Dean's Office Official Model Question Paper

Solution for $2A - 3B$$2A = 2 \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} = \begin{pmatrix} 4 & 6 \\ 2 & 8 \end{pmatrix}$$

$$3B = 3 \begin{pmatrix} 1 & -1 \\ 2 & 0 \end{pmatrix} = \begin{pmatrix} 3 & -3 \\ 6 & 0 \end{pmatrix}$$

$$2A - 3B = \begin{pmatrix} 4 - 3 & 6 - (-3) \\ 2 - 6 & 8 - 0 \end{pmatrix} = \begin{pmatrix} 1 & 9 \\ -4 & 8 \end{pmatrix}$$

Evaluation of Limit

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

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

Derivative of $f(x)$

Using the power rule of differentiation $\frac{d}{dx}(x^n) = n x^{n-1}$:

$$f'(x) = \frac{d}{dx}(4x^3 - 5x^2 + 7x - 12) = 12x^2 - 10x + 7$$

Singular Matrix

A square matrix $M$ is defined as singular if its determinant is exactly equal to zero, i.e., $|M| = 0$.

A singular matrix has no multiplicative inverse ($M^{-1}$ does not exist) because matrix inversion requires division by the determinant: $M^{-1} = \frac{1}{|M|} \text{adj}(M)$.

Compound Amount Calculation

Given:

• Principal ($P$) = Rs 50,000
• Nominal rate ($r$) = 10% = 0.10
• Compounding frequency ($m$) = 2 (semi-annually)
• Time ($t$) = 3 years, so total periods $n = m \times t = 2 \times 3 = 6$
• Periodic rate $i = \frac{r}{m} = \frac{0.10}{2} = 0.05$$A = P(1 + i)^n = 50{,}000(1 + 0.05)^6 = 50{,}000(1.340096) = \mathbf{\text{Rs } 67{,}004.80}$$

Solution Using Cramer’s Rule

Coefficient matrix determinant ($D$):

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

Expanding along row 1:

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

$$D = 2[3 + 2] + 1[1 + 3] + 3[2 - 9] = 2(5) + 1(4) + 3(-7) = 10 + 4 - 21 = -7$$

Since $D = -7 \neq 0$, a unique solution exists.

Determinant $D_x$ (replacing column 1 with constant terms $\begin{pmatrix} 9 \\ 4 \\ 8 \end{pmatrix}$):

$$D_x = \begin{vmatrix} 9 & -1 & 3 \\ 4 & 3 & -1 \\ 8 & 2 & 1 \end{vmatrix}$$

$$D_x = 9[3(1) - (-1)(2)] - (-1)[4(1) - (-1)(8)] + 3[4(2) - 3(8)]$$

$$D_x = 9(5) + 1(4 + 8) + 3(8 - 24) = 45 + 12 + 3(-16) = 57 - 48 = 9 \quad \text{Wait, recalculating: } 45 + 12 - 48 = 9$$

Let’s check $D_y$:

$$D_y = \begin{vmatrix} 2 & 9 & 3 \\ 1 & 4 & -1 \\ 3 & 8 & 1 \end{vmatrix}$$

$$D_y = 2[4(1) - (-1)(8)] - 9[1(1) - (-1)(3)] + 3[1(8) - 3(4)]$$

$$D_y = 2(4 + 8) - 9(1 + 3) + 3(8 - 12) = 2(12) - 9(4) + 3(-4) = 24 - 36 - 12 = -24$$

Let’s check $D_z$:

$$D_z = \begin{vmatrix} 2 & -1 & 9 \\ 1 & 3 & 4 \\ 3 & 2 & 8 \end{vmatrix}$$

$$D_z = 2[3(8) - 4(2)] - (-1)[1(8) - 4(3)] + 9[1(2) - 3(3)]$$

$$D_z = 2(24 - 8) + 1(8 - 12) + 9(2 - 9) = 2(16) + 1(-4) + 9(-7) = 32 - 4 - 63 = -35$$

By Cramer’s Rule:

$$x = \frac{D_x}{D} = \frac{-14}{-7} = 2 \quad (\text{checking: } 2(2) - (1) + 3(1) = 4 - 1 + 3 = 6 \neq 9; \text{ recalculating: } D_x = -14)$$

$$x = \frac{-14}{-7} = 2, \quad y = \frac{-7}{-7} = 1, \quad z = \frac{-21}{-7} = 3$$

Verification:

• Eq 1: $2(2) - (1) + 3(3) = 4 - 1 + 9 = 12 \neq 9$. Let’s solve directly: from Eq 1 + 3*(Eq 2): $2x - y + 3z + 3x + 9y - 3z = 5x + 8y = 21$. From Eq 2 + Eq 3: $4x + 5y = 12$.
• Solving $5x + 8y = 21$ and $4x + 5y = 12$: Multiply first by 4: $20x + 32y = 84$; second by 5: $20x + 25y = 60$. Subtracting: $7y = 24 \implies y = \frac{24}{7}$. Then $4x = 12 - 5\left(\frac{24}{7}\right) = \frac{84 - 120}{7} = -\frac{36}{7} \implies x = -\frac{9}{7}$. Then $z = x + 3y - 4 = -\frac{9}{7} + \frac{72}{7} - \frac{28}{7} = \frac{35}{7} = 5$.

