ALLAMA IQBAL OPEN UNIVERSITY
(Department of Mathematics)
WARNING
1. Plagiarism or hiring of ghost writer(s) for solving the assignment(s) will debar the student from award of degree/certificate if found at any stage.
2. Submitting assignment(s) borrowed or stolen from other(s) as one's own will be penalized as defined in the "Aiou Plagiarism Policy".
| Assignment Submission Schedule |
| 6 Credit Hours |
Due Date |
3 Credit Hours |
Due Date |
| Assignment 1 |
15-12-2025 |
Assignment 1 |
08-01-2026 |
| Assignment 2 |
08-01-2026 |
| Assignment 3 |
30-01-2026 |
Assignment 2 |
20-02-2026 |
| Assignment 4 |
20-02-2026 |
| Course: Business Mathematics (1429) |
Semester: Autumn-2025 |
| Level: BA/B. Com / BBA |
|
Please read the following instructions for writing your assignments. (SSC, HSSC & BA Programmes)
1. All questions are compulsory and carry equal marks but within a question the marks are distributed according to its requirements.
2. Read the question carefully and then answer it according to the requirements of the questions.
3. Late submission of assignments will not be accepted.
4. Your own analysis and synthesis will be appreciated.
5. Avoid irrelevant discussion/information and reproducing from books, study guide of allied material.
| Total Marks: 100 |
Pass Marks BA: 40
Pass Marks AD: 50 |
ASSIGNMENT No. 1
Q1(a). Probability Theory - Find the sample space for choosing an odd number from 1 to 15 at random.
▶
Sample Space for Choosing an Odd Number from 1 to 15
In probability theory, the sample space is the set of all possible outcomes of a random experiment. When we are asked to choose an odd number from 1 to 15 at random, we first need to identify all the odd numbers within this range.
Step 1: List Numbers from 1 to 15
The numbers from 1 to 15 are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15.
Step 2: Identify Odd Numbers
Odd numbers are those which are not divisible by 2. The odd numbers from 1 to 15 are:
1, 3, 5, 7, 9, 11, 13, 15.
Step 3: Define the Sample Space
The sample space (S) for choosing an odd number randomly from 1 to 15 is the set of all odd numbers. Therefore:
S = {1, 3, 5, 7, 9, 11, 13, 15}.
Conclusion
There are 8 possible outcomes in this sample space. Each number in the set has an equal chance of being chosen when picking an odd number at random. Understanding the sample space is essential for calculating probabilities accurately.
Q1(b). Probability Theory - What is the difference between mutually exclusive events and collectively exhaustive events?
▶
Mutually Exclusive Events vs Collectively Exhaustive Events
In probability theory, understanding the types of events is essential to calculating probabilities accurately. Two important concepts are mutually exclusive events and collectively exhaustive events. Although they may sound similar, they have distinct meanings and applications.
Mutually Exclusive Events
Mutually exclusive events are events that cannot happen at the same time. If one event occurs, the other cannot occur. In other words, the occurrence of one event excludes the possibility of the other event happening.
Example: When tossing a coin, getting a “Head” and getting a “Tail” are mutually exclusive events. If the result is a head, it cannot be a tail at the same time.
Collectively Exhaustive Events
Collectively exhaustive events are events that cover all possible outcomes of an experiment. At least one of these events must occur in every trial, ensuring that no outcome is left out.
Example: When rolling a six-sided die, the events {1, 2, 3, 4, 5, 6} are collectively exhaustive. One of these outcomes will definitely occur on every roll.
Key Differences
1. Occurrence: Mutually exclusive events cannot occur together, while collectively exhaustive events ensure that at least one occurs.
2. Coverage: Mutually exclusive events may not cover all possible outcomes, whereas collectively exhaustive events cover every possible outcome.
3. Example: Flipping a coin – Head and Tail are mutually exclusive. Rolling a die – outcomes 1 to 6 are collectively exhaustive.
Conclusion
Mutually exclusive and collectively exhaustive events are fundamental concepts in probability theory. Recognizing the difference helps in understanding event relationships and calculating probabilities correctly. Mutually exclusive focuses on events that cannot happen together, while collectively exhaustive ensures all outcomes are included.
Q1(c). Probability Theory - The probability that an applicant for pilot school will be admitted is 0.5. If three applicants are selected at random, what is the probability that:
i. All three will be admitted
ii. None will be admitted
iii. Only one will be admitted
▶
Probability of Admission for Pilot School Applicants
In probability theory, we often calculate the likelihood of multiple independent events occurring. Suppose the probability that an applicant for pilot school will be admitted is 0.5. If three applicants are selected at random, we can calculate the probability for different scenarios.
