A fertilizer producer finds that it can sell its product at a price of...

Question

A fertilizer producer finds that it can sell its product at a price of p=300-0.1x dollars per unit when it produces x units of fertilizer. The total production cost (in dollars) for x units is C(x)= 16,000+125x+0.025x^2.

If the production capacity of the firm is at most 1000 units of fertilizer in a specified time, how many units must be manufactured and sold in that time to maximize the profit?

Answer 1

p=300 - .1x

Revenue = price times quantity sold = px = 300x -.1x^2

Cost = 16,000 + 12x + 0.025x^2

Average Cost = 16,000/x +12 + 0.025x

Marginal Cost = 12 +0.05x

AC = MC, 12 +0.05x = 12 + .025x + 16,000/x

0.025x = 16,000/x

0.025x^2 = 16,000

x^2 = 16,000/.025 =640000

x= 800 = minimum average cost output level

but that ignores revenue and profits

Profit = R-C = 288x - 0.975x^2 -1600

Profit is maximized when marginal revenue = marginal cost, R'=C'

(R-C)' = 288 - 1.95x =0

x = 288/1.95 = about 147.7 = profit maximizing output level

Max Profit = 288(148) -0.975(148)^2 -1600 = 42,624 - 21,356.4 -16,000 = $5,267.60