Find the partial derivatives \frac{\partial z}{\partial x} and \frac{\partial z}{\partial y} for z = 3x^2y^3 - 5x^3 + 4y^2 - 7.

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

Define a first-order differential equation and state its general linear form.

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.

State Euler's Theorem for a homogeneous production function Q = f(K, L) of degree n.

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 MPK + L \cdot MPL = Q

Evaluate the definite double integral: \int0^2 \int0^1 (2x + 4y) \, dy \, dx.

Evaluation of Double Integral Inner integral with respect to y: \int0^1 (2x + 4y) \, dy = \left[ 2xy + 2y^2 \right]0^1 = (2x(1) + 2(1)^2) - 0 = 2x + 2 Outer integral with respect to x: \int0^2 (2x + 2) \, dx = \left[ x^2 + 2x \right]0^2 = (2^2 + 2(2)) - 0 = 4 + 4 = \mathbf{8}

Formulate the dual of the following Linear Programming Problem: \text{Maximize } Z = 4x1 + 6x2 \quad \text{subject to } 2x1 + x2 \le 20, \; x1 + 3x2 \le 30, \; x1, x2 \ge 0

Dual Problem Formulation Let the dual decision variables be y1 and y2 corresponding to constraints 1 and 2. Dual Objective Function: \text{Minimize } W = 20y1 + 30y2 Subject to Constraints: 2y1 + y2 \ge 4 y1 + 3y2 \ge 6 y1, y2 \ge 0

A firm produces two joint products, X and Y. The joint cost function is C(x, y) = 2x^2 + xy + 3y^2 + 500. The demand functions are Px = 100 - x and Py = 120 - 2y. a. Formulate the total profit function \pi(x, y). b. Find the profit-maximizing output levels of x and y. c. Verify the second-order conditions for maximum profit.

Multivariable Profit Maximization a. Total Profit Function \pi(x, y) Total Revenue: TR = Px \cdot x + Py \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 Since \pi{xx} 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.

Find the general solution and particular solution of the first-order differential equation: \frac{dy}{dx} + 2y = 4e^{3x} Given the initial condition y(0) = 5.

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}}

Solve the following Linear Programming Problem using the Simplex Method: \begin{aligned} \text{Maximize } & Z = 3x1 + 5x2 \\ \text{Subject to: } & x1 + 2x2 \le 20 \\ & 3x1 + x2 \le 30 \\ & x1, x2 \ge 0 \end{aligned}

Solution via Simplex Method Introducing slack variables s1 \ge 0 and s2 \ge 0: \begin{aligned} x1 + 2x2 + s1 &= 20 \\ 3x1 + x2 + s2 &= 30 \\ Z - 3x1 - 5x2 &= 0 \end{aligned} Initial Simplex Tableau: | Basis | x1 | x2 | s1 | s2 | RHS | Ratio | | :--- | :--- | :--- | :--- | :--- | :--- | :--- | | s_1 | 1 | 2 | 1 | 0 | 20 | 20/2 = 10 (Pivot) | | s_2 | 3 | 1 | 0 | 1 | 30 | 30/1 = 30 | | Z | -3 | -5 | 0 | 0 | 0 | | • 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 R1: x2 = 0.5x1 + x2 + 0.5s_1 = 10 • New R2 = R2 - 1(R1') = 2.5x1 + 0x2 - 0.5s1 + s_2 = 20 • New Z = Z + 5(R1') = -0.5x1 + 0x2 + 2.5s1 + 0s_2 = 50 Next entering variable is x_1 (indicator -0.5). • Ratios: For R1: 10 / 0.5 = 20; For R2: 20 / 2.5 = 8 (Minimum ratio; s_2 leaves). Second Iteration (Pivot on 2.5 in Row 2): • Dividing R2 by 2.5: x1 + 0x2 - 0.2s1 + 0.4s_2 = 8 • Updating R1: R1 - 0.5(R2') \implies x2 = 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 (Cj - Zj \ge 0). Optimal Solution: x1 = 8, \quad x2 = 6, \quad \mathbf{Z_{\max} = 3(8) + 5(6) = 24 + 30 = 54}

A consumer has utility function U(x, y) = x^{0.4} y^{0.6}, where x and y are quantities consumed. Price of x is Px = 2, price of y is Py = 3, and consumer income is M = 120. a. Use Lagrange Multipliers to find the utility-maximizing quantities x and y. b. Determine the marginal utility of income (\lambda).

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.

Read the following scenario and answer the questions: *A national telecommunications enterprise in Nepal operates a nationwide optical fiber data network. The monthly demand for business data subscriptions (q1) and residential subscriptions (q2) are represented by the following inverse demand equations: P1 = 200 - 2q1 P2 = 150 - q2 The firm's joint monthly total cost function is given by: C(q1, q2) = q1^2 + 2q2^2 + q1 q2 + 10q1 + 20q2 + 2{,}000 Where q1 and q2 denote subscriptions in thousands, and prices/costs are in millions of Rupees.* Questions: a. Formulate the total revenue function TR(q1, q2) and the total profit function \pi(q1, q2). b. Determine the optimal monthly subscription levels q1 and q2 that maximize total profit. c. Calculate the optimal prices P1 and P2 to charge in each market segment. d. Verify the second-order sufficiency conditions using the Hessian matrix and compute the enterprise's maximum monthly profit.

