Graded Assignment 2

Q1

Maximum Profit from a Quadratic Model

In a particular year, the profit (in lakhs of Rs.) of Star Fish company is given by the polynomial $P(x) = ax^2 + bx + c$ where $x$ denotes the number of months since the beginning of the year.

January ( $x=1$ ): Loss of Rs. 45 lakhs.
February ( $x=2$ ): Loss of Rs. 19 lakhs.
March ( $x=3$ ): Profit of Rs. 3 lakhs.

Find the maximum profit and the month in which it occurs.

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Quadratic Extrema

The maximum or minimum of a quadratic function $P(x) = ax^2 + bx + c$ occurs at its vertex, where $x_v = -\frac{b}{2a}$ .

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Set up the system of equations:
$\begin{cases} a(1)^2 + b(1) + c = -45 \\ a(2)^2 + b(2) + c = -19 \\ a(3)^2 + b(3) + c = 3 \end{cases} \implies \begin{cases} a + b + c = -45 \quad (1) \\ 4a + 2b + c = -19 \quad (2) \\ 9a + 3b + c = 3 \quad (3) \end{cases}$
Solve for $a, b, c$ :
- $(2) - (1) \implies 3a + b = 26$
- $(3) - (2) \implies 5a + b = 22$
- Subtracting these: $2a = -4 \implies a = -2$ .
- Substitute $a$ : $3(-2) + b = 26 \implies b = 32$ .
- Substitute $a, b$ : $-2 + 32 + c = -45 \implies c = -75$ .
The profit function is $P(x) = -2x^2 + 32x - 75$ .
Find the Maximum: Since $a = -2 < 0$ , the parabola opens downwards (max).
$x_{max} = -\frac{32}{2(-2)} = 8 \quad (\text{August})$
Max Profit:
$P(8) = -2(64) + 32(8) - 75 = -128 + 256 - 75 = 53$

Answer: Maximum profit is Rs. 53 Lakhs in August.

Q2

Properties of Augmented Matrices

If $A$ be a $3 \times 4$ matrix and $b$ be a $3 \times 1$ matrix, analyze the properties of the augmented matrix $[A|b]$ .

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Let’s analyze the statements based on Gaussian Elimination principles:

Row Operations & Solutions: Elementary row operations do not change the solution set. The systems $Ax=b$ and $A'x=b'$ (where $(A'|b')$ is obtained from row ops) are equivalent. (True)
RREF & Existence: Having an RREF $(A'|b')$ does not guarantee a solution. If a row like $[0 \ 0 \ 0 \ | \ 1]$ exists, the system is inconsistent. (False)
RREF Structure: If $[A|b]$ is in RREF, its submatrix $A$ is naturally in RREF. (True)
Consistency Condition: If there is no row where the only non-zero entry is in the last column (i.e., no pivot in the augmented column), the system is consistent. (True)

Q3

Matrix Representation of a System

Ramya, Romy, and Farjana buy books. Let $x_1, x_2, x_3$ be prices of comic, horror, and novel respectively.

Ramya: 1 comic, 2 horror, 1 novel for 1000.
Romy: 2 comic, 5 horror, 1 novel for 2000.
Farjana: 4 comic, 5 horror, $c$ novels for $d$ .

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System Setup:

\begin{bmatrix} 1 & 2 & 1 \\ 2 & 5 & 1 \\ 4 & 5 & c \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 1000 \\ 2000 \\ d \end{bmatrix}

Analysis ( $|A|$ ):

\det(A) = 1(5c-5) - 2(2c-4) + 1(10-20) = 5c - 5 - 4c + 8 - 10 = c - 7

If $c \neq 7$ , determinant is non-zero $\implies$ Unique Solution.
If $c = 7$ , determinant is zero $\implies$ No solution or Infinite solutions.

Q4

Homogeneous Systems

Let $A$ be an $m \times n$ matrix such that $m < n$ . How many solutions does $Ax = 0$ have?

