MTH 202

Business Mathematics II

TU BBA · Semester 2 · BBA curriculum effective from 2021

Requirement: required
Credits: 3
Past papers: 2 papers

Syllabus

What this course covers and how the teaching time is divided.

Business Mathematics II Syllabus

Official TU PDF

Tribhuvan University

Faculty of Management

Office of the Dean

Bachelor of Business Administration (BBA)

Course Title: Business Mathematics II

Course Code: MTH 202

Semester: Semester 2

Nature of Course: required

Full Marks: 100

Pass Marks: 50

Credit Hours: 3 Cr.

Curriculum: BBA curriculum effective from 2021

Course Description

This course deals on integration and applications in production, first-order differential equationsand applications, dynamics of market price, linear inequalities and linear programming, linearalgebra and applications, numerical methods for solving systems of linear equations, input/outputanalysis, functions of several variables and their applications in business and economics,difference equations and dynamic economic analysis.

Course Objective

The course introduces mathematical techniques through examples of their application toeconomic and business concepts. It also tries to get students tackling problems in economics andbusiness using these techniques as soon as possible so that they can see how useful they are.

Course Contents:

Lecture hours show the approximate classroom time allocated to each unit.

Concept of integration, Techniques of integration (Standard form, substitution method andintegration by parts), Integration of algebraic, logarithmic and exponential functions, Methods of evaluating definite integrals, Application of integration in business andeconomics ( including consumers’ surplus and producers’ surplus), Area under a curve.

Linear Inequalities in two variables, Introduction to linear programming problem,Formulation of linear programming problem, Methods of solving linear programingproblems: Graphical method, Simplex method (two variables), Duality and standardminimization linear programing problems.Online: Simplex method

Introduction to matrices and types of matrices, Operation on matrices: Addition andsubtraction of matrices, Scalar multiplication of a matrix, Multiplication of matrices,Transpose of matrix, Determinant of a square matrix, Minors and cofactors of matrix,Singular and non-singular of matrix, Adjoint and inverse matrices, Elementary rowoperations, Methods of solving linear equations: Cramer’s rule, Inverse matrix method(Application of linear equation for solving business and economics related problems), Gauss elimination method and Gauss- Jordan method, Input/output analysis, Technologycoefficient matrix, Hawkins-Simon conditions for the viability of the system.Lab. Work: Excel for linear algebraOnline: Gauss elimination method for solving system of linear equations, Gauss-Jordanmethod for solving system of linear equations and finding inverse matrices.

Functions of several variables, Applications of functions of two variables in business andeconomics, Partial differentiation, Applications of partial differentiation, Elasticity ofdemand, Utility, Production, Graphical Representations, Unconstrained optimization,constrained optimization and Lagrange multipliers.

Introduction to differential equation, Order and degree of differential equation, Solution ofdifferential equations, First-order linear differential equations with constant coefficient and constant term, Differential equation for limited and unlimited growth, Dynamics ofmarket price, First order differential equation with variable- coefficient and variable term, Exact differential equations, Nonlinear differential equations of the first order and firstdegree.

Introduction to difference equations, Solution of first order difference equations(homogeneous and non-homogeneous), Economic applications of first order differenceequation: Cobweb model, Lagged Keynesian macroeconomic model, Duopoly priceadjustment.

Chapter-wise questions

42 reviewed questions across 6 units

Open a chapter to study questions grouped by unit and syllabus topic, with verified model solutions.

Unit 1: Integration and applications

Report problem

Asked on 2024 Exam[1 marks]
$Integrate: \int \frac{x+2}{x-5} dx$
Asked on 2024 Exam[1 marks]
Find the area of the region bounded by the curve $y = \frac{3}{2}x^2$ , x axis and the two ordinates x=0 and x=2.
Asked on 2024 Exam[1 marks]
The marginal cost function of manufacturing xxx units of a product is given by MC = $3x^2 - 10x + 3$ . The total cost of producing one unit of the product is Rs. 7. Find average cost function.
Asked on 2024 Exam[3 marks]
Integrate the following:

(a) $\int \frac{\log x + 5}{x} dx $

(b) $\int (2x + 1)\sqrt{5x + 2} dx$
Asked on 2024 Exam[3 marks]
The demand function for a product is $P_d = 8 - 2x$ and the supply function is $P_s = 2x$ under pure competition. Find the consumer’s surplus and producer’s surplus.
Asked on 2024 Exam[3 marks]
Solve the differential equations: $ \frac{dy}{dx} + \frac{1}{x} \cdot y = x^3$.
Asked on 2023 Exam[2 marks]
Integrate: $\int \left( x^2 + e^x + \frac{1}{x} + 1 \right) dx$
Asked on 2023 Exam[2 marks]
Find the area of curve bounded by x-axis and ordinates of $y = 7x^2$ from $x_1 = 0$ and $x_2 = 2.$
Asked on 2023 Exam[5 marks]
Integrate the following:

(a) $\int \frac{\log x}{x} dx$

$(b) \int \frac{3x + 1}{\sqrt{5x + 2}} dx$
Asked on 2023 Exam[5 marks]
The annual rate of repair of a machine is given by $\frac{dc}{dt} = 5t + 12.5$ , where “t” is the age of machine in years, $\frac{dc}{dt}$ is Rs per year. Find total cost after 4 years.

Past exam papers

Complete papers are arranged by exam year (AD).

Business Mathematics Ii 2024 Board Question Paper

Report problem

Tribhuvan University

Faculty of Management

Office of the Dean

2024 AD / Regular Examination

Course: MTH 202 · Business Mathematics II

Level: Bachelor of Business Administration (BBA) · Semester 2

Full Marks: 100

Time: 3 hrs.

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

Section A

Brief Answer Questions .

[10*1=10]

$Integrate: \int \frac{x+2}{x-5} dx$

[1]
View model solution
Step-by-Step Solution:

Evaluate the indefinite integral:
$I = \int \frac{x+2}{x-5} \, dx$

Step 1: Perform algebraic division / rearrangement

Express the numerator in terms of $(x - 5)$ :
$\frac{x + 2}{x - 5} = \frac{(x - 5) + 7}{x - 5} = \frac{x - 5}{x - 5} + \frac{7}{x - 5} = 1 + \frac{7}{x - 5}$

Step 2: Integrate term by term

$I = \int \left( 1 + \frac{7}{x - 5} \right) dx$
$I = \int 1 \, dx + 7 \int \frac{1}{x - 5} \, dx$
$I = x + 7 \ln|x - 5| + C$
Final Answer: $x + 7 \ln|x - 5| + C$ .
Find the area of the region bounded by the curve $y = \frac{3}{2}x^2$ , x axis and the two ordinates x=0 and x=2.