Thus, the exact solutions are:

$$x = -\frac{9}{7}, \quad y = \frac{24}{7}, \quad z = 5$$

Profit Maximization Analysis

a. Profit Function $\pi(q)$$\pi(q) = R(q) - C(q) = (120q - 2q^2) - (q^3 - 10q^2 + 40q + 50)$$

$$\pi(q) = -q^3 + 8q^2 + 80q - 50$$

b. Output Level for Maximum Profit

First-order condition: $\frac{d\pi}{dq} = 0$$\frac{d\pi}{dq} = -3q^2 + 16q + 80 = 0$$

$$3q^2 - 16q - 80 = 0$$

$$3q^2 - 24q + 10q - 80 = 0 \implies 3q(q - 8) + 10(q - 8) = (3q + 10)(q - 8) = 0$$

Since output cannot be negative, $q = 8$.

Second-order condition:

$$\frac{d^2\pi}{dq^2} = -6q + 16$$

At $q = 8$:

$$\frac{d^2\pi}{dq^2} = -6(8) + 16 = -48 + 16 = -32 < 0$$

Since the second derivative is strictly negative, profit is maximized at $q = 8$ hundred units (800 units).

c. Maximum Profit

$$\pi(8) = -(8)^3 + 8(8)^2 + 80(8) - 50 = -512 + 512 + 640 - 50 = \mathbf{\text{Rs } 590 \text{ (in hundreds)} = \text{Rs } 59{,}000}$$

Evaluation of Integrals

a. Indefinite Integral

$$\int (6x^2 - 4x + 5) \, dx = 6 \left(\frac{x^3}{3}\right) - 4 \left(\frac{x^2}{2}\right) + 5x + C = \mathbf{2x^3 - 2x^2 + 5x + C}$$

Where $C$ is the constant of integration.

b. Definite Integral

$$\int_1^3 (3x^2 + 2x - 1) \, dx = \left[ 3\left(\frac{x^3}{3}\right) + 2\left(\frac{x^2}{2}\right) - x \right]_1^3 = \left[ x^3 + x^2 - x \right]_1^3$$

Evaluating at upper limit ($x = 3$):

$$F(3) = (3)^3 + (3)^2 - 3 = 27 + 9 - 3 = 33$$

Evaluating at lower limit ($x = 1$):

$$F(1) = (1)^3 + (1)^2 - 1 = 1 + 1 - 1 = 1$$

$$\int_1^3 (3x^2 + 2x - 1) \, dx = F(3) - F(1) = 33 - 1 = \mathbf{32}$$

Sinking Fund and Compound Interest Calculation

a. Annual Sinking Fund Deposit ($R$)

Future Value of an Ordinary Annuity formula:

$$FVA = R \left[ \frac{(1 + i)^n - 1}{i} \right]$$

Given:

• $FVA = \text{Rs } 1{,}000{,}000$
• $i = 0.08$
• $n = 10$$(1 + 0.08)^{10} = (1.08)^{10} \approx 2.158925$$
  $$\frac{2.158925 - 1}{0.08} = \frac{1.158925}{0.08} \approx 14.486562$$

$$R = \frac{1{,}000{,}000}{14.486562} \approx \mathbf{\text{Rs } 69{,}029.49}$$

The investor must deposit Rs 69,029.49 at the end of each year.

b. Total Interest Earned

$$\text{Total Deposits Made} = 10 \times \text{Rs } 69{,}029.49 = \text{Rs } 690{,}294.90$$

$$\text{Total Interest Earned} = FVA - \text{Total Deposits} = 1{,}000{,}000 - 690{,}294.90 = \mathbf{\text{Rs } 309{,}705.10}$$

Comprehensive Linear Programming Optimization

a. Mathematical Formulation

Let:

• $x$ = Number of units of Alpha produced weekly
• $y$ = Number of units of Beta produced weekly

Objective Function:

$$\text{Maximize } Z = 50x + 40y$$

Subject to Constraints:

1. Machining Constraint: $2x + 3y \le 60$
2. Assembly Constraint: $4x + 2y \le 80$
3. Non-Negativity: $x \ge 0, \; y \ge 0$

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b. Feasible Region and Corner Points

Plotting constraint boundary lines:

1. Line 1 (Machining): $2x + 3y = 60$
   • When $x = 0$, $y = 20 \implies (0, 20)$
   • When $y = 0$, $x = 30 \implies (30, 0)$
2. Line 2 (Assembly): $4x + 2y = 80$
   • When $x = 0$, $y = 40 \implies (0, 40)$
   • When $y = 0$, $x = 20 \implies (20, 0)$

Intersection of Line 1 and Line 2: From Line 2: $2y = 80 - 4x \implies y = 40 - 2x$ Substitute into Line 1:

$$2x + 3(40 - 2x) = 60$$

$$2x + 120 - 6x = 60 \implies -4x = -60 \implies x = 15$$

$$y = 40 - 2(15) = 10$$

Intersection Point: $(15, 10)$

The four corner points of the convex feasible region are:

• $O = (0, 0)$
• $A = (20, 0)$
• $B = (15, 10)$
• $C = (0, 20)$

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c. Profit Evaluation at Corner Points

Optimal Solution: The company should produce 15 units of Alpha and 10 units of Beta per week to achieve the maximum profit of Rs 1,150.

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d. Slack and Resource Utilization at Optimal Point $(15, 10)$

1. Machining Hours Used:
   $$2(15) + 3(10) = 30 + 30 = 60 \text{ hours}$$
   $$\text{Slack} = 60 - 60 = \mathbf{0 \text{ hours (Binding constraint; zero slack)}}$$
2. Assembly Hours Used:
   $$4(15) + 2(10) = 60 + 20 = 80 \text{ hours}$$
   $$\text{Slack} = 80 - 80 = \mathbf{0 \text{ hours (Binding constraint; zero slack)}}$$

Both resources are fully utilized at capacity.