Dean's Office Official Model Question Paper

Answer: The symmetric difference is defined as:

$$A \Delta B = (A - B) \cup (B - A)$$

• $A - B = \{1, 2\}$
• $B - A = \{5, 6\}$$A \Delta B = \{1, 2, 5, 6\}$$

Answer: Direct substitution yields the indeterminate form $\frac{0}{0}$. Factoring the numerator:

$$\lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3} = \lim_{x \to 3} (x + 3) = 3 + 3 = 6$$

Answer: Using the chain rule:

$$\frac{dy}{dx} = \frac{1}{3x^2 + 5x + 1} \cdot \frac{d}{dx}(3x^2 + 5x + 1) = \frac{6x + 5}{3x^2 + 5x + 1}$$

Answer:

$$\det(A) = |A| = (2)(5) - (4)(-3) = 10 - (-12) = 10 + 12 = 22$$

Answer:

$$\int (4x^3 - 6x + 7) \, dx = 4 \left(\frac{x^4}{4}\right) - 6 \left(\frac{x^2}{2}\right) + 7x + C = x^4 - 3x^2 + 7x + C$$

where $C$ is the arbitrary constant of integration.

Solution: Cramer’s Rule

1. Coefficient Matrix Determinant ($D$)

$$D = \begin{vmatrix} 2 & -1 & 3 \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix}$$

Expanding along the first row:

$$D = 2 \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} - (-1) \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} + 3 \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix}$$

$$D = 2[1 - (-1)] + 1[1 - 1] + 3[-1 - 1] = 2(2) + 1(0) + 3(-2) = 4 + 0 - 6 = -2$$

Since $D = -2 \neq 0$, a unique solution exists.

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2. Determinant $D_x$ (Replacing Column 1 with Constants):

$$D_x = \begin{vmatrix} 9 & -1 & 3 \\ 6 & 1 & 1 \\ 2 & -1 & 1 \end{vmatrix}$$

$$D_x = 9(1 - (-1)) + 1(6 - 2) + 3(-6 - 2) = 9(2) + 1(4) + 3(-8) = 18 + 4 - 24 = -2$$

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3. Determinant $D_y$ (Replacing Column 2 with Constants):

$$D_y = \begin{vmatrix} 2 & 9 & 3 \\ 1 & 6 & 1 \\ 1 & 2 & 1 \end{vmatrix}$$

$$D_y = 2(6 - 2) - 9(1 - 1) + 3(2 - 6) = 2(4) - 9(0) + 3(-4) = 8 - 0 - 12 = -4$$

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4. Determinant $D_z$ (Replacing Column 3 with Constants):

$$D_z = \begin{vmatrix} 2 & -1 & 9 \\ 1 & 1 & 6 \\ 1 & -1 & 2 \end{vmatrix}$$

$$D_z = 2(2 - (-6)) + 1(2 - 6) + 9(-1 - 1) = 2(8) + 1(-4) + 9(-2) = 16 - 4 - 18 = -6$$

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5. Calculate Values:

$$x = \frac{D_x}{D} = \frac{-2}{-2} = 1$$

$$y = \frac{D_y}{D} = \frac{-4}{-2} = 2$$

$$z = \frac{D_z}{D} = \frac{-6}{-2} = 3$$

Verification: $x + y + z = 1 + 2 + 3 = 6$ (Satisfied). The solution is: $x = 1, y = 2, z = 3$.

