Business Mathematics - I 2023 Board Question Paper

Step-by-Step Solution:

Given the two points:

$$(x_1, y_1) = (8, -3) \quad \text{and} \quad (x_2, y_2) = (-7, 2)$$

Step 1: Calculate the slope ($m$) of the line

Using the two-point slope formula:

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$

$$m = \frac{2 - (-3)}{-7 - 8} = \frac{2 + 3}{-15} = \frac{5}{-15} = -\frac{1}{3}$$

Step 2: Use the point-slope form to find the equation

$$y - y_1 = m(x - x_1)$$

Substitute $(x_1, y_1) = (8, -3)$ and $m = -\frac{1}{3}$:

$$y - (-3) = -\frac{1}{3}(x - 8)$$

$$y + 3 = -\frac{1}{3}(x - 8)$$

Multiply both sides by $3$:

$$3(y + 3) = -(x - 8)$$

$$3y + 9 = -x + 8$$

Rearranging into the standard linear equation form $Ax + By + C = 0$:

$$x + 3y + 9 - 8 = 0$$

$$x + 3y + 1 = 0$$

Final Answer: The equation of the straight line is $x + 3y + 1 = 0$ (or in slope-intercept form, $y = -\frac{1}{3}x - \frac{1}{3}$).

Step-by-Step Solution:

Given quadratic equation:

$$x^2 - 7x + 6 = 0$$

Method 1: Factorization Method

We find two numbers that multiply to $+6$ and add up to $-7$. These numbers are $-6$ and $-1$:

$$x^2 - 6x - x + 6 = 0$$

$$x(x - 6) - 1(x - 6) = 0$$

$$(x - 6)(x - 1) = 0$$

Setting each factor to zero:

$$x - 6 = 0 \implies x = 6$$

$$x - 1 = 0 \implies x = 1$$

Method 2: Quadratic Formula Verification

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

For $a = 1, b = -7, c = 6$:

$$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(6)}}{2(1)} = \frac{7 \pm \sqrt{49 - 24}}{2} = \frac{7 \pm \sqrt{25}}{2} = \frac{7 \pm 5}{2}$$

$$x_1 = \frac{7 + 5}{2} = \frac{12}{2} = 6, \quad x_2 = \frac{7 - 5}{2} = \frac{2}{2} = 1$$

Final Answer: The solutions to the quadratic equation are $x = 1$ and $x = 6$.

Step-by-Step Solution:

Given:

• Depreciation rate ($r$) = $5\% = 0.05$ per annum on diminishing balance
• Number of years ($n$) = $6$ years
• Book value at the end of 6 years ($V_6$) = $\text{Rs } 6,250$
• Let the original cost be $C$.

Formula for Diminishing Balance Depreciation:

$$V_n = C(1 - r)^n$$

Calculation:

$$6,250 = C(1 - 0.05)^6$$

$$6,250 = C(0.95)^6$$

Evaluating $(0.95)^6$:

$$(0.95)^6 \approx 0.73509189$$

Now, solve for $C$:

$$C = \frac{6,250}{(0.95)^6} = \frac{6,250}{0.73509189} \approx 8,502.34$$

Final Answer: The original cost of the machinery was Rs 8,502.34.

Step-by-Step Solution:

Given supply function:

$$P = 25 + 0.5Q$$

Step 1: Express Quantity ($Q$) as a function of Price ($P$)

$$0.5Q = P - 25$$

$$Q = \frac{P - 25}{0.5} = 2(P - 25) = 2P - 50$$

Step 2: Differentiate $Q$ with respect to $P$$\frac{dQ}{dP} = 2$$

Step 3: Compute Point Elasticity of Supply ($e_s$)

The formula for point price elasticity of supply is:

$$e_s = \frac{P}{Q} \cdot \frac{dQ}{dP}$$

• Case 1: If price is Rs 90 ($P = 90$):

  $$Q = 2(90) - 50 = 180 - 50 = 130$$
  $$e_s = \frac{90}{130} \times 2 = \frac{180}{130} = \frac{18}{13} \approx 1.385$$

• Case 2: If price increases by Rs 90 from the base shutdown threshold ($P_0 = 25$): Then $P = 25 + 90 = 115$.

  $$Q = 2(115) - 50 = 230 - 50 = 180$$
  $$e_s = \frac{115}{180} \times 2 = \frac{230}{180} = \frac{23}{18} \approx 1.278$$

Final Answer: At $P = \text{Rs } 90$, the point elasticity of supply is $e_s = \frac{18}{13} \approx 1.385$ (supply is price elastic, $e_s > 1$).

