Business Mathematics Ii 2023 Board Question Paper

Step-by-Step Solution:

Evaluate the indefinite integral:

$$I = \int \left( x^2 + e^x + \frac{1}{x} + 1 \right) dx$$

Step 1: Apply the sum rule of integration

$$I = \int x^2 dx + \int e^x dx + \int \frac{1}{x} dx + \int 1 dx$$

Step 2: Integrate each term individually

1. $\int x^2 dx = \frac{x^{2+1}}{2+1} = \frac{x^3}{3}$
2. $\int e^x dx = e^x$
3. $\int \frac{1}{x} dx = \ln|x|$
4. $\int 1 dx = x$

Step 3: Combine terms and add constant of integration ($C$)

$$I = \frac{x^3}{3} + e^x + \ln|x| + x + C$$

Final Answer: $\frac{x^3}{3} + e^x + \ln|x| + x + C$.

Step-by-Step Solution:

Given:

• Curve: $y = 7x^2$
• Bounded by the $x$-axis ($y = 0$) and the vertical lines (ordinates) $x_1 = 0$ and $x_2 = 2$.

Step 1: Set up the definite integral for area

Since $y = 7x^2 \ge 0$ on the interval $[0, 2]$, the area $A$ is:

$$A = \int_0^2 y \, dx = \int_0^2 7x^2 dx$$

Step 2: Evaluate the definite integral

$$A = 7 \left[ \frac{x^3}{3} \right]_0^2 = 7 \left( \frac{2^3}{3} - \frac{0^3}{3} \right)$$

$$A = 7 \left( \frac{8}{3} \right) = \frac{56}{3} \approx 18.67 \text{ sq. units}$$

Final Answer: The area is $\frac{56}{3}$ square units (or approximately $18.67$ square units).

Step-by-Step Solution:

Given first-order homogeneous difference equation:

$$Y_t = 15 Y_{t-1} \quad \text{with initial condition } Y_0 = 10$$

Step 1: Standard form and general solution

A first-order linear homogeneous difference equation of the form $Y_t = b Y_{t-1}$ has the general solution:

$$Y_t = A \cdot b^t$$

Here, $b = 15$, so:

$$Y_t = A \cdot (15)^t$$

Step 2: Apply the initial condition $Y_0 = 10$

At $t = 0$:

$$Y_0 = A \cdot (15)^0 = A \cdot 1 = A$$

Since $Y_0 = 10$:

$$A = 10$$

Step 3: Particular solution

$$Y_t = 10 \cdot (15)^t$$

Final Answer: $Y_t = 10(15)^t$.

Step-by-Step Solution:

Given multivariable function:

$$F(x, y) = 2x^3 y^2 + 7x^2 y^3$$

Step 1: Compute first-order partial derivative with respect to $x$ ($F_x$)

Treat $y$ as a constant:

$$F_x = \frac{\partial F}{\partial x} = \frac{\partial}{\partial x}(2x^3 y^2) + \frac{\partial}{\partial x}(7x^2 y^3)$$

$$F_x = 2y^2 \frac{d}{dx}(x^3) + 7y^3 \frac{d}{dx}(x^2)$$

$$F_x = 2y^2(3x^2) + 7y^3(2x) = 6x^2 y^2 + 14x y^3$$

Step 2: Compute first-order partial derivative with respect to $y$ ($F_y$)

Treat $x$ as a constant:

$$F_y = \frac{\partial F}{\partial y} = \frac{\partial}{\partial y}(2x^3 y^2) + \frac{\partial}{\partial y}(7x^2 y^3)$$

$$F_y = 2x^3 \frac{d}{dy}(y^2) + 7x^2 \frac{d}{dy}(y^3)$$

$$F_y = 2x^3(2y) + 7x^2(3y^2) = 4x^3 y + 21x^2 y^2$$

Final Answer:

• $F_x = 6x^2 y^2 + 14xy^3$
• $F_y = 4x^3 y + 21x^2 y^2$

Step-by-Step Solution:

Given differential equation:

$$\frac{dy}{dt} = 7ty$$

Step 1: Separate the variables

Divide both sides by $y$ and multiply by $dt$:

$$\frac{1}{y} dy = 7t dt$$

Step 2: Integrate both sides

$$\int \frac{1}{y} dy = \int 7t dt$$

$$\ln|y| = \frac{7t^2}{2} + C_1$$

Step 3: Solve for $y$

Exponentiate both sides:

$$|y| = e^{\frac{7}{2}t^2 + C_1} = e^{C_1} \cdot e^{\frac{7}{2}t^2}$$

Letting $A = \pm e^{C_1}$ (an arbitrary non-zero constant, or 0 if $y=0$):

$$y(t) = A e^{\frac{7}{2}t^2} = A e^{3.5t^2}$$

Final Answer: $y(t) = A e^{3.5 t^2}$ (where $A$ is an arbitrary constant).

Step-by-Step Solution:

Given function:

$$f(x, y) = x^2 + 3xy + 5$$

To find $f(2, 3)$, substitute $x = 2$ and $y = 3$:

$$f(2, 3) = 2^2 + 3(2)(3) + 5$$

$$f(2, 3) = 4 + 18 + 5$$

$$f(2, 3) = 27$$

Final Answer: $f(2, 3) = 27$.

Step-by-Step Solution:

Given differential equation:

$$\left(\frac{d^2y}{dx^2}\right)^3 - \left(\frac{dy}{dx}\right)^3 + 2y = 0$$

1. Order of the differential equation:

• The order is defined as the order of the highest derivative occurring in the equation.
• The derivatives present are $\frac{dy}{dx}$ (first derivative) and $\frac{d^2y}{dx^2}$ (second derivative).
• The highest derivative is $\frac{d^2y}{dx^2}$, so:
  $$\mathbf{\text{Order} = 2}$$

2. Degree of the differential equation:

• The degree is defined as the power (exponent) of the highest order derivative after the equation has been cleared of fractions and radicals regarding derivatives.
• The highest derivative $\frac{d^2y}{dx^2}$ is raised to the power of $3$.
• Therefore:
  $$\mathbf{\text{Degree} = 3}$$

Final Answer:

• Order = 2
• Degree = 3

Step-by-Step Solution:

Given matrices:

$$A = \begin{bmatrix} 4 & -2 \\ 3 & -1 \end{bmatrix}, \quad B = \begin{bmatrix} 4 & -2 \\ 7 & -3 \end{bmatrix}$$

Step 1: Compute matrix sum $A + B$$A + B = \begin{bmatrix} 4 + 4 & -2 + (-2) \\ 3 + 7 & -1 + (-3) \end{bmatrix} = \begin{bmatrix} 8 & -4 \\ 10 & -4 \end{bmatrix}$$

Step 2: Compute the transpose $(A + B)^T$

The transpose is obtained by interchanging rows and columns:

$$(A + B)^T = \begin{bmatrix} 8 & 10 \\ -4 & -4 \end{bmatrix}$$

(Note: By transpose properties, $(A + B)^T = A^T + B^T = \begin{bmatrix} 4 & 3 \\ -2 & -1 \end{bmatrix} + \begin{bmatrix} 4 & 7 \\ -2 & -3 \end{bmatrix} = \begin{bmatrix} 8 & 10 \\ -4 & -4 \end{bmatrix}$).

Final Answer: $(A + B)^T = \begin{bmatrix} 8 & 10 \\ -4 & -4 \end{bmatrix}$.

Step-by-Step Solution:

Evaluate the $3 \times 3$ determinant:

$$\Delta = \begin{vmatrix} 3 & 3 & 1 \\ 2 & 3 & 1 \\ 2 & 3 & 4 \end{vmatrix}$$

Method 1: Expansion along Row 1

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

Evaluating each $2 \times 2$ minor determinant:

1. $\begin{vmatrix} 3 & 1 \\ 3 & 4 \end{vmatrix} = (3)(4) - (1)(3) = 12 - 3 = 9$
2. $\begin{vmatrix} 2 & 1 \\ 2 & 4 \end{vmatrix} = (2)(4) - (1)(2) = 8 - 2 = 6$
3. $\begin{vmatrix} 2 & 3 \\ 2 & 3 \end{vmatrix} = (2)(3) - (3)(2) = 6 - 6 = 0$

Substitute back:

$$\Delta = 3(9) - 3(6) + 1(0) = 27 - 18 + 0 = 9$$

Method 2: Elementary Row Operations

Perform $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_2$:

$$\Delta = \begin{vmatrix} 3 & 3 & 1 \\ -1 & 0 & 0 \\ 0 & 0 & 3 \end{vmatrix}$$

Expanding along Row 2 gives $-(-1) \begin{vmatrix} 3 & 1 \\ 0 & 3 \end{vmatrix} = 1 \times (9 - 0) = 9$.