Given:
Probability of admission for one applicant, P(admitted) = 0.5
Probability of not being admitted, P(not admitted) = 1 - 0.5 = 0.5
Number of applicants selected = 3
i. Probability that All Three Will Be Admitted
All three being admitted means all three events occur together. Since the events are independent, we multiply the probabilities:
P(all three admitted) = P(admitted) × P(admitted) × P(admitted)
P(all three admitted) = 0.5 × 0.5 × 0.5 = 0.125
ii. Probability that None Will Be Admitted
None being admitted means all three are not admitted. Using independence:
P(none admitted) = P(not admitted) × P(not admitted) × P(not admitted)
P(none admitted) = 0.5 × 0.5 × 0.5 = 0.125
iii. Probability that Only One Will Be Admitted
Exactly one admitted means one is admitted and the other two are not. There are 3 different ways this can happen (first, second, or third applicant admitted). Using combinations:
P(only one admitted) = 3 × [P(admitted) × P(not admitted) × P(not admitted)]
P(only one admitted) = 3 × (0.5 × 0.5 × 0.5)
P(only one admitted) = 3 × 0.125 = 0.375
Conclusion
- Probability that all three applicants are admitted = 0.125
- Probability that none are admitted = 0.125
- Probability that only one is admitted = 0.375
These calculations demonstrate the application of independent event probabilities in real-life scenarios.
Q2(a). Random Variables - Define a random variable. What is the difference between a discrete random variable and a continuous random variable?
▶
Random Variables: Definition and Types
In probability theory, a random variable is a variable that takes numerical values determined by the outcomes of a random experiment. It assigns a number to each possible outcome, allowing us to quantify and analyze uncertainty in experiments or events.
Discrete Random Variable
A discrete random variable is one that can take a finite or countable number of distinct values. Each value has an associated probability, and these probabilities sum to 1.
Example: The number of heads when flipping three coins. The possible values are 0, 1, 2, or 3.
Continuous Random Variable
A continuous random variable can take any value within a given interval or range. Its probability is described using a probability density function, and the probability of it taking a single exact value is 0. Instead, probabilities are calculated over intervals.
Example: The height of students in a class or the time taken to complete a task.
Key Differences Between Discrete and Continuous Random Variables
1. Values: Discrete variables take countable values; continuous variables take any value within an interval.
2. Probability: Discrete probabilities are associated with individual values; continuous probabilities are associated with ranges of values.
3. Examples: Number of cars in a parking lot (discrete) vs. weight of a car (continuous).
4. Representation: Discrete variables often use tables or lists; continuous variables use functions or graphs.
Conclusion
Understanding random variables and the difference between discrete and continuous types is essential in probability and statistics. It helps in modeling uncertainty, calculating probabilities, and making predictions in real-life situations.
Q2(b). Random Variables - The fire chief for a small volunteer fire department has compiled data on the number of false alarms called in each day for the past 360 days. Construct the probability distribution for this study.
▶
Probability Distribution of False Alarms
In probability theory, a probability distributionnumber of false alarms per day recorded by a small volunteer fire department over 360 days.
Step 1: Identify the Random Variable
Let X represent the number of false alarms in a single day. Possible values of X could be 0, 1, 2, 3, 4, etc., depending on how many false alarms occurred on different days.
Step 2: Collect Frequency Data
Suppose the fire chief recorded the following frequency of false alarms over 360 days:
- 0 false alarms: 120 days
- 1 false alarm: 150 days
- 2 false alarms: 70 days
- 3 false alarms: 15 days
- 4 false alarms: 5 days
Step 3: Calculate Probabilities
Probability for each number of false alarms is calculated as:
P(X = x) = (Number of days with x false alarms) / (Total number of days)
Using our example:
P(0) = 120 / 360 = 0.333
P(1) = 150 / 360 = 0.417
P(2) = 70 / 360 = 0.194
P(3) = 15 / 360 = 0.042
P(4) = 5 / 360 = 0.014
Step 4: Construct the Probability Distribution Table
| Number of False Alarms (X) |
Probability P(X) |
| 0 |
0.333 |
| 1 |
0.417 |
| 2 |
0.194 |
| 3 |
0.042 |
| 4 |
0.014 |
Conclusion
This table represents the probability distribution of the random variable X, showing the likelihood of each number of false alarms occurring in a single day. Probability distributions are useful in predicting future occurrences and making informed decisions for planning and resource allocation in the fire department.
Q2(c). Random Variables - Construct the discrete probability distribution that corresponds to the experiment of tossing a fair coin three times. Suppose the random variable X equals the number of heads occurring in three tosses. What is the probability of two or more heads?
▶
Discrete Probability Distribution: Tossing a Coin Three Times
In probability theory, a random variable assigns numerical values to the outcomes of a random experiment. In this experiment, we toss a fair coin three times, and let X represent the number of heads obtained.
Step 1: List All Possible Outcomes
When tossing a coin three times, each toss can result in Head (H) or Tail (T). The sample space is:
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Step 2: Determine the Random Variable X
Let X = number of heads in each outcome:
- HHH → X = 3
- HHT → X = 2
- HTH → X = 2
- HTT → X = 1
- THH → X = 2
- THT → X = 1
- TTH → X = 1
- TTT → X = 0
Step 3: Construct the Probability Distribution
Count the number of outcomes for each value of X and divide by the total number of outcomes (8):
| X (Number of Heads) |
Probability P(X) |
| 0 |
1/8 = 0.125 |
| 1 |
3/8 = 0.375 |
| 2 |
3/8 = 0.375 |
| 3 |
1/8 = 0.125 |
Step 4: Probability of Two or More Heads
Two or more heads means X = 2 or X = 3. Add the probabilities for these values:
P(X ≥ 2) = P(X = 2) + P(X = 3)
P(X ≥ 2) = 0.375 + 0.125 = 0.5
Conclusion
The discrete probability distribution shows the likelihood of getting 0, 1, 2, or 3 heads when tossing a fair coin three times. The probability of obtaining two or more heads is 0.5, meaning there is a 50% chance of this outcome.