Step-by-Step Solution:

Given arithmetic series:

$$3 + 5 + 7 + 9 + \dots$$

Step 1: Identify the components of the series

• First term ($a$) = $3$
• Common difference ($d$) = $5 - 3 = 2$
• Number of terms ($n$) = $25$

Step 2: Apply the sum of an arithmetic progression formula

$$S_n = \frac{n}{2} \left[ 2a + (n - 1)d \right]$$

Substitute the known values:

$$S_{25} = \frac{25}{2} \left[ 2(3) + (25 - 1) \times 2 \right]$$

$$S_{25} = \frac{25}{2} \left[ 6 + 24 \times 2 \right]$$

$$S_{25} = \frac{25}{2} \left[ 6 + 48 \right] = \frac{25}{2} \times 54$$

$$S_{25} = 25 \times 27 = 675$$

Final Answer: The sum of the series to 25 terms is $675$.

Step-by-Step Solution:

Given function:

$$y = x^3 - 5x^2 - 15x + 27$$

Step 1: Find the first and second derivatives

First derivative:

$$\frac{dy}{dx} = \frac{d}{dx}(x^3 - 5x^2 - 15x + 27) = 3x^2 - 10x - 15$$

Second derivative:

$$\frac{d^2y}{dx^2} = \frac{d}{dx}(3x^2 - 10x - 15) = 6x - 10$$

Step 2: Set the second derivative equal to zero

$$\frac{d^2y}{dx^2} = 0 \implies 6x - 10 = 0$$

$$6x = 10 \implies x = \frac{10}{6} = \frac{5}{3}$$

Step 3: Check concavity change (Third derivative test)

$$\frac{d^3y}{dx^3} = 6 \neq 0$$

Since the third derivative is non-zero, the concavity changes at $x = \frac{5}{3}$, confirming a point of inflection.

Step 4: Calculate the corresponding $y$-coordinate

Substitute $x = \frac{5}{3}$ into the original equation:

$$y = \left(\frac{5}{3}\right)^3 - 5\left(\frac{5}{3}\right)^2 - 15\left(\frac{5}{3}\right) + 27$$

$$y = \frac{125}{27} - 5\left(\frac{25}{9}\right) - 25 + 27$$

$$y = \frac{125}{27} - \frac{125}{9} + 2$$

$$y = \frac{125 - 375 + 54}{27} = -\frac{196}{27} \approx -7.26$$

Final Answer: The point of inflection is $\left(\frac{5}{3}, -\frac{196}{27}\right)$ (or approximately $(1.67, -7.26)$).

Step-by-Step Solution:

Given equation:

$$4^{x+3} \times 8^{x-3} = 256$$

Step 1: Express all terms as powers of base 2

• $4 = 2^2 \implies 4^{x+3} = (2^2)^{x+3} = 2^{2(x+3)} = 2^{2x + 6}$
• $8 = 2^3 \implies 8^{x-3} = (2^3)^{x-3} = 2^{3(x-3)} = 2^{3x - 9}$
• $256 = 2^8$

Step 2: Combine the exponents on the left-hand side

$$2^{2x + 6} \times 2^{3x - 9} = 2^8$$

$$2^{(2x + 6) + (3x - 9)} = 2^8$$

$$2^{5x - 3} = 2^8$$

Step 3: Equate the exponents and solve for $x$$5x - 3 = 8$$

$$5x = 8 + 3$$

$$5x = 11 \implies x = \frac{11}{5} = 2.2$$

Verification:

$$4^{2.2 + 3} \times 8^{2.2 - 3} = 4^{5.2} \times 8^{-0.8} = 2^{10.4} \times 2^{-2.4} = 2^{8} = 256$$

Final Answer: $x = \frac{11}{5} = 2.2$.

