Dean's Office Official Model Question Paper

Partial Derivatives

Treating $y$ as a constant when differentiating with respect to $x$:

$$\frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(3x^2y^3 - 5x^3 + 4y^2 - 7) = 6xy^3 - 15x^2$$

Treating $x$ as a constant when differentiating with respect to $y$:

$$\frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(3x^2y^3 - 5x^3 + 4y^2 - 7) = 9x^2y^2 + 8y$$

First-Order Differential Equation

A first-order differential equation is an equation involving an unknown function $y(x)$ and its first derivative $\frac{dy}{dx}$ without higher-order derivatives.

Standard Linear Form:

$$\frac{dy}{dx} + P(x)y = Q(x)$$

Where $P(x)$ and $Q(x)$ are continuous functions of $x$ alone.

Euler’s Theorem for Homogeneous Functions

If a production function $Q = f(K, L)$ is homogeneous of degree $n$, then:

$$K \frac{\partial Q}{\partial K} + L \frac{\partial Q}{\partial L} = n Q$$

For a function with constant returns to scale ($n = 1$):

$$K \cdot MP_K + L \cdot MP_L = Q$$

Evaluation of Double Integral

Inner integral with respect to $y$:

$$\int_0^1 (2x + 4y) \, dy = \left[ 2xy + 2y^2 \right]_0^1 = (2x(1) + 2(1)^2) - 0 = 2x + 2$$

Outer integral with respect to $x$:

$$\int_0^2 (2x + 2) \, dx = \left[ x^2 + 2x \right]_0^2 = (2^2 + 2(2)) - 0 = 4 + 4 = \mathbf{8}$$

Dual Problem Formulation

Let the dual decision variables be $y_1$ and $y_2$ corresponding to constraints 1 and 2.

Dual Objective Function:

$$\text{Minimize } W = 20y_1 + 30y_2$$

Subject to Constraints:

$$2y_1 + y_2 \ge 4$$

$$y_1 + 3y_2 \ge 6$$

$$y_1, y_2 \ge 0$$

Multivariable Profit Maximization

a. Total Profit Function $\pi(x, y)$

Total Revenue:

$$TR = P_x \cdot x + P_y \cdot y = (100 - x)x + (120 - 2y)y = 100x - x^2 + 120y - 2y^2$$

Total Profit:

$$\pi(x, y) = TR - C = (100x - x^2 + 120y - 2y^2) - (2x^2 + xy + 3y^2 + 500)$$

$$\pi(x, y) = -3x^2 - xy - 5y^2 + 100x + 120y - 500$$

b. First-Order Necessary Conditions

$$\frac{\partial \pi}{\partial x} = -6x - y + 100 = 0 \implies 6x + y = 100 \quad \text{--- (Eq 1)}$$

$$\frac{\partial \pi}{\partial y} = -x - 10y + 120 = 0 \implies x + 10y = 120 \quad \text{--- (Eq 2)}$$

From Eq 1, $y = 100 - 6x$. Substituting into Eq 2:

$$x + 10(100 - 6x) = 120$$

$$x + 1000 - 60x = 120 \implies -59x = -880 \implies x = \frac{880}{59} \approx 14.92$$

$$y = 100 - 6(14.92) = 100 - 89.52 \approx 10.48$$

c. Second-Order Sufficiency Conditions (Hessian Matrix)

$$\pi_{xx} = \frac{\partial^2 \pi}{\partial x^2} = -6 < 0$$

$$\pi_{yy} = \frac{\partial^2 \pi}{\partial y^2} = -10 < 0$$

$$\pi_{xy} = \frac{\partial^2 \pi}{\partial x \partial y} = -1$$

Hessian determinant:

$$|H| = \pi_{xx}\pi_{yy} - (\pi_{xy})^2 = (-6)(-10) - (-1)^2 = 60 - 1 = 59 > 0$$

Since $\pi_{xx} < 0$, $\pi_{yy} < 0$, and $|H| > 0$, the Hessian matrix is negative definite, confirming that profit is strictly maximized at $x \approx 14.92$ units and $y \approx 10.48$ units.