Solution: Profit Maximization Analysis

a) Profit Function $\pi(Q)$$\pi(Q) = TR - TC = (1,200Q - 2Q^2) - (Q^3 - 61.5Q^2 + 1,500Q + 5,000)$$

$$\pi(Q) = -Q^3 + 59.5Q^2 - 300Q - 5,000$$

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b) First-Order Condition for Profit Maximization

Set the first derivative with respect to $Q$ equal to zero:

$$\frac{d\pi}{dQ} = -3Q^2 + 119Q - 300 = 0$$

$$3Q^2 - 119Q + 300 = 0$$

Using the quadratic formula: $Q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$Q = \frac{119 \pm \sqrt{(-119)^2 - 4(3)(300)}}{2(3)} = \frac{119 \pm \sqrt{14,161 - 3,600}}{6} = \frac{119 \pm \sqrt{10,561}}{6}$$

$$\sqrt{10,561} \approx 102.7667$$

$$Q_1 = \frac{119 + 102.7667}{6} = \frac{221.7667}{6} \approx 36.96 \approx 37 \text{ units}$$

$$Q_2 = \frac{119 - 102.7667}{6} = \frac{16.2333}{6} \approx 2.71 \text{ units}$$

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c) Second-Order Derivative Verification

$$\frac{d^2\pi}{dQ^2} = -6Q + 119$$

• At $Q = 2.71$:
  $$\frac{d^2\pi}{dQ^2} = -6(2.71) + 119 = -16.26 + 119 = +102.74 > 0 \quad (\text{Local Minimum})$$
• At $Q = 36.96$:
  $$\frac{d^2\pi}{dQ^2} = -6(36.96) + 119 = -221.76 + 119 = -102.76 < 0 \quad (\text{Confirmed Maximum})$$

Hence, profit is maximized at $Q^* \approx 36.96$ units (or 37 units).

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d) Maximum Profit Computation

Substitute $Q = 37$ into $\pi(Q)$:

$$\pi(37) = -(37)^3 + 59.5(37)^2 - 300(37) - 5,000$$

$$\pi(37) = -50,653 + 59.5(1,369) - 11,100 - 5,000$$

$$\pi(37) = -50,653 + 81,455.5 - 11,100 - 5,000 = \text{Rs. } 14,702.50$$

Solution: Graphical Method for Linear Programming

1. Convert Inequalities to Boundary Lines

1. Line 1: $2x_1 + x_2 = 100$
   • If $x_1 = 0 \implies x_2 = 100 \implies (0, 100)$
   • If $x_2 = 0 \implies x_1 = 50 \implies (50, 0)$
2. Line 2: $x_1 + x_2 = 80$
   • If $x_1 = 0 \implies x_2 = 80 \implies (0, 80)$
   • If $x_2 = 0 \implies x_1 = 80 \implies (80, 0)$
3. Line 3: $x_1 + 2x_2 = 120$
   • If $x_1 = 0 \implies x_2 = 60 \implies (0, 60)$
   • If $x_2 = 0 \implies x_1 = 120 \implies (120, 0)$
4. Non-negativity: $x_1 \ge 0, x_2 \ge 0$ restricts the feasible region to the first quadrant.

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2. Identify Feasible Corner Points

The feasible region is the polygon bounded by:

• Point $O$: $(0, 0)$
• Point $A$: $(0, 60)$ (from Line 3 with $y$-axis)
• Point $B$: Intersection of Line 2 ($x_1 + x_2 = 80$) and Line 3 ($x_1 + 2x_2 = 120$):
  $$(x_1 + 2x_2) - (x_1 + x_2) = 120 - 80 \implies x_2 = 40$$
  $$x_1 = 80 - 40 = 40 \implies B(40, 40)$$
• Point $C$: Intersection of Line 1 ($2x_1 + x_2 = 100$) and Line 2 ($x_1 + x_2 = 80$):
  $$(2x_1 + x_2) - (x_1 + x_2) = 100 - 80 \implies x_1 = 20$$
  $$x_2 = 80 - 20 = 60 \implies C(20, 60)$$
  Check feasibility of Point $C$ on Line 3: $x_1 + 2x_2 = 20 + 2(60) = 140 > 120$ (Infeasible!). Thus, Point $B(40, 40)$ connects directly to:
• Point $D$: Intersection of Line 1 ($2x_1 + x_2 = 100$) and Line 3 ($x_1 + 2x_2 = 120$): From Line 1: $x_2 = 100 - 2x_1$. Substitute: $x_1 + 2(100 - 2x_1) = 120 \implies x_1 + 200 - 4x_1 = 120 \implies 3x_1 = 80 \implies x_1 = 26.67, x_2 = 46.67$. Let’s re-verify intersection of Line 1 ($2x_1+x_2=100$) and Line 2 ($x_1+x_2=80$): $x_1=20, x_2=60$. Wait, at $x_1=40, x_2=40$:
  • Line 1: $2(40) + 40 = 120 > 100$ (Point $B(40, 40)$ is outside Line 1!). Let’s find the true corner points:
  1. $(0, 0)$
  2. $(0, 60)$ (on Line 3)
  3. Intersection of Line 3 ($x_1 + 2x_2 = 120$) and Line 2 ($x_1 + x_2 = 80$): $(40, 40)$ violates Line 1 ($2x_1+x_2 \le 100$). Intersection of Line 1 ($2x_1+x_2=100$) and Line 3 ($x_1+2x_2=120$): Multiply Line 1 by 2: $4x_1 + 2x_2 = 200$. Subtract Line 3: $3x_1 = 80 \implies x_1 = 80/3 \approx 26.67, x_2 = 140/3 \approx 46.67$. Check Line 2: $x_1 + x_2 = 80/3 + 140/3 = 220/3 = 73.33 \le 80$ (Feasible!).
  4. Point $(50, 0)$ (on Line 1).

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3. Evaluate Objective Function $Z = 30x_1 + 40x_2$ at Corner Points:

Conclusion: The maximum value of $Z$ is $2,666.67$, occurring at $x_1 = \frac{80}{3} \approx 26.67$ and $x_2 = \frac{140}{3} \approx 46.67$.

Solution: Consumers’ and Producers’ Surplus

a) Market Equilibrium

Set demand price equal to supply price ($P_d = P_s$):

$$100 - Q^2 = 20 + 2Q$$

$$Q^2 + 2Q - 80 = 0$$

$$(Q + 10)(Q - 8) = 0$$

Since quantity cannot be negative, $Q_e = 8$ units.

Substitute $Q_e = 8$ into the supply function to find $P_e$:

$$P_e = 20 + 2(8) = 20 + 16 = \text{Rs. } 36$$

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b) Consumers’ Surplus (CS)

$$\text{CS} = \int_0^{Q_e} [P_d(Q) - P_e] \, dQ = \int_0^8 (100 - Q^2 - 36) \, dQ = \int_0^8 (64 - Q^2) \, dQ$$

$$\text{CS} = \left[ 64Q - \frac{Q^3}{3} \right]_0^8 = 64(8) - \frac{512}{3} = 512 - 170.67 = \text{Rs. } 341.33$$

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c) Producers’ Surplus (PS)

$$\text{PS} = \int_0^{Q_e} [P_e - P_s(Q)] \, dQ = \int_0^8 [36 - (20 + 2Q)] \, dQ = \int_0^8 (16 - 2Q) \, dQ$$

$$\text{PS} = \left[ 16Q - Q^2 \right]_0^8 = 16(8) - (8)^2 = 128 - 64 = \text{Rs. } 64$$

Summary:

• Equilibrium: $Q_e = 8 \text{ units}, P_e = \text{Rs. } 36$
• Consumers’ Surplus = Rs. 341.33
• Producers’ Surplus = Rs. 64.00

Comprehensive Mathematical Case Solution

a) Matrix Formulation and Non-Singularity Proof

$$\begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 2 \\ 1 & 1 & 2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 1,800 \\ 2,800 \\ 1,200 \end{pmatrix}$$

Let $A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 2 \\ 1 & 1 & 2 \end{pmatrix}, X = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}, B = \begin{pmatrix} 1,800 \\ 2,800 \\ 1,200 \end{pmatrix}$.