Step-by-Step Solution:

Evaluate the limit:

$$L = \lim_{x \to \infty} \frac{x^2 + 9x - 11}{7x^2 + 5x - 12}$$

Step 1: Divide numerator and denominator by the highest power of $x$ in the denominator ($x^2$)

$$L = \lim_{x \to \infty} \frac{\frac{x^2}{x^2} + \frac{9x}{x^2} - \frac{11}{x^2}}{\frac{7x^2}{x^2} + \frac{5x}{x^2} - \frac{12}{x^2}}$$

$$L = \lim_{x \to \infty} \frac{1 + \frac{9}{x} - \frac{11}{x^2}}{7 + \frac{5}{x} - \frac{12}{x^2}}$$

Step 2: Apply the limit property $\lim_{x \to \infty} \frac{1}{x^k} = 0$ for $k > 0$$L = \frac{1 + 0 - 0}{7 + 0 - 0} = \frac{1}{7}$$

Final Answer: The value of the limit is $\frac{1}{7}$.

Step-by-Step Solution:

Given geometric series:

$$1 + 2 + 4 + 8 + \dots$$

Step 1: Identify the parameters

• First term ($a$) = $1$
• Common ratio ($r$) = $\frac{2}{1} = 2$
• Term number ($n$) = $10$

Step 2: Apply the $n^{\text{th}}$ term formula for a Geometric Progression

$$T_n = a \cdot r^{n-1}$$

Substitute $a = 1, r = 2, n = 10$:

$$T_{10} = 1 \cdot 2^{10 - 1} = 2^9$$

$$T_{10} = 512$$

Final Answer: The tenth term of the geometric series is $512$.

Step-by-Step Solution:

Given function:

$$f(x) = 7x^2 - 3x + 9$$

To find the value of $f(3)$, substitute $x = 3$ into the function:

$$f(3) = 7(3)^2 - 3(3) + 9$$

$$f(3) = 7(9) - 9 + 9$$

$$f(3) = 63 - 9 + 9$$

$$f(3) = 63$$

Final Answer: $f(3) = 63$.

Step-by-Step Solution:

Given:

• Principal amount ($P$) = $\text{Rs } 20,000$
• Accumulated amount ($A$) = $\text{Rs } 40,000$
• Time period ($t$) = $5$ years
• Annual compound interest rate = $r$

Step 1: Formula for Annual Compound Interest

$$A = P(1 + r)^t$$

Substitute the given values:

$$40,000 = 20,000(1 + r)^5$$

$$\frac{40,000}{20,000} = (1 + r)^5$$

$$2 = (1 + r)^5$$

Step 2: Solve for $r$

Take the 5th root of both sides:

$$1 + r = 2^{1/5} = 2^{0.2}$$

Using logarithm or calculator:

$$1 + r \approx 1.148698$$

$$r \approx 1.148698 - 1 = 0.148698$$

$$r \approx 14.87\% \text{ per annum}$$

(Note: If compounding were continuous, $A = P e^{rt} \implies 2 = e^{5r} \implies r = \frac{\ln 2}{5} \approx 13.86\%$).

Final Answer: The required rate of compound interest (compounded annually) is $14.87\%$ per annum.

Step-by-Step Solution:

Given:

• Periodic annuity payment ($R$) = $\text{Rs } 5,000$ per year
• Time period ($n$) = $15$ years
• Interest rate ($i$) = $10\% = 0.10$ per annum
• Let the lump sum payment required now be the Present Value ($PV$) of the ordinary annuity.