[1]
View model solution
Step-by-Step Solution:

Given:
- Curve: $y = \frac{3}{2}x^2$
- Bounded by the $x$ -axis and ordinates $x = 0$ and $x = 2$ .
Step 1: Set up the definite integral

Since $y \ge 0$ on $[0, 2]$ , the area $A$ is:
$A = \int_0^2 \frac{3}{2}x^2 \, dx$

Step 2: Evaluate the integral

$A = \frac{3}{2} \left[ \frac{x^3}{3} \right]_0^2 = \frac{1}{2} \left[ x^3 \right]_0^2$
$A = \frac{1}{2} (2^3 - 0^3) = \frac{1}{2}(8) = 4 \text{ sq. units}$
Final Answer: The area of the region is $4$ square units.
Find $f_{xx}$ when f(x, y) = $3x^2y^2 + 5x^2y$

[1]
View model solution
Step-by-Step Solution:

Given function:
$f(x, y) = 3x^2 y^2 + 5x^2 y$

Step 1: Compute first-order partial derivative with respect to $x$ ( $f_x$ )

Treating $y$ as a constant:
$f_x = \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(3x^2 y^2) + \frac{\partial}{\partial x}(5x^2 y)$

$f_x = 3y^2(2x) + 5y(2x) = 6xy^2 + 10xy$

Step 2: Compute second-order partial derivative $f_{xx}$

Differentiating $f_x$ again with respect to $x$ (treating $y$ as constant):
$f_{xx} = \frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}(6xy^2 + 10xy)$

$f_{xx} = 6y^2(1) + 10y(1) = 6y^2 + 10y$

Final Answer: $f_{xx} = 6y^2 + 10y$ .
$If A = \begin{bmatrix} 1 & 2 \\ 3 & 1 \end{bmatrix}, find A^2$

[1]
View model solution
Step-by-Step Solution:

Given matrix:
$A = \begin{bmatrix} 1 & 2 \\ 3 & 1 \end{bmatrix}$

Step 1: Multiply $A$ by $A$

$A^2 = A \cdot A = \begin{bmatrix} 1 & 2 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 1 \end{bmatrix}$
Evaluating each entry by row-column dot product:
- $c_{11} = (1)(1) + (2)(3) = 1 + 6 = 7$
- $c_{12} = (1)(2) + (2)(1) = 2 + 2 = 4$
- $c_{21} = (3)(1) + (1)(3) = 3 + 3 = 6$
- $c_{22} = (3)(2) + (1)(1) = 6 + 1 = 7$
$A^2 = \begin{bmatrix} 7 & 4 \\ 6 & 7 \end{bmatrix}$
Final Answer: $A^2 = \begin{bmatrix} 7 & 4 \\ 6 & 7 \end{bmatrix}$ .
Evaluate: $\begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{bmatrix} $

[1]
View model solution
Step-by-Step Solution:

Evaluate the determinant:
$\Delta = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix}$

Method 1: Expansion along Row 1

$\Delta = 1 \begin{vmatrix} 3 & 4 \\ 4 & 5 \end{vmatrix} - 2 \begin{vmatrix} 2 & 4 \\ 3 & 5 \end{vmatrix} + 3 \begin{vmatrix} 2 & 3 \\ 3 & 4 \end{vmatrix}$
Evaluate the $2 \times 2$ determinants:
- $\begin{vmatrix} 3 & 4 \\ 4 & 5 \end{vmatrix} = (3)(5) - (4)(4) = 15 - 16 = -1$
- $\begin{vmatrix} 2 & 4 \\ 3 & 5 \end{vmatrix} = (2)(5) - (4)(3) = 10 - 12 = -2$
- $\begin{vmatrix} 2 & 3 \\ 3 & 4 \end{vmatrix} = (2)(4) - (3)(3) = 8 - 9 = -1$
Substitute into the expression:
$\Delta = 1(-1) - 2(-2) + 3(-1) = -1 + 4 - 3 = 0$

Method 2: Elementary Row Operations

Perform $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_2$ :
$\Delta = \begin{vmatrix} 1 & 2 & 3 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix}$
Since row 2 and row 3 are identical, the determinant is identically $0$ .

Final Answer: The value is $0$ .
The input-output coefficient matrix of an economy of two industries is

$\begin{bmatrix} 0.8 & 0.2 \\ 0.9 & 0.7 \end{bmatrix}$

Test whether the system is viable as per Hawkins-Simon conditions.

[1]
View model solution
Step-by-Step Solution:

Given input-output coefficient matrix:
$A = \begin{bmatrix} 0.8 & 0.2 \\ 0.9 & 0.7 \end{bmatrix}$

Step 1: Form the Leontief matrix $(I - A)$

$I - A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 0.8 & 0.2 \\ 0.9 & 0.7 \end{bmatrix} = \begin{bmatrix} 1 - 0.8 & -0.2 \\ -0.9 & 1 - 0.7 \end{bmatrix} = \begin{bmatrix} 0.2 & -0.2 \\ -0.9 & 0.3 \end{bmatrix}$
Step 2: Check Hawkins-Simon Condition 1

The principal diagonal elements must be positive:
$(1 - a_{11}) = 0.2 > 0 \quad (\text{Satisfied})$

$(1 - a_{22}) = 0.3 > 0 \quad (\text{Satisfied})$

Step 3: Check Hawkins-Simon Condition 2

The determinant $|I - A|$ must be strictly positive:
$|I - A| = \begin{vmatrix} 0.2 & -0.2 \\ -0.9 & 0.3 \end{vmatrix} = (0.2)(0.3) - (-0.2)(-0.9)$

$|I - A| = 0.06 - 0.18 = -0.12$

Since $|I - A| = -0.12 < 0$ , the second Hawkins-Simon condition fails.

Economic Interpretation:

The economy requires more inputs than it can produce as final outputs, meaning the system is incapable of sustaining positive net production.

