Question

A vegetable producer wants to maximize his profit by growing cabbage. Suppose the demand function is \(\mathrm{P} = 40 - 0.02\mathrm{Q}\) where P is the price in Rupees per unit of cabbage and the cost function is \(\mathrm{C} = 20\mathrm{Q} + 400\) where Q is the quantity daily produced. His profit will be maximum at the output level of:

• A. 200 units
• B. 500 units
• C. 1000 units
• D. 400 units

Hint:

Exam Tip:
Profit Maximization Rule: MR = MC.
For a linear demand curve \(P = a - bQ\), the MR curve is \(MR = a - 2bQ\).
This is a standard result used in many optimization problems.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question involves maximizing profit given a demand function and a cost function. Profit maximization occurs where Marginal Revenue (MR) equals Marginal Cost (MC).

Step 2: Key Formula or Approach:
Profit (\(\pi\)) is given by Total Revenue (TR) minus Total Cost (TC). \[ \pi = TR - TC \] We need to find the quantity \(Q\) that maximizes profit.

Step 3: Detailed Explanation:

• Find Total Revenue (TR):
TR = Price \(\times\) Quantity = \(P \times Q\)
Given \(P = 40 - 0.02Q\), we have: \[ TR = (40 - 0.02Q) \times Q = 40Q - 0.02Q^2 \]
• Find Marginal Revenue (MR):
MR is the derivative of TR with respect to Q: \[ MR = \frac{d(TR)}{dQ} = 40 - 0.04Q \]
• Find Marginal Cost (MC):
MC is the derivative of TC with respect to Q:
Given \(TC = 20Q + 400\), we have: \[ MC = \frac{d(TC)}{dQ} = 20 \]
• Set MR = MC for Profit Maximization: \[ 40 - 0.04Q = 20 \]
• Solve for Q: \[ 40 - 20 = 0.04Q \] \[ 20 = 0.04Q \] \[ Q = \frac{20}{0.04} = 500 \]
• Verify Profit Maximization:
The second-order condition requires that the slope of MR is less than the slope of MC.
\(\frac{d(MR)}{dQ} = -0.04\) and \(\frac{d(MC)}{dQ} = 0\).
Since \(-0.04 < 0\), profit is maximized at \(Q = 500\).

Step 4: Final Answer:
The profit will be maximum at an output level of 500 units. Therefore, option (B) is correct.