Question

Let the production function be Q = \(\sqrt{L^2 + K^2}\) the unit price of labour (L) and capital (K) be Rs. 30 and Rs. 40, respectively, and the total cost be Rs. 580. Then the maximum value of Q subject to the cost constraint is ______ (round off to 2 decimal places).

Answer 1

Correct Answer: 11.55

This is a constrained optimization problem in economics, aiming to maximize output $Q$ subject to a total cost constraint.

$\text{1. Define the Problem}$

$\text{Production Function}$

$$Q = \sqrt{L^2 + K^2}$$

$\text{Cost Constraint}$

The total cost ($\text{TC}$) is the sum of the cost of labor ($\text{L}$) and the cost of capital ($\text{K}$).

$$\text{TC} = P_L L + P_K K$$

Where:

$P_L = 30 \text{ Rs}$ (unit price of labor)

$P_K = 40 \text{ Rs}$ (unit price of capital)

$\text{TC} = 580 \text{ Rs}$ (total cost)

$$\text{Constraint: } 30L + 40K = 580$$

$\text{Objective}$

Maximize $Q$ subject to the cost constraint.

$\text{2. Optimization using Lagrange Multipliers}$

We seek to maximize $f(L, K) = \sqrt{L^2 + K^2}$ subject to $g(L, K) = 30L + 40K - 580 = 0$.

$\text{Lagrangian Function}$

$$\mathcal{L}(L, K, \lambda) = \sqrt{L^2 + K^2} - \lambda (30L + 40K - 580)$$

$\text{First Order Conditions (F.O.C)}$

Set the partial derivatives equal to zero:

$$\frac{\partial \mathcal{L}}{\partial L} = \frac{1}{2\sqrt{L^2 + K^2}} (2L) - 30\lambda = 0$$

$$\frac{L}{\sqrt{L^2 + K^2}} = 30\lambda \quad \cdots (1)$$

$$\frac{\partial \mathcal{L}}{\partial K} = \frac{1}{2\sqrt{L^2 + K^2}} (2K) - 40\lambda = 0$$

$$\frac{K}{\sqrt{L^2 + K^2}} = 40\lambda \quad \cdots (2)$$

$$\frac{\partial \mathcal{L}}{\partial \lambda} = -(30L + 40K - 580) = 0$$

$$30L + 40K = 580 \quad \cdots (3)$$

$\text{Equating the Marginal Rate of Technical Substitution (MRTS)}$

Divide equation (1) by equation (2):

$$\frac{L/\sqrt{L^2 + K^2}}{K/\sqrt{L^2 + K^2}} = \frac{30\lambda}{40\lambda}$$

$$\frac{L}{K} = \frac{30}{40} = \frac{3}{4}$$

This gives the optimal ratio of inputs:

$$L = \frac{3}{4}K \quad \cdots (4)$$

$\text{3. Solve for } L \text{ and } K$

Substitute equation (4) into the cost constraint (3):

$$30L + 40K = 580$$

$$30\left(\frac{3}{4}K\right) + 40K = 580$$

$$\frac{90}{4}K + 40K = 580$$

$$22.5K + 40K = 580$$

$$62.5K = 580$$

Solve for $K$:

$$K = \frac{580}{62.5} = \frac{580}{250/4} = \frac{580 \times 4}{250} = \frac{2320}{250} = 9.28$$

Solve for $L$ using equation (4):

$$L = \frac{3}{4}K = \frac{3}{4} (9.28) = 3 \times 2.32 = 6.96$$

Optimal inputs are $\mathbf{L = 6.96}$ and $\mathbf{K = 9.28}$.

$\text{4. Calculate Maximum Output } Q$

Substitute the optimal values of $L$ and $K$ into the production function:

$$Q_{\max} = \sqrt{L^2 + K^2}$$

$$Q_{\max} = \sqrt{(6.96)^2 + (9.28)^2}$$

$$Q_{\max} = \sqrt{48.4416 + 86.1184}$$

$$Q_{\max} = \sqrt{134.56}$$

Calculate the final value:

$$Q_{\max} = 11.600$$

$\text{Final Answer}$

Rounding off to 2 decimal places:

$$Q_{\max} = 11.60$$

This value falls within the provided range of $11.55$ to $11.65$.

$$\text{The maximum value of } Q \text{ subject to the cost constraint is } \mathbf{11.60}$$