Final Answer: The system is NOT viable as per Hawkins-Simon conditions because $|I - A| = -0.12 < 0$ .
Determine the order and degree of the differential equations: $\left( \frac{dy}{dx} \right)^2 - \left( \frac{dy}{dx} \right)^5 + 7xy = 0. .$

[1]
View model solution
Step-by-Step Solution:

Given differential equation:
$\left( \frac{dy}{dx} \right)^2 - \left( \frac{dy}{dx} \right)^5 + 7xy = 0$

1. Determine the Order:
- The order of a differential equation is the order of the highest derivative present.
- The only derivative present is the first derivative $\frac{dy}{dx}$ .
- Therefore:
  $\mathbf{\text{Order} = 1}$
2. Determine the Degree:
- The degree is the exponent of the highest power of the highest-order derivative occurring in polynomial form.
- The highest-order derivative is $\frac{dy}{dx}$ , and its highest power is $5$ .
- Therefore:
  $\mathbf{\text{Degree} = 5}$
Final Answer:
- Order = 1
- Degree = 5
Solve the differential equation: $\frac{dy}{dx} = \frac{1}{x}$

[1]
View model solution
Step-by-Step Solution:

Given differential equation:
$\frac{dy}{dx} = \frac{1}{x}$

Step 1: Separate variables

$dy = \frac{1}{x} \, dx$
Step 2: Integrate both sides

$\int dy = \int \frac{1}{x} \, dx$
$y(x) = \ln|x| + C$
Final Answer: $y = \ln|x| + C$ (where $C$ is an arbitrary constant of integration).
Solve the first order homogeneous difference equations: $y_{t+1} - 2y_t = 0, y_0 = 3.$

[1]
View model solution
Step-by-Step Solution:

Given first-order homogeneous difference equation:
$y_{t+1} - 2y_t = 0 \quad \text{with } y_0 = 3$

Step 1: Rewrite in standard recursive form

$y_{t+1} = 2y_t$
or with index shifted:
$y_t = 2y_{t-1}$

Step 2: General solution

The general solution of a homogeneous first-order difference equation $y_t = b y_{t-1}$ is:
$y_t = A \cdot b^t$
Here $b = 2$ , so:
$y_t = A \cdot 2^t$

Step 3: Apply the initial condition $y_0 = 3$

At $t = 0$ :
$y_0 = A \cdot 2^0 = A = 3$

Thus, the particular solution is:
$y_t = 3 \cdot 2^t$

Final Answer: $y_t = 3 \cdot 2^t$ .
The marginal cost function of manufacturing xxx units of a product is given by MC = $3x^2 - 10x + 3$ . The total cost of producing one unit of the product is Rs. 7. Find average cost function.

[1]
View model solution
Step-by-Step Solution:

Given:
- Marginal Cost function: $MC = 3x^2 - 10x + 3$
- Total cost of producing $1$ unit: $TC(1) = \text{Rs } 7$
Step 1: Find the Total Cost function $TC(x)$ by integration

$TC(x) = \int MC \, dx = \int (3x^2 - 10x + 3) \, dx$
$TC(x) = 3\left(\frac{x^3}{3}\right) - 10\left(\frac{x^2}{2}\right) + 3x + C$
$TC(x) = x^3 - 5x^2 + 3x + C$
Step 2: Determine the integration constant ( $C$ ) using $TC(1) = 7$

$7 = 1^3 - 5(1^2) + 3(1) + C$
$7 = 1 - 5 + 3 + C$
$7 = -1 + C \implies C = 7 + 1 = 8$
Thus, the Total Cost function is:
$TC(x) = x^3 - 5x^2 + 3x + 8$

Step 3: Find the Average Cost function $AC(x)$

$AC(x) = \frac{TC(x)}{x} = \frac{x^3 - 5x^2 + 3x + 8}{x}$
$AC(x) = x^2 - 5x + 3 + \frac{8}{x}$
Final Answer: The Average Cost function is $AC(x) = x^2 - 5x + 3 + \frac{8}{x}$ .

Section B

Short Answer Questions : (Attempt any FIVE Questions )

[5*3=15]

Integrate the following:

(a) $\int \frac{\log x + 5}{x} dx $

(b) $\int (2x + 1)\sqrt{5x + 2} dx$

[3]
View model solution
Step-by-Step Solution:

(a) Evaluate $\int \frac{\log x + 5}{x} \, dx$ :

Let $u = \log x + 5 = \ln x + 5$ . Differentiating with respect to $x$ :
$du = \frac{1}{x} \, dx$

Substitute into the integral:
$I = \int u \, du = \frac{u^2}{2} + C$

Substitute back $u = \log x + 5$ :
$I = \frac{(\log x + 5)^2}{2} + C$
(Alternatively: $\int \frac{\ln x}{x} dx + 5 \int \frac{1}{x} dx = \frac{(\ln x)^2}{2} + 5\ln x + C_1$ , which is algebraically equivalent up to a constant).

(b) Evaluate $\int (2x + 1)\sqrt{5x + 2} \, dx$ :

Let $u = 5x + 2 \implies 5x = u - 2 \implies x = \frac{u - 2}{5}$ . Differentiating gives:
$dx = \frac{1}{5} \, du$

Express the term $(2x + 1)$ in terms of $u$ :
$2x + 1 = 2\left(\frac{u - 2}{5}\right) + 1 = \frac{2u - 4 + 5}{5} = \frac{2u + 1}{5}$

Substitute into the integral:
$I = \int \left( \frac{2u + 1}{5} \right) \sqrt{u} \cdot \left( \frac{1}{5} \, du \right) = \frac{1}{25} \int (2u + 1) u^{1/2} \, du$

$I = \frac{1}{25} \int \left( 2u^{3/2} + u^{1/2} \right) du$

Integrate term by term:
$I = \frac{1}{25} \left[ 2 \cdot \frac{u^{5/2}}{5/2} + \frac{u^{3/2}}{3/2} \right] + C$

$I = \frac{1}{25} \left[ \frac{4}{5} u^{5/2} + \frac{2}{3} u^{3/2} \right] + C$
Factor out common terms $\frac{2}{15} u^{3/2}$ :
$I = \frac{1}{25} \cdot \frac{2}{15} u^{3/2} (6u + 5) + C = \frac{2}{375} u^{3/2} (6u + 5) + C$

Substitute back $u = 5x + 2$ :
$6u + 5 = 6(5x + 2) + 5 = 30x + 12 + 5 = 30x + 17$

$I = \frac{2}{375} (30x + 17)(5x + 2)^{3/2} + C$

Final Answer:
- (a) $\frac{(\log x + 5)^2}{2} + C$
- (b) $\frac{2}{375}(30x + 17)(5x + 2)^{3/2} + C$
The demand function for a product is $P_d = 8 - 2x$ and the supply function is $P_s = 2x$ under pure competition. Find the consumer’s surplus and producer’s surplus.

