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The Apex Widget Company has determined that its marginal cost function is C\'(x)

ID: 1208125 • Letter: T

Question

The Apex Widget Company has determined that its marginal cost function is C'(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. Suppose that plant capacity is such that not more than 200 boxes of widgets can be produced each week. Find the following answers below.

(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 fuction.

(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.

Explanation / Answer

Ans:

a)

D(50) = 350

D(75) = 300

Direct (y = deal cost)

D(y) = Ay + b

A = - 2

b = 450

D(y) = 450 - 2y

b) revenue function

Income would be yD(y)

R(y) = 450y - 2y^2

R'(y) = 450 - 4y

R'(y) = 0 @ 112.5

Max income @ 112.5 boxes

Max income + 450(112.5) - 2(112.5)^2

Max income = 25,313

So if C'(x) = 0.5x + 50 then you coordinate to get C(x) which is the thing that you truly think about.

C(x) = 0.25x^2 + 50x + c

C(200) = 22,500 = 0.25(200)^2 + 50(200) + c

c = 2500

c)

C(x) = 0.25x^2 + 50x + 2500

Normal expense would be C(x)/x

AvgC(x) = 0.25x + 50 + 2500/x

Normal expense has no worldwide least, it diminishes consistently as generation increments.

d)

Benefit is income - cost

P(y) = 450y - 2y^2 - (C(D(y))

P(y) = 450y - 2y^2 - [0.25(450-2y)^2 + 50(450-2y) + 2500] which lessens to

P(y) = - 3y^2 + 1000y - 75625

e)

P'(y) = - 6y + 1000 which is zero @ y = 167

As characterized from above, y = deal cost

f)

D(167) = 450 - 2(167) = 116 boxes

Max Profit = - 3(167)^2 + 1000(167) - 75625

Max Profit = $7708 @ $167/box