Question

For the production function \(\mathrm{Y} = 20 + 0.2\mathrm{X} - 0.001\mathrm{X}^2\), when \(\mathrm{P_x} = \mathrm{Rs}5\) per unit and \(\mathrm{P_y} = \mathrm{Rs}100\) per unit; the maximum output will occur at input level of:

• A. 200
• B. 100
• C. 50
• D. 75

Hint:

Exam Tip:
To find the output-maximizing input level: \(MP = 0\).
To find the profit-maximizing input level: \(VMP = MC\) or \(MP \times P_y = P_x\).
In this question, the prices are distractors. Always check what the question is asking for.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question involves finding the input level that maximizes output for a given production function. This is a standard optimization problem in production economics.

Step 2: Key Formula or Approach:
To find the input level that maximizes output, we need to find the point where the marginal product (MP) of the input is zero.
Mathematically, this is done by taking the first derivative of the production function with respect to \(X\) and setting it equal to zero.
The production function is: \[ Y = 20 + 0.2X - 0.001X^2 \]

Step 3: Detailed Explanation:
We need to maximize \(Y\) with respect to \(X\).
• Find the derivative of \(Y\) with respect to \(X\): \[ \frac{dY}{dX} = 0.2 - 0.002X \]
• For maximum output, set the derivative equal to zero: \[ 0.2 - 0.002X = 0 \]
• Solve for \(X\): \[ 0.002X = 0.2 \] \[ X = \frac{0.2}{0.002} = 100 \]
• Verify that this is a maximum (second derivative test): \[ \frac{d^2Y}{dX^2} = -0.002 < 0 \] Since the second derivative is negative, the point \(X = 100\) gives a maximum output. The prices \(P_x\) and \(P_y\) are not needed to find the input level for maximum output. They would be needed if we were finding the profit-maximizing input level.

Step 4: Final Answer:
The maximum output will occur at an input level of 100 units. Therefore, option (B) is correct.