Solution of Linear First-Order Differential Equation

The equation is in standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$, with $P(x) = 2$ and $Q(x) = 4e^{3x}$.

1. Integrating Factor (IF)

$$IF = e^{\int P(x) \, dx} = e^{\int 2 \, dx} = e^{2x}$$

2. General Solution

$$y \cdot (IF) = \int Q(x) \cdot (IF) \, dx + C$$

$$y e^{2x} = \int 4e^{3x} \cdot e^{2x} \, dx + C = \int 4e^{5x} \, dx + C$$

$$y e^{2x} = 4 \left( \frac{e^{5x}}{5} \right) + C = \frac{4}{5}e^{5x} + C$$

Dividing through by $e^{2x}$:

$$y(x) = \frac{4}{5}e^{3x} + C e^{-2x} \quad \text{(General Solution)}$$

3. Particular Solution with $y(0) = 5$$5 = \frac{4}{5}e^{0} + C e^{0} = 0.8 + C \implies C = 5 - 0.8 = 4.2 = \frac{21}{5}$$

$$\mathbf{y(x) = \frac{4}{5}e^{3x} + \frac{21}{5}e^{-2x}}$$

Solution via Simplex Method

Introducing slack variables $s_1 \ge 0$ and $s_2 \ge 0$:

$$\begin{aligned} x_1 + 2x_2 + s_1 &= 20 \\ 3x_1 + x_2 + s_2 &= 30 \\ Z - 3x_1 - 5x_2 &= 0 \end{aligned}$$

Initial Simplex Tableau:

• Most negative indicator in Z-row is $-5 \implies x_2$ enters.
• Minimum positive ratio is $10 \implies s_1$ leaves. Pivot element is 2.

First Iteration (Dividing Row 1 by 2 and eliminating $x_2$):

• New $R_1$: $x_2 = 0.5x_1 + x_2 + 0.5s_1 = 10$
• New $R_2 = R_2 - 1(R_1') = 2.5x_1 + 0x_2 - 0.5s_1 + s_2 = 20$
• New $Z = Z + 5(R_1') = -0.5x_1 + 0x_2 + 2.5s_1 + 0s_2 = 50$

Next entering variable is $x_1$ (indicator $-0.5$).

• Ratios: For $R_1$: $10 / 0.5 = 20$; For $R_2$: $20 / 2.5 = 8$ (Minimum ratio; $s_2$ leaves).

Second Iteration (Pivot on $2.5$ in Row 2):

• Dividing $R_2$ by $2.5$: $x_1 + 0x_2 - 0.2s_1 + 0.4s_2 = 8$
• Updating $R_1$: $R_1 - 0.5(R_2') \implies x_2 = 10 - 0.5(8) = 6$
• Updating $Z$: $Z + 0.5(R_2') \implies Z = 50 + 0.5(8) = 54$

All indicators in the Z-row are non-negative ($C_j - Z_j \ge 0$). Optimal Solution:

$$x_1 = 8, \quad x_2 = 6, \quad \mathbf{Z_{\max} = 3(8) + 5(6) = 24 + 30 = 54}$$

Utility Maximization via Lagrange Multipliers

a. Lagrangian Formulation

$$\mathcal{L}(x, y, \lambda) = x^{0.4} y^{0.6} + \lambda(120 - 2x - 3y)$$

First-order conditions:

$$\frac{\partial \mathcal{L}}{\partial x} = 0.4 x^{-0.6} y^{0.6} - 2\lambda = 0 \implies \lambda = 0.2 x^{-0.6} y^{0.6} \quad \text{--- (1)}$$

$$\frac{\partial \mathcal{L}}{\partial y} = 0.6 x^{0.4} y^{-0.4} - 3\lambda = 0 \implies \lambda = 0.2 x^{0.4} y^{-0.4} \quad \text{--- (2)}$$

$$\frac{\partial \mathcal{L}}{\partial \lambda} = 120 - 2x - 3y = 0 \quad \text{--- (3)}$$

Equating (1) and (2):

$$0.2 x^{-0.6} y^{0.6} = 0.2 x^{0.4} y^{-0.4}$$

$$\frac{y^{0.6}}{y^{-0.4}} = \frac{x^{0.4}}{x^{-0.6}} \implies y = x$$

Substitute $y = x$ into budget constraint (3):

$$2x + 3(x) = 120 \implies 5x = 120 \implies \mathbf{x = 24}$$

$$\mathbf{y = 24}$$

b. Marginal Utility of Income ($\lambda$)

$$\lambda = 0.2 (24)^{-0.6} (24)^{0.6} = 0.2(1) = \mathbf{0.2}$$

This indicates that for each additional rupee of income, total maximum utility increases by approximately 0.2 units.