[3]
View model solution
Step-by-Step Solution:

Given market functions under pure competition:
- Demand function: $P_d = 8 - 2x$
- Supply function: $P_s = 2x$ where $x$ denotes quantity and $P$ denotes price.
Step 1: Find Market Equilibrium Quantity ( $x^*$ ) and Price ( $P^*$ )

At equilibrium, $P_d = P_s$ :
$8 - 2x = 2x$

$8 = 4x \implies x^* = 2 \text{ units}$

Substitute $x^* = 2$ into either equation:
$P^* = 2(2) = 4$

Step 2: Compute Consumer’s Surplus ( $CS$ )

Consumer’s surplus is the integral of $(P_d - P^*)$ from $0$ to $x^*$ :
$CS = \int_0^{x^*} (P_d - P^*) \, dx = \int_0^2 (8 - 2x - 4) \, dx = \int_0^2 (4 - 2x) \, dx$

$CS = \left[ 4x - x^2 \right]_0^2 = 4(2) - (2)^2 = 8 - 4 = 4$

(Geometrically: Triangle area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times (8 - 4) = 4$ ).

Step 3: Compute Producer’s Surplus ( $PS$ )

Producer’s surplus is the integral of $(P^* - P_s)$ from $0$ to $x^*$ :
$PS = \int_0^{x^*} (P^* - P_s) \, dx = \int_0^2 (4 - 2x) \, dx$

$PS = \left[ 4x - x^2 \right]_0^2 = 4(2) - (2)^2 = 8 - 4 = 4$

(Geometrically: Triangle area $= \frac{1}{2} \times 2 \times (4 - 0) = 4$ ).

Step 4: Total Surplus

$TS = CS + PS = 4 + 4 = 8$
Final Answer:
- Equilibrium: $x^* = 2 \text{ units}$ , $P^* = 4$
- Consumer’s Surplus: $4$
- Producer’s Surplus: $4$
Solve the following system of linear equations using Cramer’s rule or matrix method:

4x+6z=100

3x+6y+z=100

3x+4y+3z=100

[3]
View model solution
Step-by-Step Solution:

Given system of linear equations:
$4x + 0y + 6z = 100 \quad \text{--- (1)}$

$3x + 6y + z = 100 \quad \text{--- (2)}$

$3x + 4y + 3z = 100 \quad \text{--- (3)}$

Method: Cramer’s Rule

Step 1: Calculate the Coefficient Determinant ( $D$ )

$D = \begin{vmatrix} 4 & 0 & 6 \\ 3 & 6 & 1 \\ 3 & 4 & 3 \end{vmatrix}$
Expand along Row 1:
$D = 4 \begin{vmatrix} 6 & 1 \\ 4 & 3 \end{vmatrix} - 0 + 6 \begin{vmatrix} 3 & 6 \\ 3 & 4 \end{vmatrix}$

$D = 4(18 - 4) + 6(12 - 18) = 4(14) + 6(-6) = 56 - 36 = 20$

Step 2: Calculate $D_x$

Replace Column 1 with constant terms:
$D_x = \begin{vmatrix} 100 & 0 & 6 \\ 100 & 6 & 1 \\ 100 & 4 & 3 \end{vmatrix} = 100 \begin{vmatrix} 1 & 0 & 6 \\ 1 & 6 & 1 \\ 1 & 4 & 3 \end{vmatrix}$
Expand along Row 1:
$D_x = 100 \left[ 1(18 - 4) - 0 + 6(4 - 6) \right] = 100 [ 14 + 6(-2) ] = 100 [ 14 - 12 ] = 100(2) = 200$

$x = \frac{D_x}{D} = \frac{200}{20} = 10$

Step 3: Calculate $D_y$

Replace Column 2 with constant terms:
$D_y = \begin{vmatrix} 4 & 100 & 6 \\ 3 & 100 & 1 \\ 3 & 100 & 3 \end{vmatrix} = 100 \begin{vmatrix} 4 & 1 & 6 \\ 3 & 1 & 1 \\ 3 & 1 & 3 \end{vmatrix}$
Expand along Row 1:
$D_y = 100 \left[ 4(3 - 1) - 1(9 - 3) + 6(3 - 3) \right] = 100 [ 4(2) - 6 + 0 ] = 100 [ 8 - 6 ] = 100(2) = 200$

$y = \frac{D_y}{D} = \frac{200}{20} = 10$

Step 4: Calculate $D_z$

Replace Column 3 with constant terms:
$D_z = \begin{vmatrix} 4 & 0 & 100 \\ 3 & 6 & 100 \\ 3 & 4 & 100 \end{vmatrix} = 100 \begin{vmatrix} 4 & 0 & 1 \\ 3 & 6 & 1 \\ 3 & 4 & 1 \end{vmatrix}$
Expand along Row 1:
$D_z = 100 \left[ 4(6 - 4) - 0 + 1(12 - 18) \right] = 100 [ 4(2) + (-6) ] = 100 [ 8 - 6 ] = 100(2) = 200$

$z = \frac{D_z}{D} = \frac{200}{20} = 10$

Verification:
- $4(10) + 6(10) = 40 + 60 = 100$ (Matches)
- $3(10) + 6(10) + 10 = 30 + 60 + 10 = 100$ (Matches)
- $3(10) + 4(10) + 3(10) = 30 + 40 + 30 = 100$ (Matches)
Final Answer: $x = 10$ , $y = 10$ , $z = 10$ .
A firm’s production function is given by the equation Q = $100L^{0.5}K^{0.5} $ . Use partial differentiation to find the approximate change in output Q when the firm’s manager increases capital by 5% and decreases labour by 3%.

[3]
View model solution
Step-by-Step Solution:

Given production function:
$Q = 100 L^{0.5} K^{0.5}$
where $L$ = labor and $K$ = capital.

Step 1: Compute the partial derivatives
- With respect to labor ( $L$ ):
  $\frac{\partial Q}{\partial L} = 100(0.5 L^{-0.5} K^{0.5}) = 50 L^{-0.5} K^{0.5} = 0.5 \frac{Q}{L}$
- With respect to capital ( $K$ ):
  $\frac{\partial Q}{\partial K} = 100(0.5 L^{0.5} K^{-0.5}) = 50 L^{0.5} K^{-0.5} = 0.5 \frac{Q}{K}$
Step 2: Use the Total Differential formula

The total differential $dQ$ approximates the change in output:
$dQ = \frac{\partial Q}{\partial L} dL + \frac{\partial Q}{\partial K} dK$

$dQ = \left( 0.5 \frac{Q}{L} \right) dL + \left( 0.5 \frac{Q}{K} \right) dK$

Divide both sides by $Q$ to express in terms of percentage/proportional changes:
$\frac{dQ}{Q} = 0.5 \left(\frac{dL}{L}\right) + 0.5 \left(\frac{dK}{K}\right)$

Step 3: Substitute the given percentage changes
- Labor decreases by $3\%$ :
  $\frac{dL}{L} = -3\% = -0.03$
- Capital increases by $5\%$ :
  $\frac{dK}{K} = +5\% = +0.05$
Substitute these values:
$\frac{dQ}{Q} = 0.5(-0.03) + 0.5(+0.05)$

$\frac{dQ}{Q} = -0.015 + 0.025 = +0.010 = +1.0\%$

Final Answer: The approximate change in output $Q$ is an increase of $1\%$ .
Solve the differential equations: $ \frac{dy}{dx} + \frac{1}{x} \cdot y = x^3$.

