The inverse demand function and total cost function for a manufacturer are given as follows: P = 2000 – 10Q TC = 5000 + 500Q Find the manufacturer’s total revenue function and profit function in terms of quantity. [2 marks] Using optimization, find the level of output and price at which the total revenue is maximized; and calculate the maximum revenue that the manufacturer can achieve. Demonstrate that it is maximum using second-order conditions. [6 marks] Explain why the turning point is defined as maximum when the value of the second-order condition is negative. [2 marks]
The Correct Answer and Explanation is:
Solution:
1. Total Revenue (TR) and Profit Function (π)
Given:
- Inverse demand function: P=2000−10QP = 2000 – 10QP=2000−10Q
- Total cost function: TC=5000+500QTC = 5000 + 500QTC=5000+500Q
Total Revenue (TR):TR=P⋅Q=(2000−10Q)⋅Q=2000Q−10Q2TR = P \cdot Q = (2000 – 10Q) \cdot Q = 2000Q – 10Q^2TR=P⋅Q=(2000−10Q)⋅Q=2000Q−10Q2
Profit Function (π):π(Q)=TR−TC=(2000Q−10Q2)−(5000+500Q)\pi(Q) = TR – TC = (2000Q – 10Q^2) – (5000 + 500Q)π(Q)=TR−TC=(2000Q−10Q2)−(5000+500Q)π(Q)=2000Q−10Q2−5000−500Q=1500Q−10Q2−5000\pi(Q) = 2000Q – 10Q^2 – 5000 – 500Q = 1500Q – 10Q^2 – 5000π(Q)=2000Q−10Q2−5000−500Q=1500Q−10Q2−5000
2. Revenue Maximization
To maximize total revenue:TR(Q)=2000Q−10Q2TR(Q) = 2000Q – 10Q^2TR(Q)=2000Q−10Q2
First-order condition:d(TR)dQ=2000−20Q=0⇒Q=100\frac{d(TR)}{dQ} = 2000 – 20Q = 0 \Rightarrow Q = 100dQd(TR)=2000−20Q=0⇒Q=100
Price at this output:P=2000−10(100)=1000P = 2000 – 10(100) = 1000P=2000−10(100)=1000
Maximum revenue:TR=2000(100)−10(100)2=200000−100000=100000TR = 2000(100) – 10(100)^2 = 200000 – 100000 = 100000TR=2000(100)−10(100)2=200000−100000=100000
Second-order condition:d2(TR)dQ2=−20<0⇒Maximum\frac{d^2(TR)}{dQ^2} = -20 < 0 \Rightarrow \text{Maximum}dQ2d2(TR)=−20<0⇒Maximum
Explanation
In microeconomic analysis, understanding how revenue and profit vary with output is critical for managerial decision-making. Here, we are provided with the inverse demand function P=2000−10QP = 2000 – 10QP=2000−10Q and total cost function TC=5000+500QTC = 5000 + 500QTC=5000+500Q. The first step in revenue and profit analysis is to derive the total revenue function, which is the product of price and quantity:TR=P⋅Q=(2000−10Q)Q=2000Q−10Q2.TR = P \cdot Q = (2000 – 10Q)Q = 2000Q – 10Q^2.TR=P⋅Q=(2000−10Q)Q=2000Q−10Q2.
Profit is defined as the difference between total revenue and total cost:π(Q)=TR−TC=(2000Q−10Q2)−(5000+500Q)=1500Q−10Q2−5000.\pi(Q) = TR – TC = (2000Q – 10Q^2) – (5000 + 500Q) = 1500Q – 10Q^2 – 5000.π(Q)=TR−TC=(2000Q−10Q2)−(5000+500Q)=1500Q−10Q2−5000.
To find the level of output that maximizes total revenue, we apply optimization by taking the first derivative of the revenue function with respect to quantity and setting it equal to zero:d(TR)dQ=2000−20Q=0⇒Q=100.\frac{d(TR)}{dQ} = 2000 – 20Q = 0 \Rightarrow Q = 100.dQd(TR)=2000−20Q=0⇒Q=100.
At Q=100Q = 100Q=100, substituting back into the demand function yields P=1000P = 1000P=1000, and maximum revenue is calculated as:TR=2000(100)−10(100)2=100000.TR = 2000(100) – 10(100)^2 = 100000.TR=2000(100)−10(100)2=100000.
To confirm this is a maximum, we examine the second derivative:d2(TR)dQ2=−20<0,\frac{d^2(TR)}{dQ^2} = -20 < 0,dQ2d2(TR)=−20<0,
indicating a concave function, confirming a maximum point.
The second-order condition being negative implies that the slope of the function is decreasing at the critical point, characteristic of a peak in the revenue curve. Thus, when the second derivative is negative, it confirms the presence of a local maximum, ensuring optimal output and pricing for maximum revenue.