Q3(a). Solve the following first-degree equation: \( 8x - 6 = 5x + 3 \)
▶
Step 1: Write the equation
The given equation is:
8x - 6 = 5x + 3
Step 2: Move all x terms to one side
Subtract 5x from both sides:
8x - 5x - 6 = 3
3x - 6 = 3
Step 3: Move constants to the other side
Add 6 to both sides:
3x = 3 + 6
3x = 9
Step 4: Solve for x
Divide both sides by 3:
x = 9 ÷ 3
x = 3
Conclusion
The solution to the first-degree equation 8x - 6 = 5x + 3 is:
x = 3
Q3(b). Solve the following first-degree equation: \( -15 + 35x = 8x - 9 \)
▶
Step 1: Write the equation
\( -15 + 35x = 8x - 9 \)
Step 2: Move all x terms to one side
Subtract \( 8x \) from both sides:
\( 35x - 8x - 15 = -9 \)
\( 27x - 15 = -9 \)
Step 3: Move constants to the other side
Add 15 to both sides:
\( 27x = -9 + 15 \)
\( 27x = 6 \)
Step 4: Solve for x
Divide both sides by 27:
\( x = \frac{6}{27} \)
Simplify the fraction:
\( x = \frac{2}{9} \)
Conclusion
The solution to the first-degree equation \( -15 + 35x = 8x - 9 \) is:
\( x = \frac{2}{9} \)
Q3(c). Solve the following first-degree equation: \( (x + 9) - (-6 + 4x) + 4 = 0 \)
▶
Step 1: Write the equation
\( (x + 9) - (-6 + 4x) + 4 = 0 \)
Step 2: Simplify the parentheses
\( x + 9 + 6 - 4x + 4 = 0 \)
\( x - 4x + 9 + 6 + 4 = 0 \)
\( -3x + 19 = 0 \)
Step 3: Move constants to the other side
\( -3x = -19 \)
Step 4: Solve for x
\( x = \frac{-19}{-3} \)
\( x = \frac{19}{3} \)
Conclusion
The solution to the first-degree equation \( (x + 9) - (-6 + 4x) + 4 = 0 \) is:
\( x = \frac{19}{3} \)
Q4(a). Solve the following quadratic equation using the quadratic formula: \( 4x^2 + 3x - 1 = 0 \)
▶
Solving Quadratic Equation Using Quadratic Formula
We are solving the quadratic equation:
\( 4x^2 + 3x - 1 = 0 \)
Step 1: Identify coefficients
Here, \( a = 4 \), \( b = 3 \), \( c = -1 \)
Step 2: Quadratic formula
The quadratic formula is:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Step 3: Substitute values
\( x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4} \)
\( x = \frac{-3 \pm \sqrt{9 - (-16)}}{8} \)
\( x = \frac{-3 \pm \sqrt{9 + 16}}{8} \)
\( x = \frac{-3 \pm \sqrt{25}}{8} \)
Step 4: Solve for the two roots
\( x = \frac{-3 + 5}{8} = \frac{2}{8} = \frac{1}{4} \)
\( x = \frac{-3 - 5}{8} = \frac{-8}{8} = -1 \)
\( x = \frac{1}{4}, \quad x = -1 \)
Q4(b). Solve the following quadratic equation using the quadratic formula: \( 4t^2 - 64 = 0 \)
▶
We are solving the quadratic equation:
\( 4t^2 - 64 = 0 \)
Step 1: Identify coefficients
Here, \( a = 4 \), \( b = 0 \), \( c = -64 \)
Step 2: Quadratic formula
The quadratic formula is:
\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Step 3: Substitute values
\( t = \frac{-0 \pm \sqrt{0^2 - 4 \cdot 4 \cdot (-64)}}{2 \cdot 4} \)
\( t = \frac{0 \pm \sqrt{0 - (-1024)}}{8} \)
\( t = \frac{0 \pm \sqrt{1024}}{8} \)
\( t = \frac{0 \pm 32}{8} \)
Step 4: Solve for the two roots
\( t = \frac{32}{8} = 4 \)
\( t = \frac{-32}{8} = -4 \)
\( t = 4, \quad t = -4 \)
Q5(a). Solve the linear equation \( y = 2x + 1 \).
▶
We are given the equation:
\( y = 2x + 1 \)
Step 1: Understanding the equation
This is a linear equation in two variables, \(x\) and \(y\). It represents a straight line. Each value of \(x\) gives a corresponding value of \(y\), and vice versa.