[3]
View model solution
Step-by-Step Solution:

Given first-order linear differential equation:
$\frac{dy}{dx} + \frac{1}{x} \cdot y = x^3$

Step 1: Identify standard form and integrating factor

Standard linear form: $\frac{dy}{dx} + P(x)y = Q(x)$ Here:
$P(x) = \frac{1}{x}, \quad Q(x) = x^3$

The integrating factor $I(x)$ is:
$I(x) = e^{\int P(x) \, dx} = e^{\int \frac{1}{x} \, dx} = e^{\ln x} = x$

Step 2: Multiply the differential equation by $I(x)$

$x \left( \frac{dy}{dx} + \frac{1}{x} y \right) = x(x^3)$
$\frac{d}{dx}[y \cdot x] = x^4$
Step 3: Integrate both sides with respect to $x$

$\int \frac{d}{dx}[y \cdot x] \, dx = \int x^4 \, dx$
$y \cdot x = \frac{x^5}{5} + C$
Step 4: Solve explicitly for $y$

Divide both sides by $x$ :
$y(x) = \frac{x^4}{5} + \frac{C}{x}$

Final Answer: $y(x) = \frac{x^4}{5} + \frac{C}{x}$ (where $C$ is an arbitrary constant of integration).
The reaction functions for the two duopolists, firms X and Y are given by the equations $ P_t^X = 45 + 0.8 , P_{t-1}^Y$ and $P_t^Y = 45 + 0.8P_{t-1}^X$ , respectively. If the assumptions of the Bertrand model hold, derive a difference equation for $P_t^X$ and calculate what $ P_t^X$ will be in time period 10 if firm X starts off in time period 0 by setting a price of 300.

[3]
View model solution
Step-by-Step Solution:

Given Bertrand Duopoly Reaction Functions:

$P_t^X = 45 + 0.8 P_{t-1}^Y \quad \text{--- (1)}$
$P_t^Y = 45 + 0.8 P_{t-1}^X \quad \text{--- (2)}$
with initial price set by Firm X in period 0: $P_0^X = 300$ .

Step 1: Derive a single difference equation for $P_t^X$

Lag equation (2) by one time period to express $P_{t-1}^Y$ :
$P_{t-1}^Y = 45 + 0.8 P_{t-2}^X$

Substitute this into equation (1):
$P_t^X = 45 + 0.8 (45 + 0.8 P_{t-2}^X)$

$P_t^X = 45 + 36 + 0.64 P_{t-2}^X$

$P_t^X = 81 + 0.64 P_{t-2}^X$

$P_t^X - 0.64 P_{t-2}^X = 81$

This is a linear second-order difference equation with two-period step dynamics.

Step 2: Solve the difference equation for even periods ( $t = 2k$ )

Let $k = t / 2$ . Then:
$P_{2k}^X = 81 + 0.64 P_{2(k-1)}^X$
1. Equilibrium price ( $P^*$ ):
  $P^* = 81 + 0.64 P^* \implies 0.36 P^* = 81 \implies P^* = \frac{81}{0.36} = 225$
2. General solution for even periods:
  $P_{2k}^X = P^* + (P_0^X - P^*)(0.64)^k$
  
  $P_{2k}^X = 225 + (300 - 225)(0.64)^k = 225 + 75(0.64)^k$
Step 3: Calculate $P_{10}^X$ (Period 10, so $k = 10/2 = 5$ )

$P_{10}^X = 225 + 75(0.64)^5$
Compute $(0.64)^5$ :
$(0.64)^5 \approx 0.107374$

$P_{10}^X = 225 + 75(0.107374) = 225 + 8.053 = 233.05$

(Sequential check: $P_0^X=300 \implies P_1^Y=285 \implies P_2^X=273 \implies P_4^X=255.72 \implies P_6^X=244.66 \implies P_8^X=237.58 \implies P_{10}^X=233.05$ ).

Final Answer:
- Difference Equation: $P_t^X = 81 + 0.64 P_{t-2}^X$
- Price in period 10 ( $P_{10}^X$ ): $233.05$
Find the time path of the national income $y_t$ of the following data and comment on the stability of the time path. $y_t = C_t + I_t, $ $C_t = 200 + 0.75y_{t-1},$ , $ I_t = 50 + 0.15y_{t-1}$ and $y_0 = 3000.$

[3]
View model solution
Step-by-Step Solution:

Given dynamic macroeconomic data:
- National income: $y_t = C_t + I_t$
- Consumption function: $C_t = 200 + 0.75y_{t-1}$
- Investment function: $I_t = 50 + 0.15y_{t-1}$
- Initial national income: $y_0 = 3,000$
Step 1: Formulate the First-Order Difference Equation

Substitute $C_t$ and $I_t$ into $y_t$ :
$y_t = (200 + 0.75y_{t-1}) + (50 + 0.15y_{t-1})$

$y_t = (200 + 50) + (0.75 + 0.15)y_{t-1}$

$y_t = 250 + 0.90 y_{t-1}$

$y_t - 0.90 y_{t-1} = 250$

Step 2: Determine Particular Solution ( $y^*$ / Equilibrium Level)

Set $y_t = y_{t-1} = y^*$ :
$y^* - 0.90 y^* = 250$

$0.10 y^* = 250 \implies y^* = \frac{250}{0.10} = 2,500$

Step 3: Determine Complementary Function ( $y_c$ ) and General Time Path

Homogeneous equation: $y_t - 0.90 y_{t-1} = 0 \implies y_c = A(0.90)^t$ . General solution:
$y_t = y^* + A(0.90)^t = 2,500 + A(0.90)^t$

Apply the initial condition $y_0 = 3,000$ :
$3,000 = 2,500 + A(0.90)^0 \implies A = 3,000 - 2,500 = 500$

Thus, the exact time path is:
$y_t = 2,500 + 500(0.90)^t$

Step 4: Comment on Stability of the Time Path

In the solution $y_t = y^* + A(b)^t$ , the base is $b = 0.90$ :
1. Since $|b| = 0.90 < 1$ , the term $(0.90)^t \to 0$ as $t \to \infty$ .
2. Since $b > 0$ , the deviation converges without oscillation (non-oscillatory convergence).
3. Therefore, the time path is dynamically stable and converges monotonically to the intertemporal equilibrium income of $2,500$ .
Final Answer:
- Time path: $y_t = 2,500 + 500(0.90)^t$
- Stability: Dynamically stable (monotonically converges to $2,500$ as $t \to \infty$ ).