Case Analysis: Joint Segment Profit Optimization

a. Revenue and Profit Formulations

$$\begin{aligned} TR(q_1, q_2) &= P_1 q_1 + P_2 q_2 = (200 - 2q_1)q_1 + (150 - q_2)q_2 \\ &= 200q_1 - 2q_1^2 + 150q_2 - q_2^2 \end{aligned}$$

$$\begin{aligned} \pi(q_1, q_2) &= TR - C \\ &= (200q_1 - 2q_1^2 + 150q_2 - q_2^2) - (q_1^2 + 2q_2^2 + q_1 q_2 + 10q_1 + 20q_2 + 2{,}000) \\ &= -3q_1^2 - 3q_2^2 - q_1 q_2 + 190q_1 + 130q_2 - 2{,}000 \end{aligned}$$

---

b. First-Order Necessary Conditions

$$\frac{\partial \pi}{\partial q_1} = -6q_1 - q_2 + 190 = 0 \implies 6q_1 + q_2 = 190 \quad \text{--- (1)}$$

$$\frac{\partial \pi}{\partial q_2} = -q_1 - 6q_2 + 130 = 0 \implies q_1 + 6q_2 = 130 \quad \text{--- (2)}$$

From (1), $q_2 = 190 - 6q_1$. Substituting into (2):

$$q_1 + 6(190 - 6q_1) = 130$$

$$q_1 + 1140 - 36q_1 = 130 \implies -35q_1 = -1010 \implies q_1 = \frac{1010}{35} = \frac{202}{7} \approx \mathbf{28.86 \text{ thousand}}$$

$$q_2 = 190 - 6\left(\frac{202}{7}\right) = \frac{1330 - 1212}{7} = \frac{118}{7} \approx \mathbf{16.86 \text{ thousand}}$$

---

c. Optimal Prices in Each Segment

$$P_1 = 200 - 2\left(\frac{202}{7}\right) = 200 - 57.71 = \mathbf{\text{Rs } 142.29}$$

$$P_2 = 150 - \frac{118}{7} = 150 - 16.86 = \mathbf{\text{Rs } 133.14}$$

---

d. Second-Order Sufficiency Conditions and Maximum Profit

Hessian Matrix of Second Partials:

$$H = \begin{pmatrix} \pi_{11} & \pi_{12} \\ \pi_{21} & \pi_{22} \end{pmatrix}$$

$$\pi_{11} = \frac{\partial^2 \pi}{\partial q_1^2} = -6$$

$$\pi_{22} = \frac{\partial^2 \pi}{\partial q_2^2} = -6$$

$$\pi_{12} = \pi_{21} = -1$$

Principal minors:

1. $H_1 = \pi_{11} = -6 < 0$
2. $H_2 = |H| = (-6)(-6) - (-1)^2 = 36 - 1 = 35 > 0$

Since $H_1 < 0$ and $H_2 > 0$, the Hessian matrix is strictly negative definite, proving that the solution is a unique global maximum.

Maximum Profit Computation:

$$\pi_{\max} = -3(28.86)^2 - 3(16.86)^2 - (28.86)(16.86) + 190(28.86) + 130(16.86) - 2{,}000$$

$$\pi_{\max} = -3(832.90) - 3(284.26) - 486.58 + 5483.40 + 2191.80 - 2000$$

$$\pi_{\max} = -2498.70 - 852.78 - 486.58 + 7675.20 - 2000 = \mathbf{\text{Rs } 1{,}837.14 \text{ Million}}$$