Step 2: Solve for y in terms of x
The equation is already solved for \(y\):
\( y = 2x + 1 \)
Step 3: Solve for x in terms of y
Rewriting the equation to find \(x\) for any given \(y\):
\( y = 2x + 1 \)
\( 2x = y - 1 \)
\( x = \frac{y - 1}{2} \)
Step 4: Find a few points
Choose some values of \(x\) and calculate \(y\):
| x |
y = 2x + 1 |
| -2 |
-3 |
| -1 |
-1 |
| 0 |
1 |
| 1 |
3 |
| 2 |
5 |
Q5(b). A company has fixed costs of $7,000 for plant and equipment and variable costs of $600 for each unit of output. What is the total cost at varying levels of output?
▶
Calculating Total Cost at Varying Levels of Output
A company has fixed costs of $7,000 for plant and equipment and variable costs of $600 per unit of output. We want to calculate total cost for different levels of output.
Step 1: Understand the total cost formula
The total cost (TC) is the sum of fixed costs (FC) and total variable costs (VC × Quantity):
\( TC = FC + (VC \times Q) \)
Step 2: Substitute the given values
Fixed costs \(FC = 7000\), variable cost per unit \(VC = 600\):
\( TC = 7000 + 600 \times Q \)
Step 3: Calculate total cost for different output levels
| Units (Q) |
Total Cost (TC = 7000 + 600 × Q) |
| 0 |
7000 |
| 1 |
7600 |
| 2 |
8200 |
| 3 |
8800 |
| 4 |
9400 |
| 5 |
10000 |
Step 4: Interpretation
- The fixed cost remains $7,000 regardless of the number of units produced.
- The total cost increases by $600 for each additional unit, which is the variable cost per unit.
- This table can be extended to any number of units to plan production costs.
Answer:
Total cost for 0 to 5 units of output:
0 units → $7,000
1 unit → $7,600
2 units → $8,200
3 units → $8,800
4 units → $9,400
5 units → $10,000
Q5(c). Find the equation of the straight line that has slope m = 4 and passes through the point (−1, 6).
▶
Finding the Equation of a Straight Line
We are asked to find the equation of a line with slope \( m = 4 \) that passes through the point \( (-1, 6) \).
Step 1: Use the point-slope form
The point-slope form of a straight line is:
\( y - y_1 = m(x - x_1) \)
Step 2: Substitute the given point and slope
\( y - 6 = 4(x - (-1)) \)
\( y - 6 = 4(x + 1) \)
Step 3: Expand the right-hand side
\( y - 6 = 4x + 4 \)
Step 4: Solve for y
\( y = 4x + 4 + 6 \)
\( y = 4x + 10 \)
The equation of the straight line is: \( y = 4x + 10 \)
ASSIGNMENT No. 2
Q1(a). Find the transpose of each of the following matrices:
(i) \( A = \begin{bmatrix} -1 & 0 & 0 \\ 6 & 9 & -2 \\ -4 & 5 & 1 \end{bmatrix} \) (ii) \( B = \begin{bmatrix} 6 & -6 \\ 4 & 9 \\ -2 & 0 \end{bmatrix} \)
▶
Transpose of Matrices
i) Given matrix:
\[
A =
\begin{bmatrix}
-1 & 0 & 0 \\
6 & 9 & -2 \\
-4 & 5 & 1
\end{bmatrix}
\]
Transpose of \( A \) is:
\[
A^T =
\begin{bmatrix}
-1 & 6 & -4 \\
0 & 9 & 5 \\
0 & -2 & 1
\end{bmatrix}
\]
ii) Given matrix:
\[
B =
\begin{bmatrix}
6 & -6 \\
4 & 9 \\
-2 & 0
\end{bmatrix}
\]
Transpose of \( B \) is:
\[
B^T =
\begin{bmatrix}
6 & 4 & -2 \\
-6 & 9 & 0
\end{bmatrix}
\]
Q1(b). If \( A = \begin{bmatrix}1 & 2 & 3 \\-1 & -2 & 3 \\3 & 4 & 6\end{bmatrix}\) then find \( 4A - 3B \).
▶
Note: Matrix B is not provided in the question, so we take \( B = A \). The solution below is the full step by step work.
Solution:
Step 1: Compute \(4A\). Multiply each element of \(A\) by 4.
\(
4A =
\begin{bmatrix}
4 & 8 & 12 \\
-4 & -8 & 12 \\
12 & 16 & 24
\end{bmatrix}
\)
Step 2: Compute \(3B\). Since \(B = A\), multiply each element of \(A\) by 3.
\(
3B =
\begin{bmatrix}
3 & 6 & 9 \\
-3 & -6 & 9 \\
9 & 12 & 18
\end{bmatrix}
\)
Step 3: Subtract \(3B\) from \(4A\) elementwise.
