) the average cost function. (g) the production level (number of boxes) which yields minimum average cost (h) the minimum average cost (i) the profit function (G) the number of boxes which they should produce and sell to maximize profit (k) the maximum profit (1) the price which should be charged in order to maximize the profit. 2. Suppose that boxes of widgets cannot be opened and divided before being shipped to customers. Which of your answers to the questions in Part 1. will be changed? Explain how. Find what the new answers to these questions will be.

Explanation / Answer

Hi :)

For part a:

D(50) = 350

D(75) = 300
Linear (y = sale price)
D(y) = ay + b
a = solope=(350-300)/(50-75)

a=-2
350=-2(50)+b

b=450

Demand equation:

D(y) = 450 - 2y

b) Revenue would be yD(y)
R(y) = 450y - 2y^2
c) To maximize
R'(y) = 450 - 4y=0
y=112.5 boxes

we reach the max revenue at 112.5 boxes so:

d) Max revenue + 450(112.5) - 2(112.5)^2
Max revenue = 25,313

e) So if C'(x) = 0.5x + 50 then you integrate to get C(x) (total cost function)
C(x) = 0.25x^2 + 50x + c
C(200) = 22,500 = 0.25(200)^2 + 50(200) + c
c = 2500
C(x) = 0.25x^2 + 50x + 2500

f) Average cost would be C(x)/x
AvgC(x) = (0.25x + 50 + 2500)/x
Average cost has no global minimum, it decreases steadily as production increases.

i) Profit is revenue - cost
P(y) = 450y - 2y^2 - (C(D(y))
P(y) = 450y - 2y^2 - [0.25(450-2y)^2 + 50(450-2y) + 2500] which reduces to
P(y) = -3y^2 + 1000y - 75625
j) To maximize the profit:

P'(y) = -6y + 1000=0

y=167
As defined from above, y = sale price
D(167) = 450 - 2(167) = 116 boxes
k,l) Max Profit = -3(167)^2 + 1000(167) - 75625
Max Profit = $7708 at $167/box

For part 2 and 3 you have basically to do this whole process again with another considerations, if you have any questions for those parts you can post them in another question and you will get the explanation.

I hope it helps :)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.