Section C

Long Answer Questions : (Attempt any THREE Questions )

[3*5=15]

Solve the following LP problem using the simplex method or graphic method:

Maximize Z = $5x_1 + 3x_2 $

.Subject to the constraints

$ 2x_1 + x_2 \leq 5$

$ x_1 + x_2 \leq 4 $

and x1,x2 ≥ 0

[5]

View model solution

Step-by-Step Solution:

Problem Formulation:

\text{Maximize } Z = 5x_1 + 3x_2

\text{Subject to the constraints:}

2x_1 + x_2 \le 5 \quad \text{--- (1)}

x_1 + x_2 \le 4 \quad \text{--- (2)}

x_1 \ge 0, \quad x_2 \ge 0

Method 1: Graphical Method

Step 1: Plot the constraint boundary lines

Line 1 ( $2x_1 + x_2 = 5$ ):
- When $x_1 = 0 \implies x_2 = 5 \implies (0, 5)$
- When $x_2 = 0 \implies 2x_1 = 5 \implies x_1 = 2.5 \implies (2.5, 0)$
Line 2 ( $x_1 + x_2 = 4$ ):
- When $x_1 = 0 \implies x_2 = 4 \implies (0, 4)$
- When $x_2 = 0 \implies x_1 = 4 \implies (4, 0)$

Step 2: Find intersection point of the two constraint lines

Subtract equation (2) from equation (1):

(2x_1 + x_2) - (x_1 + x_2) = 5 - 4 \implies x_1 = 1

Substitute

x_1 = 1

into equation (2):

1 + x_2 = 4 \implies x_2 = 3

The intersection point is

(1, 3)

Step 3: Evaluate the objective function $Z = 5x_1 + 3x_2$ at all corner points

The feasible region is bounded by the convex polygon with vertices:

Corner Point $(x_1, x_2)$	$Z = 5x_1 + 3x_2$	Value
$(0, 0)$	$5(0) + 3(0)$	$0$
$(2.5, 0)$	$5(2.5) + 3(0)$	$12.5$
$(1, 3)$	$5(1) + 3(3) = 5 + 9$	$14$ (Optimal Maximum)
$(0, 4)$	$5(0) + 3(4)$	$12$

Method 2: Simplex Method

Step 1: Standard Canonical Form

Introduce non-negative slack variables $s_1 \ge 0, s_2 \ge 0$ :

2x_1 + x_2 + s_1 = 5

x_1 + x_2 + s_2 = 4

Z - 5x_1 - 3x_2 + 0s_1 + 0s_2 = 0

Tableau 1 (Initial):

Basic	$x_1$	$x_2$	$s_1$	$s_2$	RHS	Ratio
$s_1$	2	1	1	0	5	$5/2 = 2.5$ (Pivot)
$s_2$	1	1	0	1	4	$4/1 = 4$
$Z$	-5	-3	0	0	0

Entering variable: $x_1$ (most negative indicator $-5$ ).
Leaving variable: $s_1$ (minimum ratio $2.5$ ).
Pivot element: $2$ .
Perform row operations:
- $R_1 \leftarrow R_1 / 2$ : $[1, 0.5, 0.5, 0 \mid 2.5]$
- $R_2 \leftarrow R_2 - R_1$ : $[0, 0.5, -0.5, 1 \mid 1.5]$
- $R_Z \leftarrow R_Z + 5R_1$ : $[0, -0.5, 2.5, 0 \mid 12.5]$

Tableau 2:

Basic	$x_1$	$x_2$	$s_1$	$s_2$	RHS	Ratio
$x_1$	1	0.5	0.5	0	2.5	$2.5 / 0.5 = 5$
$s_2$	0	0.5	-0.5	1	1.5	$1.5 / 0.5 = 3$ (Pivot)
$Z$	0	-0.5	2.5	0	12.5

Entering variable: $x_2$ (indicator $-0.5$ ).
Leaving variable: $s_2$ (minimum ratio $3$ ).
Pivot element: $0.5$ .
Perform row operations:
- $R_2 \leftarrow R_2 / 0.5$ : $[0, 1, -1, 2 \mid 3]$
- $R_1 \leftarrow R_1 - 0.5 R_2$ : $[1, 0, 1, -1 \mid 1]$
- $R_Z \leftarrow R_Z + 0.5 R_2$ : $[0, 0, 2, 1 \mid 14]$

Tableau 3 (Optimal):

Basic	$x_1$	$x_2$	$s_1$	$s_2$	RHS
$x_1$	1	0	1	-1	1
$x_2$	0	1	-1	2	3
$Z$	0	0	2	1	14

All objective row coefficients are $\ge 0$ . The optimal solution is reached:

$x_1 = 1$
$x_2 = 3$
$s_1 = 0, s_2 = 0$
$Z_{\text{max}} = 14$

Final Answer: The optimal solution is $x_1 = 1$ , $x_2 = 3$ , with maximum value $Z = 14$ .

For the following transaction matrix of two sector economy consisting two industries P and Q, calculate the gross output for each industry if the final demand changes to 18 units for P and 44 units for Q.

Producer User Final demand Total output

P Q

P 16 20 4 40

Q 8 40 32 80

[5]
View model solution
Step-by-Step Solution:

Given Transaction Matrix:

Producer User: P User: Q Final Demand ( $d$ ) Total Output ( $X$ )

P 16 20 4 40

Q 8 40 32 80

Step 1: Derive Technical Coefficient Matrix ( $A$ )

The elements of technical matrix $A = [a_{ij}]$ are given by $a_{ij} = \frac{X_{ij}}{X_j}$ :
$a_{11} = \frac{16}{40} = 0.40, \quad a_{12} = \frac{20}{80} = 0.25$

$a_{21} = \frac{8}{40} = 0.20, \quad a_{22} = \frac{40}{80} = 0.50$

$A = \begin{bmatrix} 0.40 & 0.25 \\ 0.20 & 0.50 \end{bmatrix}$

Step 2: Determine the Leontief Matrix $(I - A)$

$I - A = \begin{bmatrix} 1 - 0.40 & -0.25 \\ -0.20 & 1 - 0.50 \end{bmatrix} = \begin{bmatrix} 0.60 & -0.25 \\ -0.20 & 0.50 \end{bmatrix}$
Step 3: Compute the Determinant $|I - A|$

$|I - A| = (0.60)(0.50) - (-0.25)(-0.20) = 0.30 - 0.05 = 0.25$
Since $|I - A| = 0.25 > 0$ and $1 - a_{ii} > 0$ , the system satisfies the Hawkins-Simon conditions.

