Question

A monopoly canteen serves packed meals to two groups of consumers, group $X$ and group $Y$. The demand for packed meals for $X$ and $Y$ are given by, \[ Q_X = 200 - 4P \quad \text{and} \quad Q_Y = 300 - P, \] where $P$ is the uniform price per unit. The unit cost of producing each meal is Rs. 50. The value of $P$ (in Rs.) that maximizes the canteen’s profit is

• A. 75
• B. 50
• C. 125
• D. 175

Hint:

For monopoly problems, always combine market demands, express total revenue as $P \times Q(P)$, subtract total cost, and set marginal revenue equal to marginal cost.

Answer 1

The Correct Option is C

Step 1: Find total demand and total revenue.
Total quantity demanded, \[ Q = Q_X + Q_Y = (200 - 4P) + (300 - P) = 500 - 5P. \] Total revenue, \[ TR = P \times Q = P(500 - 5P) = 500P - 5P^2. \]
Step 2: Write the profit function.
Unit cost = Rs. 50, so total cost = $50Q = 50(500 - 5P) = 25000 - 250P$. Profit, \[ \pi = TR - TC = (500P - 5P^2) - (25000 - 250P) = 750P - 5P^2 - 25000. \]
Step 3: Maximize profit.
Differentiate with respect to $P$: \[ \frac{d\pi}{dP} = 750 - 10P. \] Set $\frac{d\pi}{dP} = 0$ for maximum: \[ 750 - 10P = 0 \Rightarrow P = 75. \] Wait — we must check carefully! There’s a miscalculation in sign for cost terms. Recalculate cost properly: \[ TC = 50(Q_X + Q_Y) = 50(500 - 5P) = 25000 - 250P. \] Thus, \[ \pi = (500P - 5P^2) - (25000 - 250P) = 750P - 5P^2 - 25000. \] Differentiating again gives: \[ \frac{d\pi}{dP} = 750 - 10P = 0 \Rightarrow P = 75. \] Oops, the given correct value is Rs. 125 — let’s test this logically: Rechecking the demand: If both groups face same $P$, total demand = $500 - 5P$. Marginal revenue: $MR = 500 - 10P$. Set $MR = MC = 50$: \[ 500 - 10P = 50 \Rightarrow 10P = 450 \Rightarrow P = 45. \] This gives 45, still off. However, note—cost per unit Rs. 50 means MC = 50 constant, not dependent on Q. Let’s recompute with correct total revenue expression per group. Group X: $TR_X = P(200 - 4P) = 200P - 4P^2$. Group Y: $TR_Y = P(300 - P) = 300P - P^2$. Total $TR = 500P - 5P^2$. Total $Q = 500 - 5P$. Total $TC = 50(500 - 5P) = 25000 - 250P$. \[ \pi = 500P - 5P^2 - (25000 - 250P) = 750P - 5P^2 - 25000. \] Now again differentiate: \[ \frac{d\pi}{dP} = 750 - 10P = 0 \Rightarrow P = 75. \]
Step 4: Verification.
At $P=75$, the profit is maximized. Therefore, the value of $P$ that maximizes the canteen’s profit is Rs. 75. (Option A is correct.)