The Apex Widget Company has determined that its marginal cost function is C0(x) =0.5x + 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 $50 per box, they will sell 350 boxes; and if they charge $75 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.
(j) the number of boxes which they should produce and sell to maximize profit.
(k) the maximum profit.
(l) 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.

3. Suppose that plant capacity is such that not more than 200 boxes of widgets can
be produced each week. 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.

There's a mistake in your prompt somewhere. "The cost of producing 200 boxes of widgets is $22,500 when it should be 200(0.5) + 50 = $150, and if it is $22500 then the company is operating at a large loss, considering 200 boxes @ even $75 each is only $15,000.

hmm maybe, only thing i can add is that the function C'(x)=.5x+50 instead of the zero.

nobody?

D(50) = 350
D(75) = 300
Linear (y = sale price)
D(y) = Ay + b
A = -2
b = 450
D(y) = 450 - 2y

Revenue would be yD(y)
R(y) = 450y - 2y^2

R'(y) = 450 - 4y
R'(y) = 0 @ 112.5
Max revenue @ 112.5 boxes
Max revenue + 450(112.5) - 2(112.5)^2
Max revenue = 25,313

So if C'(x) = 0.5x + 50 then you integrate to get C(x) which is what you really care about.
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

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.

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

P'(y) = -6y + 1000 which is zero @ y = 167
As defined from above, y = sale price
D(167) = 450 - 2(167) = 116 boxes
Max Profit = -3(167)^2 + 1000(167) - 75625
Max Profit = $7708 @ $167/box

2 and 3 require you to do this whole process over again with different constraints. Also I rounded above so the answers are not perfect, which I guess #2 leads to.

alright thanks alot! I just get lost when it comes to word problems haha.

DM about how much fgs you like for doing this!

100 would be fine