Step 4: Compute the Leontief Inverse $(I - A)^{-1}$

$(I - A)^{-1} = \frac{1}{|I - A|} \operatorname{adj}(I - A) = \frac{1}{0.25} \begin{bmatrix} 0.50 & 0.25 \\ 0.20 & 0.60 \end{bmatrix}$
$(I - A)^{-1} = 4 \begin{bmatrix} 0.50 & 0.25 \\ 0.20 & 0.60 \end{bmatrix} = \begin{bmatrix} 2.0 & 1.0 \\ 0.8 & 2.4 \end{bmatrix}$

Step 5: Calculate New Gross Output Vector ( $X$ )

Given new final demand:
$D_{\text{new}} = \begin{bmatrix} 18 \\ 44 \end{bmatrix}$

Using the Leontief equation $X = (I - A)^{-1} D_{\text{new}}$ :
$X = \begin{bmatrix} 2.0 & 1.0 \\ 0.8 & 2.4 \end{bmatrix} \begin{bmatrix} 18 \\ 44 \end{bmatrix}$

$X_P = (2.0 \times 18) + (1.0 \times 44) = 36 + 44 = 80 \text{ units}$

$X_Q = (0.8 \times 18) + (2.4 \times 44) = 14.4 + 105.6 = 120 \text{ units}$

Final Answer: The gross output required for each industry is:
- Industry P: $80$ units
- Industry Q: $120$ units
Solve the difference equation: , $y_t = 0.6y_{t-1} + 21$ , $y_0 = 50$ . Find $y_5 $ and $ y_1 $ ?

[5]
View model solution
Step-by-Step Solution:

Given difference equation:
$y_t = 0.6y_{t-1} + 21 \quad \text{with } y_0 = 50$

Step 1: Find Particular Solution ( $y^*$ / Equilibrium Level)

Set $y_t = y_{t-1} = y^*$ :
$y^* = 0.6y^* + 21$

$y^* - 0.6y^* = 21$

$0.4y^* = 21$

$y^* = \frac{21}{0.4} = 52.5$

Step 2: Find General Solution

The homogeneous equation $y_t - 0.6y_{t-1} = 0$ has solution:
$y_c = A(0.6)^t$

General solution:
$y_t = y^* + A(0.6)^t = 52.5 + A(0.6)^t$

Step 3: Apply Initial Condition $y_0 = 50$

$50 = 52.5 + A(0.6)^0 = 52.5 + A$
$A = 50 - 52.5 = -2.5$
Thus, the closed-form time path is:
$y_t = 52.5 - 2.5(0.6)^t$

Step 4: Calculate $y_1$ and $y_5$
1. Calculate $y_1$ :
  $y_1 = 52.5 - 2.5(0.6)^1 = 52.5 - 1.5 = \mathbf{51}$
  (Or by direct recurrence: $y_1 = 0.6(y_0) + 21 = 0.6(50) + 21 = 30 + 21 = 51$ ).
2. Calculate $y_5$ :
  $(0.6)^5 = 0.07776$
  
  $y_5 = 52.5 - 2.5(0.07776) = 52.5 - 0.1944 = \mathbf{52.3056}$
Final Answer:
- Time path: $y_t = 52.5 - 2.5(0.6)^t$
- $y_1 = 51$
- $y_5 = 52.3056$
In a competitive market price where $Q_d = 500 - 5P$ and $Q_s = 40 + 20P$ , the initial price P(0) is Rs 50.

(a) Derive a function for the time-path of P and use it to predict price in time period 5 given that price adjusts in proportion to excess demand at the rate $\frac{dP}{dt} = 0.01 (Q_d - Q_s)$ .

(b) How many time periods would you have to wait for the price to drop by Rs. 30?

[5]
View model solution
Step-by-Step Solution:

Given:
- Demand function: $Q_d = 500 - 5P$
- Supply function: $Q_s = 40 + 20P$
- Initial price: $P(0) = \text{Rs } 50$
- Price adjustment equation: $\frac{dP}{dt} = 0.01 (Q_d - Q_s)$
(a) Derive the time-path function $P(t)$ and predict price at $t = 5$ :

Step 1: Compute excess demand $(Q_d - Q_s)$

$Q_d - Q_s = (500 - 5P) - (40 + 20P) = 500 - 5P - 40 - 20P = 460 - 25P$
Step 2: Formulate and solve the differential equation

$\frac{dP}{dt} = 0.01 (460 - 25P) = 4.6 - 0.25P$
$\frac{dP}{dt} + 0.25P = 4.6$
Integrating factor $I(t)$ :
$I(t) = e^{\int 0.25 \, dt} = e^{0.25t}$

Multiply both sides by $e^{0.25t}$ :
$\frac{d}{dt} [P e^{0.25t}] = 4.6 e^{0.25t}$
Integrate both sides:
$P e^{0.25t} = \int 4.6 e^{0.25t} \, dt = \frac{4.6}{0.25} e^{0.25t} + C = 18.4 e^{0.25t} + C$
Divide by $e^{0.25t}$ :
$P(t) = 18.4 + C e^{-0.25t}$

(Notice the long-run equilibrium price is $P_e = 18.4$ ).

Step 3: Apply the initial condition $P(0) = 50$

$50 = 18.4 + C e^0 \implies C = 50 - 18.4 = 31.6$
Thus, the time path of price is:
$P(t) = 18.4 + 31.6 e^{-0.25t}$

Step 4: Predict price in time period $t = 5$

$P(5) = 18.4 + 31.6 e^{-0.25(5)} = 18.4 + 31.6 e^{-1.25}$
Since $e^{-1.25} \approx 0.2865048$ :
$P(5) = 18.4 + 31.6(0.2865048) = 18.4 + 9.0536 \approx \text{Rs } 27.45$

(b) Time periods required for price to drop by Rs 30:

The price drops by Rs 30 from its initial value of Rs 50, so the target price is:
$P(t) = 50 - 30 = \text{Rs } 20$

Substitute $P(t) = 20$ into the time-path function:
$18.4 + 31.6 e^{-0.25t} = 20$

$31.6 e^{-0.25t} = 20 - 18.4 = 1.6$

$e^{-0.25t} = \frac{1.6}{31.6} = \frac{4}{79} \approx 0.0506329$

Take natural logarithms on both sides:
$-0.25t = \ln(0.0506329) \approx -2.98315$

$t = \frac{-2.98315}{-0.25} \approx 11.9326 \text{ time periods}$

Final Answer:
- (a) Time-path function: $P(t) = 18.4 + 31.6 e^{-0.25t}$
  
  Predicted price at $t = 5$ : Rs 27.45
- (b) Waiting time for price to drop by Rs 30: $11.93$ time periods (approx. $12$ periods).