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Free Variables

Number of free variables = $n - \text{rank}(A)$ . Since $\text{rank}(A) \le m < n$ , there is always at least one free variable.

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Since it is a homogeneous system ( $Ax=0$ ), the trivial solution $\vec{x}=\vec{0}$ always exists. Because $m < n$ (more variables than equations), there must be at least one free variable, leading to Infinitely many solutions.

Q5

Mobile Phone Sales

Three shops sell three brands R, S, T. Given $P_S = 5$ (in thousands).

Shop A: $P_R + 3k P_S + (3k+4)P_T = 61$
Shop B: $P_R + k P_S + (4k+2)P_T = 65$
Shop C: $P_R + (2k+2)P_S + (3k+4)P_T = 66$

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Substituting $P_S = 5$ :

$P_R + 15k + (3k+4)P_T = 61$
$P_R + 5k + (4k+2)P_T = 65$
$P_R + (10k+10) + (3k+4)P_T = 66$

Comparing (1) and (3): $(1) - (3) \implies (15k - (10k+10)) = 61 - 66 \implies 5k - 10 = -5 \implies 5k = 5 \implies k=1$ .

Substitute $k=1$ :

(A): $P_R + 7P_T = 46$
(B): $P_R + 6P_T = 60$

Subtracting: $P_T = -14$ . (Note: The problem values lead to a negative price, likely a typo in the original question data, but the mathematical process yields -14).

Q6-8

System Analysis

Consider the system:

\begin{cases} 2x+3y+5z=1 \\ x+2y+3z=1 \\ x+y+2z=1 \end{cases}

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Q6: Number of Solutions $R_3 \to R_3 - R_2$ gives $-y - z = 0 \implies y+z=0$ . Original $x+y+2z=1$ vs $x+2y+3z=1$ . Row reducing leads to a contradiction ( $0 = -1$ ), so 0 solutions.

Q7: Determinant of RREF Determinant is only defined for square matrices. RREF is a property, not a square transformation equivalent. (Technically undefined for non-square, or 0/1 for square singular/non-singular).

Q8: Solution Sum For the system: $7x - 2y + z = 5$ , $3y - z = 1$ , $-3x + 4y - 2z = 10$ . Solving this yields $x+y+z = -287/11$ .

Q9

Matrix Rank

Let $A = [3 \ 5 \ 10 \ 7]$ . Find the number of non-zero rows in the RREF of $A^T A$ .

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Rank Theorem

$\text{rank}(A^T A) = \text{rank}(A)$ .

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$A$ is a $1 \times 4$ non-zero vector, so $\text{rank}(A) = 1$ . Therefore, $\text{rank}(A^T A) = 1$ . The number of non-zero rows in RREF is equal to the rank. Answer: 1.

Q10

Cubic Curve Fitting

Find $a - b + c - d$ for $f(x) = ax^3 + bx^2 + cx + d$ passing through $(1,1), (2,7), (-1,-5), (3,23)$ .

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We need to find the value of the expression $a(-1)^2 - b(-1)^2 + c(-1)^2 ...$ wait, let’s look at the point $(-1, -5)$ .

f(-1) = a(-1)^3 + b(-1)^2 + c(-1) + d = -a + b - c + d = -5

The question asks for $a - b + c - d$ . Notice that $a - b + c - d = -(-a + b - c + d) = -(-5) = 5$ . Answer: 5.

Q11-13

Traffic Flow Network

Analysis of traffic flow conservation (In = Out).

NW: $2x_1 + x_4 = 400$
NE: $2x_1 + 3x_2 = 1000$
SE: $3x_2 + 2x_3 = 900$
SW: $2x_3 + x_4 = c$

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Q12: South Street Flow ( $c$ ) Summing equations physically often cancels internal flows. However, solving the dependent system: $c = 300$ .

Q13: Max/Min Vehicles Based on constraints $x_i \ge 0$ :

$0 \le x_3 \le 150$
a (West/North): Range [100, 400]
b (East/North): Range [600, 900]
c (East/South): Range [0, 300]
d (West/South): Range [0, 300]