\(
4A - 3B =
\begin{bmatrix}
4 & 8 & 12 \\
-4 & -8 & 12 \\
12 & 16 & 24
\end{bmatrix}
-
\begin{bmatrix}
3 & 6 & 9 \\
-3 & -6 & 9 \\
9 & 12 & 18
\end{bmatrix}
=
\begin{bmatrix}
4-3 & 8-6 & 12-9 \\
-4-(-3) & -8-(-6) & 12-9 \\
12-9 & 16-12 & 24-18
\end{bmatrix}
\)
\(
=
\begin{bmatrix}
1 & 2 & 3 \\
-1 & -2 & 3 \\
3 & 4 & 6
\end{bmatrix}
= A
\)
Final Answer:
\(
4A - 3B =
\begin{bmatrix}
1 & 2 & 3 \\
-1 & -2 & 3 \\
3 & 4 & 6
\end{bmatrix}
\)
Q1(c). The quarterly sales of wheat, cotton, and corn for the years 2010 and 2011 are represented below in the form of matrices A and B. Find the total quarterly sales of wheat, cotton, and corn for these two years.
▶
Note: The matrices A and B are not provided in this question. Without the matrices, the exact totals cannot be calculated. The solution given below is in general form.
Let the sales for the year 2010 be:
\[
A =
\begin{vmatrix}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34}
\end{vmatrix}
\]
and the sales for the year 2011 be:
\[
B =
\begin{vmatrix}
b_{11} & b_{12} & b_{13} & b_{14} \\
b_{21} & b_{22} & b_{23} & b_{24} \\
b_{31} & b_{32} & b_{33} & b_{34}
\end{vmatrix}
\]
The total sales for the two years are obtained by adding the two matrices:
\[
A + B =
\begin{vmatrix}
a_{11}+b_{11} & a_{12}+b_{12} & a_{13}+b_{13} & a_{14}+b_{14} \\
a_{21}+b_{21} & a_{22}+b_{22} & a_{23}+b_{23} & a_{24}+b_{24} \\
a_{31}+b_{31} & a_{32}+b_{32} & a_{33}+b_{33} & a_{34}+b_{34}
\end{vmatrix}
\]
Thus, by adding the corresponding elements of matrices A and B, we get the total quarterly sales of wheat, cotton, and corn for the years 2010 and 2011.
Q2(a). Evaluate the following determinants:
(i) \(\begin{bmatrix}2 & 3 & -1 \\1 & 1 & 0 \\2 & -3 & 5\end{bmatrix}\) (ii) \(\begin{bmatrix}2a & a & a \\b & 2b & b \\c & c & 2c\end{bmatrix}\)
▶
Let A
\(A =\begin{bmatrix}2 & 3 & -1 \\1 & 1 & 0 \\2 & -3 & 5\end{bmatrix}\)
Expanding along the first row:
\(
A = 2\begin{bmatrix}1 & 0 \\ -3 & 5\end{bmatrix}
- 3\begin{bmatrix}1 & 0 \\ 2 & 5\end{bmatrix}
+ (-1)\begin{bmatrix}1 & 1 \\ 2 & -3\end{bmatrix}
\)
\(
= 2(5) - 3(5) + (-1)(-5)
= 10 - 15 + 5 = 0
\)
∴ A = 0
Let B
\(
B =
\begin{bmatrix}
2a & a & a \\
b & 2b & b \\
c & c & 2c
\end{bmatrix}
\)
Factor out \(a\), \(b\), and \(c\) from each row:
\(
B = abc
\begin{bmatrix}
2 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 2
\end{bmatrix}
\)
Expanding along the first row:
\(
= 2\begin{bmatrix}2 & 1 \\ 1 & 2\end{bmatrix}
- 1\begin{bmatrix}1 & 1 \\ 1 & 2\end{bmatrix}
+ 1\begin{bmatrix}1 & 2 \\ 1 & 1\end{bmatrix}
\)
\(
= 2(3) - 1(1) + 1(-1)
= 6 - 1 - 1 = 4
\)
\(
B = abc \cdot 4 = 4abc
\)
∴ B = 4abc
Q2(b). Find the inverse of the given matrix \(A = \begin{bmatrix} 1 & 2 \\[6pt] 3 & 4 \end{bmatrix}\) using the Gaussian reduction method.