Case Analysis: Joint Segment Profit Optimization a. Revenue and Profit Formulations \begin{aligned} TR(q1, q2) &= P1 q1 + P2 q2 = (200 - 2q1)q1 + (150 - q2)q2 \\ &= 200q1 - 2q1^2 + 150q2 - q2^2 \end{aligned} \begin{aligned} \pi(q1, q2) &= TR - C \\ &= (200q1 - 2q1^2 + 150q2 - q2^2) - (q1^2 + 2q2^2 + q1 q2 + 10q1 + 20q2 + 2{,}000) \\ &= -3q1^2 - 3q2^2 - q1 q2 + 190q1 + 130q2 - 2{,}000 \end{aligned} b. First-Order Necessary Conditions \frac{\partial \pi}{\partial q1} = -6q1 - q2 + 190 = 0 \implies 6q1 + q_2 = 190 \quad \text{--- (1)} \frac{\partial \pi}{\partial q2} = -q1 - 6q2 + 130 = 0 \implies q1 + 6q_2 = 130 \quad \text{--- (2)} From (1), q2 = 190 - 6q1. Substituting into (2): q1 + 6(190 - 6q1) = 130 q1 + 1140 - 36q1 = 130 \implies -35q1 = -1010 \implies q1 = \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 q1^2} = -6 \pi{22} = \frac{\partial^2 \pi}{\partial q2^2} = -6 \pi{12} = \pi{21} = -1 Principal minors: 1. H1 = \pi{11} = -6 0 Since H1 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}}

MTH 202 Business Mathematics II 2080 Model Questions with Solutions

Tribhuvan University

Faculty of Management

Office of the Dean

Official Model Question Paper / Dean's Office Blueprint

Course: MTH 202 · Business Mathematics II

Level: Bachelor of Business Management (BBM) · Semester 2

Full Marks: 60

Time: 3 hrs.

Candidates are required to give their answers in their own words as far as practicable. Figures in the margin indicate full marks.

Group A

Brief Answer Questions. Attempt ALL questions. (5 × 2 = 10)

[5*2=10]

[2]
Find the partial derivatives $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ for $z = 3x^2y^3 - 5x^3 + 4y^2 - 7$ .

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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 $y$ :
$\frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(3x^2y^3 - 5x^3 + 4y^2 - 7) = 9x^2y^2 + 8y$
[2]
Define a first-order differential equation and state its general linear form.

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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.
[2]
State Euler’s Theorem for a homogeneous production function $Q = f(K, L)$ of degree $n$ .

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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 ( $n = 1$ ):
$K \cdot MP_K + L \cdot MP_L = Q$
[2]
Evaluate the definite double integral: $\int_0^2 \int_0^1 (2x + 4y) \, dy \, dx$ .

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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 $x$ :
$\int_0^2 (2x + 2) \, dx = \left[ x^2 + 2x \right]_0^2 = (2^2 + 2(2)) - 0 = 4 + 4 = \mathbf{8}$
[2]
Formulate the dual of the following Linear Programming Problem:
$\text{Maximize } Z = 4x_1 + 6x_2 \quad \text{subject to } 2x_1 + x_2 \le 20, \; x_1 + 3x_2 \le 30, \; x_1, x_2 \ge 0$

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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:
$2y_1 + y_2 \ge 4$

$y_1 + 3y_2 \ge 6$

$y_1, y_2 \ge 0$

Group B

Short Answer Questions. Attempt any THREE questions. (3 × 10 = 30)

[3*10=30]

[10]
A firm produces two joint products, $X$ and $Y$ . The joint cost function is $C(x, y) = 2x^2 + xy + 3y^2 + 500$ . The demand functions are $P_x = 100 - x$ and $P_y = 120 - 2y$ . a. Formulate the total profit function $\pi(x, y)$ . b. Find the profit-maximizing output levels of $x$ and $y$ . c. Verify the second-order conditions for maximum profit.

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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.
[10]
Find the general solution and particular solution of the first-order differential equation:
$\frac{dy}{dx} + 2y = 4e^{3x}$
Given the initial condition $y(0) = 5$ .

