Monopoly Profit Maximization: Exercise and Step-by-Step Solution

Monopoly Profit Maximization Exercise

A monopolist operates with the following total revenue function:

$$TR = 1000Q – 15Q^2$$

where $P$ is the price per unit and $Q$ is the quantity produced. The total cost function is:

$$TC = Q^2 + 80Q + 2000$$

The production limit is 40 units.

Questions:

1. (a) Find the quantity that maximizes the profit and calculate the profit at this quantity.
2. (b) Derive the marginal cost (MC) and marginal revenue (MR) functions. Compute the marginal cost and marginal revenue for the profit-maximizing quantity.
3. (c) Identify the quantity at which the company breaks even (i.e., where total revenue equals total cost).

Solutions

(a) Maximizing Profit

Profit is given by:

$$\pi(Q) = TR – TC$$ $$\pi(Q) = (1000Q – 15Q^2) – (Q^2 + 80Q + 2000)$$ $$\pi(Q) = 1000Q – 15Q^2 – Q^2 – 80Q – 2000$$ $$\pi(Q) = 920Q – 16Q^2 – 2000$$

To maximize profit, take the derivative and set it to zero:

$$\frac{d\pi}{dQ} = 920 – 32Q = 0$$

Solving for $Q$:

$$Q = \frac{920}{32} = 28.75$$

Since $Q$ must be an integer, we check $Q = 28$ and $Q = 29$.

For $Q = 28$:

$$TR = 1000(28) – 15(28)^2 = 28000 – 11760 = 16240$$ $$TC = (28)^2 + 80(28) + 2000 = 784 + 2240 + 2000 = 5024$$ $$\pi(28) = 16240 – 5024 = 11216$$

For $Q = 29$:

$$TR = 1000(29) – 15(29)^2 = 29000 – 12615 = 16385$$ $$TC = (29)^2 + 80(29) + 2000 = 841 + 2320 + 2000 = 5161$$ $$\pi(29) = 16385 – 5161 = 11224$$

Since $Q = 29$ gives a slightly higher profit, the monopoly maximizes profit at $Q = 29$ with a profit of 11,224.

(b) Marginal Cost and Marginal Revenue

Marginal Revenue:

$$MR = \frac{d(TR)}{dQ} = 1000 – 30Q$$

Marginal Cost:

$$MC = \frac{d(TC)}{dQ} = 2Q + 80$$

At $Q = 29$:

$$MR = 1000 – 30(29) = 1000 – 870 = 130$$ $$MC = 2(29) + 80 = 58 + 80 = 138$$

Since $MR \approx MC$, the solution is verified.

(c) Break-even Point

Break-even occurs when:

$$TR = TC$$ $$1000Q – 15Q^2 = Q^2 + 80Q + 2000$$

Rearrange:

$$1000Q – 15Q^2 – Q^2 – 80Q – 2000 = 0$$ $$920Q – 16Q^2 – 2000 = 0$$

Solve using the quadratic formula:

$$Q = \frac{-920 \pm \sqrt{920^2 – 4(-16)(2000)}}{2(-16)}$$ $$Q = \frac{-920 \pm \sqrt{846400 + 128000}}{-32}$$ $$Q = \frac{-920 \pm \sqrt{974400}}{-32}$$

Approximating:

$$Q = \frac{-920 \pm 987}{-32}$$ $$Q = \frac{-920 + 987}{-32} \quad \text{or} \quad Q = \frac{-920 – 987}{-32}$$ $$Q = \frac{67}{-32} \approx -2.1 \quad \text{(not valid)}$$ $$Q = \frac{-1907}{-32} \approx 39.8$$

Thus, the monopoly breaks even at approximately $Q = 40$, which is within the production limit.

Final Answers

• Profit-maximizing quantity: $Q = 29$, with profit 11,224.
• Marginal cost and marginal revenue at $Q = 29$:
  • $MC = 138$
  • $MR = 130$
• Break-even quantity: $Q \approx 40$.