Step 1: Formula for Present Value of an Ordinary Annuity

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

Step 2: Compute $(1 + i)^{-n}$$(1 + 0.10)^{-15} = (1.10)^{-15} = \frac{1}{(1.10)^{15}} \approx \frac{1}{4.177248} \approx 0.239392$$

Step 3: Compute $PV$$PV = 5,000 \left[ \frac{1 - 0.239392}{0.10} \right]$$

$$PV = 5,000 \left[ \frac{0.760608}{0.10} \right]$$

$$PV = 5,000 \times 7.60608 = 38,030.40$$

(Note: If payments are made at the beginning of each year (annuity due), $PV_{\text{due}} = PV \times (1 + i) = 38,030.40 \times 1.10 = \text{Rs } 41,833.44$).

Final Answer: To secure the annuity, an amount of Rs 38,030.40 should be paid now.

Step-by-Step Solution:

Given consumption function:

$$C = 700(1 - e^{-0.4y})$$

where $y$ represents income and $C$ represents consumption expenditure.

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(i) Level of consumption at $y = 5, 10, 15, 20, 25$:

1. For $y = 5$:

   $$C(5) = 700(1 - e^{-0.4 \times 5}) = 700(1 - e^{-2.0})$$
   Since $e^{-2.0} \approx 0.135335$:
   $$C(5) = 700(1 - 0.135335) = 700(0.864665) \approx 605.27$$

2. For $y = 10$:

   $$C(10) = 700(1 - e^{-0.4 \times 10}) = 700(1 - e^{-4.0})$$
   Since $e^{-4.0} \approx 0.018316$:
   $$C(10) = 700(1 - 0.018316) = 700(0.981684) \approx 687.18$$

3. For $y = 15$:

   $$C(15) = 700(1 - e^{-0.4 \times 15}) = 700(1 - e^{-6.0})$$
   Since $e^{-6.0} \approx 0.002479$:
   $$C(15) = 700(1 - 0.002479) = 700(0.997521) \approx 698.26$$

4. For $y = 20$:

   $$C(20) = 700(1 - e^{-0.4 \times 20}) = 700(1 - e^{-8.0})$$
   Since $e^{-8.0} \approx 0.000335$:
   $$C(20) = 700(1 - 0.000335) = 700(0.999665) \approx 699.77$$

5. For $y = 25$:

   $$C(25) = 700(1 - e^{-0.4 \times 25}) = 700(1 - e^{-10.0})$$
   Since $e^{-10.0} \approx 0.000045$:
   $$C(25) = 700(1 - 0.000045) = 700(0.999955) \approx 699.97$$

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(ii) Plotting table for graph paper:

Description of Graph: The curve passes through the origin $(0,0)$, increases sharply at initial income levels, exhibits diminishing marginal propensity to consume (concave downwards), and levels off horizontally towards its asymptote $C = 700$.

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(iii) Saturation Level:

The saturation level of consumption occurs as income becomes arbitrarily large ($y \to \infty$):

$$\text{Saturation Level} = \lim_{y \to \infty} C(y) = \lim_{y \to \infty} 700(1 - e^{-0.4y})$$

Since $\lim_{y \to \infty} e^{-0.4y} = 0$:

$$\text{Saturation Level} = 700(1 - 0) = 700$$

Final Answer:

• Levels of consumption: $C(5) \approx 605.27$, $C(10) \approx 687.18$, $C(15) \approx 698.26$, $C(20) \approx 699.77$, $C(25) \approx 699.97$.
• Saturation level is $700$.

Step-by-Step Solution:

Given:

• Demand function: $P_d = 80 - 0.3Q_d$
• Supply function: $P_s = 30 + 0.2Q_s$

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a) Calculate Equilibrium Price ($P^*$) and Quantity ($Q^*$):

At market equilibrium:

$$P_d = P_s$$

$$80 - 0.3Q = 30 + 0.2Q$$

$$80 - 30 = 0.2Q + 0.3Q$$

$$50 = 0.5Q$$

$$Q^* = \frac{50}{0.5} = 100 \text{ units}$$

Now substitute $Q^* = 100$ into either function to find $P^*$:

$$P^* = 80 - 0.3(100) = 80 - 30 = \text{Rs } 50$$

$$\text{Check with supply: } P^* = 30 + 0.2(100) = 30 + 20 = \text{Rs } 50$$

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b) Calculate Consumer’s Surplus, Producer’s Surplus, and Total Surplus:

1. Consumer’s Surplus ($CS$): The choke price (price intercept where $Q = 0$) for demand is $P_{\text{max}} = 80$.

   $$CS = \int_0^{Q^*} (P_d - P^*) dQ = \int_0^{100} (80 - 0.3Q - 50) dQ = \int_0^{100} (30 - 0.3Q) dQ$$
   $$CS = \left[ 30Q - \frac{0.3Q^2}{2} \right]_0^{100} = \left[ 30Q - 0.15Q^2 \right]_0^{100}$$
   $$CS = 30(100) - 0.15(100^2) = 3,000 - 1,500 = 1,500$$
   (Or geometrically: $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 100 \times (80 - 50) = 1,500$).

2. Producer’s Surplus ($PS$): The minimum supply price (price intercept where $Q = 0$) is $P_{\text{min}} = 30$.

   $$PS = \int_0^{Q^*} (P^* - P_s) dQ = \int_0^{100} (50 - (30 + 0.2Q)) dQ = \int_0^{100} (20 - 0.2Q) dQ$$
   $$PS = \left[ 20Q - \frac{0.2Q^2}{2} \right]_0^{100} = \left[ 20Q - 0.10Q^2 \right]_0^{100}$$
   $$PS = 20(100) - 0.10(100^2) = 2,000 - 1,000 = 1,000$$
   (Or geometrically: $\frac{1}{2} \times 100 \times (50 - 30) = 1,000$).

3. Total Surplus ($TS$):

   $$TS = CS + PS = 1,500 + 1,000 = 2,500$$

Final Answer:

• Equilibrium: $P^* = \text{Rs } 50$, $Q^* = 100 \text{ units}$
• Consumer’s Surplus: $1,500$
• Producer’s Surplus: $1,000$
• Total Surplus: $2,500$

Step-by-Step Solution:

Given system of three linear equations:

$$2x - y + z = 30 \quad \text{--- (1)}$$

$$3x + 2y - z = 20 \quad \text{--- (2)}$$

$$4x + 2y + 3z = 61 \quad \text{--- (3)}$$

Method: Elimination / Cramer’s Rule

Step 1: Eliminate $z$ between pairs of equations

Add equation (1) and equation (2):

$$(2x - y + z) + (3x + 2y - z) = 30 + 20$$

$$5x + y = 50 \implies y = 50 - 5x \quad \text{--- (4)}$$

Multiply equation (2) by 3 and add to equation (3):

$$3(3x + 2y - z) + (4x + 2y + 3z) = 3(20) + 61$$

$$(9x + 6y - 3z) + (4x + 2y + 3z) = 60 + 61$$

$$13x + 8y = 121 \quad \text{--- (5)}$$

Step 2: Solve the 2-variable system for $x$ and $y$

Substitute equation (4) into equation (5):

$$13x + 8(50 - 5x) = 121$$

$$13x + 400 - 40x = 121$$

$$-27x + 400 = 121$$

$$-27x = 121 - 400$$

$$-27x = -279$$

$$x = \frac{-279}{-27} = \frac{31}{3}$$

Step 3: Find $y$

Using equation (4):

$$y = 50 - 5\left(\frac{31}{3}\right) = \frac{150 - 155}{3} = -\frac{5}{3}$$

Step 4: Find $z$

Substitute $x$ and $y$ into equation (1):

$$2\left(\frac{31}{3}\right) - \left(-\frac{5}{3}\right) + z = 30$$

$$\frac{62}{3} + \frac{5}{3} + z = 30$$

$$\frac{67}{3} + z = 30$$

$$z = 30 - \frac{67}{3} = \frac{90 - 67}{3} = \frac{23}{3}$$

Verification:

Substitute into equation (3):

$$4\left(\frac{31}{3}\right) + 2\left(-\frac{5}{3}\right) + 3\left(\frac{23}{3}\right) = \frac{124 - 10 + 69}{3} = \frac{183}{3} = 61 \quad (\text{Confirmed})$$

Final Answer:

$$x = \frac{31}{3} \approx 10.33, \quad y = -\frac{5}{3} \approx -1.67, \quad z = \frac{23}{3} \approx 7.67$$

Step-by-Step Solution:

(i) Find $\frac{dy}{dx}$ from $3x^3 + 2y^2 = 9$:

Differentiating both sides with respect to $x$ using implicit differentiation:

$$\frac{d}{dx}(3x^3) + \frac{d}{dx}(2y^2) = \frac{d}{dx}(9)$$

$$9x^2 + 4y \frac{dy}{dx} = 0$$

Isolate $\frac{dy}{dx}$:

$$4y \frac{dy}{dx} = -9x^2$$

$$\frac{dy}{dx} = -\frac{9x^2}{4y}$$

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(ii) Find $\frac{dy}{dx}$ from parametric equations $x = t^2 + 2$ and $y = t^3 - 3t$:

Differentiate both parametric equations with respect to parameter $t$:

$$\frac{dx}{dt} = \frac{d}{dt}(t^2 + 2) = 2t$$

$$\frac{dy}{dt} = \frac{d}{dt}(t^3 - 3t) = 3t^2 - 3$$

Using the parametric derivative chain rule:

$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

$$\frac{dy}{dx} = \frac{3t^2 - 3}{2t} = \frac{3(t^2 - 1)}{2t}$$

Final Answer:

• (i) $\frac{dy}{dx} = -\frac{9x^2}{4y}$
• (ii) $\frac{dy}{dx} = \frac{3(t^2 - 1)}{2t}$

Step-by-Step Solution:

Given quadratic function:

$$y = x^2 - 5x + 6$$

Step 1: Find Intercepts

• $y$-intercept: Set $x = 0$:
  $$y = 0^2 - 5(0) + 6 = 6 \implies (0, 6)$$
• $x$-intercepts: Set $y = 0$:
  $$x^2 - 5x + 6 = 0 \implies (x - 2)(x - 3) = 0 \implies x = 2, x = 3$$
  Intercept points are $(2, 0)$ and $(3, 0)$.

Step 2: Find Vertex and Axis of Symmetry

The vertex $x$-coordinate is:

$$x_v = -\frac{b}{2a} = -\frac{-5}{2(1)} = \frac{5}{2} = 2.5$$

The vertex $y$-coordinate is:

$$y_v = (2.5)^2 - 5(2.5) + 6 = 6.25 - 12.5 + 6 = -0.25 = -\frac{1}{4}$$

Vertex: $(2.5, -0.25)$. Since the coefficient of $x^2$ is $a = 1 > 0$, the parabola opens upward, and the vertex represents a global minimum.

Step 3: Table of Values for Graphing

Step 4: Determine Domain and Range

• Domain: The quadratic function is defined for all real numbers:
  $$\text{Domain} = \mathbb{R} = (-\infty, \infty)$$
• Range: Since the parabola opens upward from the minimum point $y = -0.25$:
  $$\text{Range} = [-0.25, \infty) = \left[ -\frac{1}{4}, \infty \right)$$

Final Answer:

• Vertex: $(2.5, -0.25)$, Axis of symmetry: $x = 2.5$
• Domain: $(-\infty, \infty)$
• Range: $\left[ -\frac{1}{4}, \infty \right)$

Step-by-Step Solution:

Given data:

• Independent variable ($x$) = Year of service
• Dependent variable ($y$) = Monthly income (in thousands Rs)
• Number of observations ($N$) = $7$

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Step 1: Compute the means $\bar{x}$ and $\bar{y}$$\bar{x} = \frac{\sum x}{N} = \frac{62}{7} \approx 8.8571$$

$$\bar{y} = \frac{\sum y}{N} = \frac{70}{7} = 10$$

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Step 2: Compute regression slope ($b$) and intercept ($a$)