Section D

Comprehensive Answer / Case/ Situation Analysis Questions :

[20]

Producer	User	Final demand	Total output
	P	Q
P	16	20	4	40
Q	8	40	32	80

Producer	User: P	User: Q	Final Demand ( $d$ )	Total Output ( $X$ )
P	16	20	4	40
Q	8	40	32	80

Syllabus

Business Mathematics II Syllabus

Course Description

Course Objective

Course Contents:

Chapter-wise questions

Unit 1: Integration and applications

Past exam papers

Business Mathematics Ii 2024 Board Question Paper

Step-by-Step Solution:

Step 1: Perform algebraic division / rearrangement

Step 2: Integrate term by term

Step-by-Step Solution:

Given:

Step 1: Set up the definite integral

Step 2: Evaluate the integral

Step-by-Step Solution:

Step 1: Compute first-order partial derivative with respect to xxx (fxf_xfx​)

Step 2: Compute second-order partial derivative fxxf_{xx}fxx​

Step-by-Step Solution:

Step 1: Multiply AAA by AAA

Step-by-Step Solution:

Method 1: Expansion along Row 1

Method 2: Elementary Row Operations

Step-by-Step Solution:

Step 1: Form the Leontief matrix (I−A)(I - A)(I−A)

Step 2: Check Hawkins-Simon Condition 1

Step 3: Check Hawkins-Simon Condition 2

Economic Interpretation:

Step-by-Step Solution:

1. Determine the Order:

2. Determine the Degree:

Step-by-Step Solution:

Step 1: Separate variables

Step 2: Integrate both sides

Step-by-Step Solution:

Step 1: Rewrite in standard recursive form

Step 2: General solution

Step 3: Apply the initial condition y0=3y_0 = 3y0​=3

Step-by-Step Solution:

Given:

Step 1: Find the Total Cost function TC(x)TC(x)TC(x) by integration

Step 2: Determine the integration constant (CCC) using TC(1)=7TC(1) = 7TC(1)=7

Step 3: Find the Average Cost function AC(x)AC(x)AC(x)

Step-by-Step Solution:

(a) Evaluate ∫log⁡x+5x dx\int \frac{\log x + 5}{x} \, dx∫xlogx+5​dx:

(b) Evaluate ∫(2x+1)5x+2 dx\int (2x + 1)\sqrt{5x + 2} \, dx∫(2x+1)5x+2​dx:

Step-by-Step Solution:

Step 1: Find Market Equilibrium Quantity (x∗x^*x∗) and Price (P∗P^*P∗)

Step 2: Compute Consumer’s Surplus (CSCSCS)

Step 3: Compute Producer’s Surplus (PSPSPS)

Step 4: Total Surplus

Step-by-Step Solution:

Method: Cramer’s Rule

Step 1: Calculate the Coefficient Determinant (DDD)

Step 2: Calculate DxD_xDx​

Step 3: Calculate DyD_yDy​

Step 4: Calculate DzD_zDz​

Verification:

Step-by-Step Solution:

Step 1: Compute the partial derivatives

Step 2: Use the Total Differential formula

Step 3: Substitute the given percentage changes

Step-by-Step Solution:

Step 1: Identify standard form and integrating factor

Step 2: Multiply the differential equation by I(x)I(x)I(x)

Step 3: Integrate both sides with respect to xxx

Step 4: Solve explicitly for yyy

Step-by-Step Solution:

Given Bertrand Duopoly Reaction Functions:

Step 1: Derive a single difference equation for PtXP_t^XPtX​

Step 2: Solve the difference equation for even periods (t=2kt = 2kt=2k)

Step 3: Calculate P10XP_{10}^XP10X​ (Period 10, so k=10/2=5k = 10/2 = 5k=10/2=5)

Step-by-Step Solution:

Step 1: Formulate the First-Order Difference Equation

Step 2: Determine Particular Solution (y∗y^*y∗ / Equilibrium Level)

Step 3: Determine Complementary Function (ycy_cyc​) and General Time Path

Step 4: Comment on Stability of the Time Path

Step-by-Step Solution:

Problem Formulation:

Step 1: Compute first-order partial derivative with respect to $x$ ( $f_x$ )

Step 2: Compute second-order partial derivative $f_{xx}$

Step 1: Multiply $A$ by $A$

Step 1: Form the Leontief matrix $(I - A)$

Step 3: Apply the initial condition $y_0 = 3$

Step 1: Find the Total Cost function $TC(x)$ by integration

Step 2: Determine the integration constant ( $C$ ) using $TC(1) = 7$

Step 3: Find the Average Cost function $AC(x)$

(a) Evaluate $\int \frac{\log x + 5}{x} \, dx$ :

(b) Evaluate $\int (2x + 1)\sqrt{5x + 2} \, dx$ :

Step 1: Find Market Equilibrium Quantity ( $x^$ ) and Price ( $P^$ )

Step 2: Compute Consumer’s Surplus ( $CS$ )

Step 3: Compute Producer’s Surplus ( $PS$ )

Step 1: Calculate the Coefficient Determinant ( $D$ )

Step 2: Calculate $D_x$

Step 3: Calculate $D_y$

Step 4: Calculate $D_z$

Step 2: Multiply the differential equation by $I(x)$

Step 3: Integrate both sides with respect to $x$

Step 4: Solve explicitly for $y$

Step 1: Derive a single difference equation for $P_t^X$

Step 2: Solve the difference equation for even periods ( $t = 2k$ )

Step 3: Calculate $P_{10}^X$ (Period 10, so $k = 10/2 = 5$ )

Step 2: Determine Particular Solution ( $y^*$ / Equilibrium Level)

Step 3: Determine Complementary Function ( $y_c$ ) and General Time Path

Step 3: Evaluate the objective function $Z = 5x_1 + 3x_2$ at all corner points

Step 1: Derive Technical Coefficient Matrix ( $A$ )

Step 2: Determine the Leontief Matrix $(I - A)$

Step 3: Compute the Determinant $|I - A|$

Step 4: Compute the Leontief Inverse $(I - A)^{-1}$

Step 5: Calculate New Gross Output Vector ( $X$ )

Step 1: Find Particular Solution ( $y^*$ / Equilibrium Level)

Step 3: Apply Initial Condition $y_0 = 50$

Step 4: Calculate $y_1$ and $y_5$

(a) Derive the time-path function $P(t)$ and predict price at $t = 5$ :

Step 1: Compute excess demand $(Q_d - Q_s)$

Step 3: Apply the initial condition $P(0) = 50$

Step 4: Predict price in time period $t = 5$