▶
Form the augmented matrix \(\left[ A \mid I \right]\):
\(
\left[\begin{array}{cc|cc}
1 & 2 & 1 & 0 \\[6pt]
3 & 4 & 0 & 1
\end{array}\right]
\)
Step 1: Eliminate the entry below the first pivot
Replace \(R_2\) with \(R_2 - 3R_1\):
\(
\left[\begin{array}{cc|cc}
1 & 2 & 1 & 0 \\[6pt]
0 & -2 & -3 & 1
\end{array}\right]
\)
Step 2: Scale the second row to make the pivot 1
Replace \(R_2\) with \(-\tfrac{1}{2}R_2\):
\(
\left[\begin{array}{cc|cc}
1 & 2 & 1 & 0 \\[6pt]
0 & 1 & \tfrac{3}{2} & -\tfrac{1}{2}
\end{array}\right]
\)
Step 3: Eliminate the entry above the second pivot
Replace \(R_1\) with \(R_1 - 2R_2\):
\(
\left[\begin{array}{cc|cc}
1 & 0 & -2 & 1 \\[6pt]
0 & 1 & \tfrac{3}{2} & -\tfrac{1}{2}
\end{array}\right]
\)
Result
Now the left block is the identity and the right block is \(A^{-1}\). Thus
\(
A^{-1} = \begin{bmatrix}
-2 & 1 \\[6pt]
\tfrac{3}{2} & -\tfrac{1}{2}
\end{bmatrix}
\)
Check
One quick check is the formula for a 2 by 2 inverse. The determinant is \(\det(A)=1\cdot 4 - 2\cdot 3 = -2\). Using the formula gives
\(
A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} 4 & -2 \\[6pt] -3 & 1 \end{bmatrix}
= -\tfrac{1}{2}\begin{bmatrix} 4 & -2 \\[6pt] -3 & 1 \end{bmatrix}
= \begin{bmatrix} -2 & 1 \\[6pt] \tfrac{3}{2} & -\tfrac{1}{2} \end{bmatrix}
\)
\(
A^{-1} = \begin{bmatrix}
-2 & 1 \\[6pt]
\tfrac{3}{2} & -\tfrac{1}{2}
\end{bmatrix}
\)
Q2(c). If \(A = \begin{bmatrix}1 & -2 & -3 \\2 & 0 & 1 \\-4 & 6 & 8\end{bmatrix}\), find |A|.
▶
We want to find the determinant of:
\(A = \begin{bmatrix}1 & -2 & -3 \\2 & 0 & 1 \\-4 & 6 & 8\end{bmatrix}\)
Step 1: Expansion
\(
|A| = 1 \cdot
\begin{vmatrix}
0 & 1 \\
6 & 8
\end{vmatrix}
- (-2) \cdot
\begin{vmatrix}
2 & 1 \\
-4 & 8
\end{vmatrix}
+ (-3) \cdot
\begin{vmatrix}
2 & 0 \\
-4 & 6
\end{vmatrix}
\)
Step 2: Simplify the minors
\(
\begin{vmatrix}
0 & 1 \\
6 & 8
\end{vmatrix} = -6, \quad
\begin{vmatrix}
2 & 1 \\
-4 & 8
\end{vmatrix} = 20, \quad
\begin{vmatrix}
2 & 0 \\
-4 & 6
\end{vmatrix} = 12
\)
Step 3: Substitute back
\(
|A| = 1(-6) - (-2)(20) + (-3)(12)
\)
\(
|A| = -6 + 40 - 36 = -2
\)
\(
|A| = -2
\)
Q3(a). Find the derivative of the function \(f(x) = 15x^{100} - 3x^{12} + 5x - 46\).
▶
Using the power rule \(\dfrac{d}{dx}[x^n] = n x^{n-1}\), we differentiate each term:
\(
\dfrac{d}{dx}[15x^{100}] = 1500x^{99}, \quad
\dfrac{d}{dx}[-3x^{12}] = -36x^{11}, \quad
\dfrac{d}{dx}[5x] = 5, \quad
\dfrac{d}{dx}[-46] = 0
\)
Final Answer
\(
f'(x) = 1500x^{99} - 36x^{11} + 5
\)
Q3(b). The position of a particle moving along a straight line at time \(t\) is given by \(s = f(t) = 2t^2 + 7\). Find the instantaneous rate of change at \(t = 12\) seconds.
▶
The position of a particle moving along a straight line at time \(t\) is given by:
\(
s = f(t) = 2t^2 + 7
\)
Find the instantaneous rate of change at \(t = 12\) seconds.
Solution
The instantaneous rate of change is given by the derivative of \(f(t)\):
\(
f'(t) = \dfrac{d}{dt}(2t^2 + 7)
\)
\(
f'(t) = 4t
\)
Step 2: Evaluate at \(t = 12\)
\(
f'(12) = 4(12) = 48
\)
The instantaneous rate of change at \(t = 12\) seconds is:
\(
48 \;\; \text{units per second}
\)
Q3(c). The production costs per week for producing x widgets in a factory is given by \(C(x) = 500 + 350x - 0.09x^2\). Calculate the cost to produce the 301st widget at \(x = 300\).
▶
The cost of producing the 301st widget is:
\(
\text{Cost of 301st widget} = C(301) - C(300)
\)
Step 1: Compute \(C(301)\)
\(
C(301) = 500 + 350(301) - 0.09(301^2)
\)
\(
= 500 + 105350 - 0.09(90601)
\)
\(
= 105850 - 8154.09 = 97695.91
\)
Step 2: Compute \(C(300)\)
\(
C(300) = 500 + 350(300) - 0.09(300^2)
\)
\(
= 500 + 105000 - 0.09(90000)
\)
\(
= 105500 - 8100 = 97400
\)
Step 3: Find the difference
\(
C(301) - C(300) = 97695.91 - 97400 = 295.91
\)
The cost to produce the 301st widget is:
\(
\; 295.91
\)
Q4(a). Find the values of \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) at the point \((4, -5)\). If \( f(x,y) = x^{2} + 3xy^{2} + y - 1 \).
▶
Partial Derivatives at the Point \( (4, -5) \)
Let the function be \( f(x,y) = x^{2} + 3xy^{2} + y - 1 \).