The least squares regression line is $y = a + bx$.

$$b = \frac{N \sum xy - (\sum x)(\sum y)}{N \sum x^2 - (\sum x)^2}$$

$$b = \frac{7(658) - (62)(70)}{7(584) - (62)^2} = \frac{4,606 - 4,340}{4,088 - 3,844} = \frac{266}{244} = \frac{133}{122} \approx 1.09016$$

Now find intercept $a$:

$$a = \bar{y} - b\bar{x} = 10 - \left(\frac{133}{122}\right)\left(\frac{62}{7}\right) = 10 - \frac{1,178}{122} = \frac{1,220 - 1,178}{122} = \frac{42}{122} = \frac{21}{61} \approx 0.34426$$

Thus, the least squares trend line is:

$$y = 0.3443 + 1.0902 x$$

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Step 3: Estimate monthly income for 13 years and 15 years

1. For $x = 13$ years:

   $$y_{13} = 0.34426 + 1.09016(13) = 0.34426 + 14.17208 = 14.51634 \text{ (thousand Rs)}$$
   $$\text{Estimated Income} \approx \text{Rs } 14,516.34$$

2. For $x = 15$ years:

   $$y_{15} = 0.34426 + 1.09016(15) = 0.34426 + 16.3524 = 16.69666 \text{ (thousand Rs)}$$
   $$\text{Estimated Income} \approx \text{Rs } 16,696.66$$

Final Answer:

• Equation of straight line: $y = 0.3443 + 1.0902 x$
• Estimated income for 13 years: Rs 14,516.34
• Estimated income for 15 years: Rs 16,696.66

Step-by-Step Solution:

Given market equations:

• Supply: $P_s = 3Q_s + 12$
• Demand: $P_d = 92 - 2Q_d$
• Specific per-unit sales tax imposed on producers = $\text{Rs } t$.

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Step 1: Formulate the market equilibrium with tax

With a specific tax of $t$ per unit, the supply price becomes:

$$P_{st} = P_s + t = 3Q + 12 + t$$

At equilibrium, the price paid by consumers equals the price required by producers including tax ($P_d = P_{st}$):

$$92 - 2Q = 3Q + 12 + t$$

$$92 - 12 - t = 3Q + 2Q$$

$$80 - t = 5Q$$

$$Q = \frac{80 - t}{5} = 16 - 0.2t$$

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Step 2: Formulate the Total Tax Revenue function ($T$)

The government’s total tax revenue is given by:

$$T(t) = t \times Q = t(16 - 0.2t) = 16t - 0.2t^2$$

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Step 3: Maximize the tax revenue function

Differentiate $T(t)$ with respect to $t$:

$$\frac{dT}{dt} = 16 - 0.4t$$

Set the first derivative to zero for critical points:

$$16 - 0.4t = 0 \implies 0.4t = 16 \implies t = \frac{16}{0.4} = 40$$

Second Derivative Test:

$$\frac{d^2T}{dt^2} = -0.4 < 0$$

Since the second derivative is strictly negative, $t = 40$ gives the absolute maximum total tax revenue.

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Step 4: Calculate the maximum tax revenue and equilibrium values

• Tax rate ($t$): $\text{Rs } 40$ per unit.
• Equilibrium quantity ($Q^*$):
  $$Q^* = 16 - 0.2(40) = 16 - 8 = 8 \text{ units}$$
• Equilibrium price ($P^*$):
  $$P^* = 92 - 2(8) = 92 - 16 = \text{Rs } 76$$
• Maximum Tax Revenue ($T_{\text{max}}$):
  $$T_{\text{max}} = t \times Q^* = 40 \times 8 = \text{Rs } 320$$

Final Answer:

• The tax rate that maximizes tax revenue is $t = \text{Rs } 40$ per unit.
• The maximum total tax revenue is Rs 320 (with $Q = 8$ units, $P = \text{Rs } 76$).