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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 $y(0) = 5$

$5 = \frac{4}{5}e^{0} + C e^{0} = 0.8 + C \implies C = 5 - 0.8 = 4.2 = \frac{21}{5}$
$\mathbf{y(x) = \frac{4}{5}e^{3x} + \frac{21}{5}e^{-2x}}$
[10]
Solve the following Linear Programming Problem using the Simplex Method:
$\begin{aligned} \text{Maximize } & Z = 3x_1 + 5x_2 \\ \text{Subject to: } & x_1 + 2x_2 \le 20 \\ & 3x_1 + x_2 \le 30 \\ & x_1, x_2 \ge 0 \end{aligned}$
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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:

Basis $x_1$ $x_2$ $s_1$ $s_2$ RHS Ratio

$s_1$ 1 2 1 0 20 $20/2 = 10$ (Pivot)

$s_2$ 3 1 0 1 30 $30/1 = 30$

$Z$ -3 -5 0 0 0
- 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:
$x_1 = 8, \quad x_2 = 6, \quad \mathbf{Z_{\max} = 3(8) + 5(6) = 24 + 30 = 54}$
[10]
A consumer has utility function $U(x, y) = x^{0.4} y^{0.6}$ , where $x$ and $y$ are quantities consumed. Price of $x$ is $P_x = 2$ , price of $y$ is $P_y = 3$ , and consumer income is $M = 120$ . a. Use Lagrange Multipliers to find the utility-maximizing quantities $x$ and $y$ . b. Determine the marginal utility of income ( $\lambda$ ).

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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.

Basis	$x_1$	$x_2$	$s_1$	$s_2$	RHS	Ratio
$s_1$	1	2	1	0	20	$20/2 = 10$ (Pivot)
$s_2$	3	1	0	1	30	$30/1 = 30$
$Z$	-3	-5	0	0	0

Group C

Comprehensive Answer / Case Analysis Question. (1 × 20 = 20)

[1*20=20]

[20]
Read the following scenario and answer the questions:

A national telecommunications enterprise in Nepal operates a nationwide optical fiber data network. The monthly demand for business data subscriptions ( $q_1$ ) and residential subscriptions ( $q_2$ ) are represented by the following inverse demand equations:
$P_1 = 200 - 2q_1$

$P_2 = 150 - q_2$
The firm’s joint monthly total cost function is given by:
$C(q_1, q_2) = q_1^2 + 2q_2^2 + q_1 q_2 + 10q_1 + 20q_2 + 2{,}000$
Where $q_1$ and $q_2$ denote subscriptions in thousands, and prices/costs are in millions of Rupees.

Questions: a. Formulate the total revenue function $TR(q_1, q_2)$ and the total profit function $\pi(q_1, q_2)$ . b. Determine the optimal monthly subscription levels $q_1$ and $q_2$ that maximize total profit. c. Calculate the optimal prices $P_1$ and $P_2$ to charge in each market segment. d. Verify the second-order sufficiency conditions using the Hessian matrix and compute the enterprise’s maximum monthly profit.
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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:
$\pi_{\max} = -3(28.86)^2 - 3(16.86)^2 - (28.86)(16.86) + 190(28.86) + 130(16.86) - 2{,}000$

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

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

Business Mathematics II

Past exam papers

Business Mathematics II 2080 Official Model Question Paper

Faculty of Management

Brief Answer Questions. Attempt ALL questions. (5 × 2 = 10)

Partial Derivatives

First-Order Differential Equation

Euler’s Theorem for Homogeneous Functions

Evaluation of Double Integral

Dual Problem Formulation

Short Answer Questions. Attempt any THREE questions. (3 × 10 = 30)

Multivariable Profit Maximization

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

b. First-Order Necessary Conditions

c. Second-Order Sufficiency Conditions (Hessian Matrix)

Solution of Linear First-Order Differential Equation

1. Integrating Factor (IF)

2. General Solution

3. Particular Solution with $y(0) = 5$

Solution via Simplex Method

Initial Simplex Tableau:

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

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

Utility Maximization via Lagrange Multipliers

a. Lagrangian Formulation

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

Comprehensive Answer / Case Analysis Question. (1 × 20 = 20)

Case Analysis: Joint Segment Profit Optimization

a. Revenue and Profit Formulations

b. First-Order Necessary Conditions

c. Optimal Prices in Each Segment

d. Second-Order Sufficiency Conditions and Maximum Profit

Past exam papers

Business Mathematics II 2080 Official Model Question Paper

Partial Derivatives

First-Order Differential Equation

Euler’s Theorem for Homogeneous Functions

Evaluation of Double Integral

Dual Problem Formulation

Multivariable Profit Maximization

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

b. First-Order Necessary Conditions

c. Second-Order Sufficiency Conditions (Hessian Matrix)

Solution of Linear First-Order Differential Equation

1. Integrating Factor (IF)

2. General Solution

3. Particular Solution with y(0)=5y(0) = 5y(0)=5

Solution via Simplex Method

Initial Simplex Tableau:

First Iteration (Dividing Row 1 by 2 and eliminating x2x_2x2​):

Second Iteration (Pivot on 2.52.52.5 in Row 2):

Utility Maximization via Lagrange Multipliers

a. Lagrangian Formulation

b. Marginal Utility of Income (λ\lambdaλ)

Case Analysis: Joint Segment Profit Optimization

a. Revenue and Profit Formulations

b. First-Order Necessary Conditions

c. Optimal Prices in Each Segment

d. Second-Order Sufficiency Conditions and Maximum Profit

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

3. Particular Solution with $y(0) = 5$

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

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

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