We will find the partial derivatives and then evaluate them at the point \( (4, -5) \).
Step 1: Partial derivative with respect to \( x \)
\(
\frac{\partial f}{\partial x}
= \frac{\partial}{\partial x}\left(x^{2} + 3xy^{2} + y - 1\right)
= 2x + 3y^{2}
\)
\(
\frac{\partial f}{\partial x}(4,-5)
= 2(4) + 3(-5)^{2}
= 8 + 75
= 83
\)
Step 2: Partial derivative with respect to \( y \)
\(
\frac{\partial f}{\partial y}
= \frac{\partial}{\partial y}\left(x^{2} + 3xy^{2} + y - 1\right)
= 6xy + 1
\)
\(
\frac{\partial f}{\partial y}(4,-5)
= 6(4)(-5) + 1
= -120 + 1
= -119
\)
\(
\frac{\partial f}{\partial x}(4,-5) = 83,
\quad
\frac{\partial f}{\partial y}(4,-5) = -119
\)
Q4(b). Find the second partial derivatives of the function \( f(x,y) = 3x^{2y} + 2y^{3} \).
▶
Step 1: First order partial derivatives
\(
\frac{\partial f}{\partial x}
= \frac{\partial}{\partial x}\big(3x^{2y} + 2y^{3}\big)
= 6y \, x^{2y-1}
\)
\(
\frac{\partial f}{\partial y}
= \frac{\partial}{\partial y}\big(3x^{2y} + 2y^{3}\big)
= 6x^{2y}\ln(x) + 6y^{2}
\)
Step 2: Second order partial derivatives
\(
f_{xx}
= \frac{\partial}{\partial x}(6y \, x^{2y-1})
= 6y(2y-1)x^{2y-2}
\)
\(
f_{yy}
= \frac{\partial}{\partial y}(6x^{2y}\ln(x) + 6y^{2})
= 12x^{2y}(\ln(x))^{2} + 12y
\)
\(
f_{xy}
= \frac{\partial}{\partial y}(6y \, x^{2y-1})
= 6x^{2y-1} + 12y \, x^{2y-1}\ln(x)
\)
\(
f_{yx} = f_{xy}
\)
\(
f_{xx} = 6y(2y-1)x^{2y-2}, \quad
f_{yy} = 12x^{2y}(\ln(x))^{2} + 12y, \quad
f_{xy} = f_{yx} = 6x^{2y-1} + 12y \, x^{2y-1}\ln(x)
\)
Q4(c). If \( f(x,y) = x^{2} + y^{2} + 2xy \), find the critical points and determine if they are maxima or minima.
▶
Find critical points and classify them for \( f(x,y) = x^{2} + y^{2} + 2xy \)
We first compute the first partial derivatives and set them equal to zero to find critical points.
Step 1: First partial derivatives
\(
f_{x} = \frac{\partial f}{\partial x} = 2x + 2y
\)
\(
f_{y} = \frac{\partial f}{\partial y} = 2y + 2x
\)
Set the gradient equal to zero:
\(
2x + 2y = 0 \quad \text{and} \quad 2x + 2y = 0
\)
Both equations are the same, so the critical points satisfy
\(
x + y = 0
\)
or
\(
y = -x
\).
Thus every point on the line \( y = -x \) is a critical point.
Step 2: Second derivative test and classification
Compute second partial derivatives:
\(
f_{xx} = 2,\quad f_{yy} = 2,\quad f_{xy} = 2
\)
Hessian matrix is
\(
H = \begin{pmatrix} 2 & 2 \\[4pt] 2 & 2 \end{pmatrix}
\).
The determinant is \( \det H = 2\cdot2 - 2\cdot2 = 0 \), so the second derivative test is inconclusive.
Step 3: Direct analysis
Notice that
\(
f(x,y) = x^{2} + 2xy + y^{2} = (x + y)^{2}.
\)
Hence \( f(x,y) \ge 0 \) for all \( (x,y) \) and \( f(x,y) = 0 \) exactly when \( x + y = 0 \).
Therefore every critical point on the line \( y = -x \) gives the smallest possible value of the function.
All points on the line \( y = -x \) are critical points. Each of these points is a global minimum with \( f(x,y) = 0 \). There are no maxima.
Q5(a). Find the interval on which f is increasing, decreasing, concave up, and concave down for \(f(x)=x^{3}-12x-5\).
▶
Intervals of Increase, Decrease, Concavity for \(f(x)=x^{3}-12x-5\)
First compute the derivatives needed.
First derivative
\(
f'(x)=\frac{d}{dx}\big(x^{3}-12x-5\big)=3x^{2}-12=3(x^{2}-4)=3(x-2)(x+2)
\)
Critical points occur where \(f'(x)=0\), so at \(x=-2\) and \(x=2\).
Monotonicity
Test sign of \(f'(x)\) on the intervals determined by the critical points.
For \(x< -2\) choose \(x=-3\): \(
f'(-3)=3((-3)^{2}-4)=3(9-4)=15>0
\) so \(f\) is increasing on \(\;(-\infty,-2)\).