Compute $|A|$ expanding along the first row:

$$|A| = 1 \begin{vmatrix} 4 & 2 \\ 1 & 2 \end{vmatrix} - 2 \begin{vmatrix} 2 & 2 \\ 1 & 2 \end{vmatrix} + 3 \begin{vmatrix} 2 & 4 \\ 1 & 1 \end{vmatrix}$$

$$|A| = 1(8 - 2) - 2(4 - 2) + 3(2 - 4) = 1(6) - 2(2) + 3(-2) = 6 - 4 - 6 = -4$$

Since $|A| = -4 \neq 0$, matrix $A$ is non-singular and its inverse $A^{-1}$ exists.

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b) Finding $A^{-1}$ via Adjoint Method

The cofactors $C_{ij}$ of matrix $A$:

• $C_{11} = +(8 - 2) = 6$
• $C_{12} = -(4 - 2) = -2$
• $C_{13} = +(2 - 4) = -2$
• $C_{21} = -(4 - 3) = -1$
• $C_{22} = +(2 - 3) = -1$
• $C_{23} = -(1 - 2) = 1$
• $C_{31} = +(4 - 12) = -8$
• $C_{32} = -(2 - 6) = 4$
• $C_{33} = +(4 - 4) = 0$

Cofactor matrix $C = \begin{pmatrix} 6 & -2 & -2 \\ -1 & -1 & 1 \\ -8 & 4 & 0 \end{pmatrix}$.

Adjoint matrix $\text{Adj}(A) = C^T$:

$$\text{Adj}(A) = \begin{pmatrix} 6 & -1 & -8 \\ -2 & -1 & 4 \\ -2 & 1 & 0 \end{pmatrix}$$

Inverse Matrix $A^{-1} = \frac{1}{|A|} \text{Adj}(A)$:

$$A^{-1} = -\frac{1}{4} \begin{pmatrix} 6 & -1 & -8 \\ -2 & -1 & 4 \\ -2 & 1 & 0 \end{pmatrix} = \begin{pmatrix} -1.5 & 0.25 & 2.0 \\ 0.5 & 0.25 & -1.0 \\ 0.5 & -0.25 & 0 \end{pmatrix}$$

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c) Optimal Output Vector $X = A^{-1} B$ and Gross Profit

$$\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = -\frac{1}{4} \begin{pmatrix} 6 & -1 & -8 \\ -2 & -1 & 4 \\ -2 & 1 & 0 \end{pmatrix} \begin{pmatrix} 1,800 \\ 2,800 \\ 1,200 \end{pmatrix}$$

1. Calculate $x_1$ (Shirts):

   $$x_1 = -\frac{1}{4} [6(1,800) - 1(2,800) - 8(1,200)] = -\frac{1}{4} [10,800 - 2,800 - 9,600] = -\frac{1}{4} [-1,600] = 400$$

2. Calculate $x_2$ (Trousers):

   $$x_2 = -\frac{1}{4} [-2(1,800) - 1(2,800) + 4(1,200)] = -\frac{1}{4} [-3,600 - 2,800 + 4,800] = -\frac{1}{4} [-1,600] = 400$$

3. Calculate $x_3$ (Jackets):

   $$x_3 = -\frac{1}{4} [-2(1,800) + 1(2,800) + 0] = -\frac{1}{4} [-3,600 + 2,800] = -\frac{1}{4} [-800] = 200$$

Verification:

• Cutting: $1(400) + 2(400) + 3(200) = 400 + 800 + 600 = 1,800$ hours.
• Sewing: $2(400) + 4(400) + 2(200) = 800 + 1,600 + 400 = 2,800$ hours.
• Finishing: $1(400) + 1(400) + 2(200) = 400 + 400 + 400 = 1,200$ hours. (All capacity exactly consumed!)

Total Monthly Gross Profit:

$$\text{Profit} = 300(400) + 500(400) + 800(200) = 120,000 + 200,000 + 160,000 = \text{Rs. } 480,000$$