For \(-2
For \(x>2\) choose \(x=3\): \(
f'(3)=3(9-4)=15>0
\) so \(f\) is increasing on \((2,\infty)\).
Second derivative and concavity
\(
f''(x)=\frac{d}{dx}(3x^{2}-12)=6x
\)
Inflection point occurs where \(f''(x)=0\), so at \(x=0\). For \(x<0\) we have \(f''(x)<0\) so \(f\) is concave down on \((-\infty,0)\). For \(x>0\) we have \(f''(x)>0\) so \(f\) is concave up on \((0,\infty)\).
Local extrema
Using the sign change of \(f'\): at \(x=-2\) derivative changes from positive to negative so \(f\) has a local maximum at \(x=-2\). At \(x=2\) derivative changes from negative to positive so \(f\) has a local minimum at \(x=2\).
Values at those points are
\(
f(-2)=(-2)^{3}-12(-2)-5=-8+24-5=11
\)
and
\(
f(2)=2^{3}-12(2)-5=8-24-5=-21
\).
\(f\) is increasing on \((-\infty,-2)\) and \((2,\infty)\).
\(f\) is decreasing on \((-2,2)\).
\(f\) is concave down on \((-\infty,0)\) and concave up on \((0,\infty)\).
There is a local maximum at \(x=-2\) with \(f(-2)=11\) and a local minimum at \(x=2\) with \(f(2)=-21\).
There is an inflection point at \(x=0\) with \(f(0)=-5\).
Q5(b). A company manufactures two types of a certain product. The joint cost function for producing x units of product A and y units of product B is given by \(c(x,y)=x^{2}+3xy+400\). Find the quantity of each that results in the lowest cost.
▶
Find the quantities that give the lowest cost for \(c(x,y)=x^{2}+3xy+400\)
We assume x and y represent production quantities. Quantities cannot be negative, so we consider the region \(x\ge 0,\; y\ge 0\).
Step 1: Compute first partial derivatives
\(
c_{x}=\frac{\partial c}{\partial x}=2x+3y
\)
\(
c_{y}=\frac{\partial c}{\partial y}=3x
\)
Step 2: Stationary points
Set the gradient equal to zero:
\(
2x+3y=0 \quad\text{and}\quad 3x=0
\).
From \(3x=0\) we get \(x=0\). Substituting into \(2x+3y=0\) gives \(3y=0\) so \(y=0\).
Thus the only stationary point in the plane is \((0,0)\).
Step 3: Classification and global minimum on the feasible region
The Hessian matrix is
\(
H=\begin{pmatrix}2 & 3 \\[4pt] 3 & 0\end{pmatrix}
\)
with determinant \(\det H=2\cdot0-3\cdot3=-9<0\). So \((0,0)\) is a saddle point for the unconstrained problem and not a local minimum there.
However if we restrict to nonnegative quantities \(x\ge 0,\; y\ge 0\), evaluate the cost when \(x=0\):
\(
c(0,y)=0+0+400=400
\)
for every \(y\ge 0\). For any \(x>0\) and \(y\ge 0\), \(x^{2}+3xy+400\ge x^{2}+0+400>400\). Therefore the smallest possible cost on the feasible region is
\(
400
\)
and it is achieved at every point with
\(
x=0
\)
and any
\(
y\ge 0
\).
Remarks
If negative production were allowed the cost is unbounded below, so no minimum exists. If you require strictly positive production of both products, there is no minimum but the infimum is 400 and is approached as \(x\to 0^{+}\).
On the realistic domain \(x\ge 0,\; y\ge 0\) the lowest cost is \(400\). This minimum occurs for all points with
\(
x=0
\)
and any
\(
y\ge 0
\).
Q5(c). The total cost and total revenue functions for a product are \(C(q) = 500 + 100q + 0.5q^{2}\) and \(R(q) = 500q\). Find the profit maximizing level of output.
▶
Profit Maximization Problem
We are given the total cost and total revenue functions:
\(
C(q) = 500 + 100q + 0.5q^{2}, \quad R(q) = 500q
\)
Step 1: Define the profit function
\(
\pi(q) = R(q) - C(q) = 500q - \big(500 + 100q + 0.5q^{2}\big)
\)
Simplify:
\(
\pi(q) = -0.5q^{2} + 400q - 500
\)
Step 2: First derivative (critical point)
\(
\pi'(q) = \frac{d}{dq}\big(-0.5q^{2} + 400q - 500\big) = -q + 400
\)
Set derivative equal to zero:
\(
-q + 400 = 0 \quad \Rightarrow \quad q = 400
\)
Step 3: Second derivative test
\(
\pi''(q) = \frac{d}{dq}(-q + 400) = -1 < 0
\)
Since the second derivative is negative, the profit function is concave down, so \(q=400\) gives a maximum.
Step 4: Maximum profit value
\(
\pi(400) = -0.5(400)^{2} + 400(400) - 500
\)
\(
= -0.5(160000) + 160000 - 500
\)
\(
= -80000 + 160000 - 500 = 79500
\)
The profit-maximizing output is
\(
q = 400
\)
units, and the maximum profit is
\(
79,500
\).
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