Final Answer: The value of the determinant is $9$.

Step-by-Step Solution:

Given input-output technology coefficient matrix:

$$A = \begin{bmatrix} 0.2 & 0.3 \\ 0.4 & 0.1 \end{bmatrix}$$

Step 1: Form the Leontief matrix $(I - A)$$I - A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 0.2 & 0.3 \\ 0.4 & 0.1 \end{bmatrix} = \begin{bmatrix} 1 - 0.2 & -0.3 \\ -0.4 & 1 - 0.1 \end{bmatrix} = \begin{bmatrix} 0.8 & -0.3 \\ -0.4 & 0.9 \end{bmatrix}$$

Step 2: Test Hawkins-Simon Condition 1

The principal diagonal elements of $(I - A)$ must be positive:

$$(1 - a_{11}) = 0.8 > 0 \quad (\text{Satisfied})$$

$$(1 - a_{22}) = 0.9 > 0 \quad (\text{Satisfied})$$

Step 3: Test Hawkins-Simon Condition 2

The determinant of $(I - A)$ must be strictly positive:

$$|I - A| = \begin{vmatrix} 0.8 & -0.3 \\ -0.4 & 0.9 \end{vmatrix} = (0.8)(0.9) - (-0.3)(-0.4)$$

$$|I - A| = 0.72 - 0.12 = 0.60 > 0 \quad (\text{Satisfied})$$

Conclusion:

Since both principal diagonal elements are positive and the determinant $|I - A| = 0.60 > 0$, the Hawkins-Simon conditions are fully satisfied, guaranteeing that the economic system is productive and viable.

Final Answer: Hawkins-Simon conditions are satisfied ($1 - a_{11} = 0.8 > 0$ and $|I - A| = 0.60 > 0$).

Step-by-Step Solution:

Given first-order linear differential equation:

$$\frac{dy}{dt} + \frac{y}{t} = t^2$$

Step 1: Identify standard form and integrating factor

Standard linear form: $\frac{dy}{dt} + P(t)y = Q(t)$ Here:

$$P(t) = \frac{1}{t}, \quad Q(t) = t^2$$

Integrating factor $I(t)$:

$$I(t) = e^{\int P(t) dt} = e^{\int \frac{1}{t} dt} = e^{\ln t} = t$$

Step 2: Multiply the differential equation by $I(t)$$t \left( \frac{dy}{dt} + \frac{y}{t} \right) = t(t^2)$$

$$\frac{d}{dt} [y \cdot t] = t^3$$

Step 3: Integrate both sides

$$\int \frac{d}{dt}[y \cdot t] \, dt = \int t^3 \, dt$$

$$y \cdot t = \frac{t^4}{4} + C$$

Dividing by $t$:

$$y(t) = \frac{t^3}{4} + \frac{C}{t}$$

Step 4: Apply Initial Condition

• Standard Interpretation ($y(1) = 1$): In textbook problems where $t=0$ is a singular point for $\frac{1}{t}$, the initial condition is typically stated at $t=1$:
  $$1 = \frac{1^3}{4} + \frac{C}{1} \implies 1 = \frac{1}{4} + C \implies C = \frac{3}{4}$$
  $$y(t) = \frac{t^3}{4} + \frac{3}{4t}$$
• Boundary Behavior at $t \to 0$: If evaluated strictly at $t=0$, the solution remains finite only if $C = 0$, giving $y(t) = \frac{t^3}{4}$.

Final Answer: The general solution is $y(t) = \frac{t^3}{4} + \frac{C}{t}$ (for $y(1) = 1$, $y(t) = \frac{t^3}{4} + \frac{3}{4t}$).

Step-by-Step Solution:

Given difference equation:

$$Y_{t-1} - 0.7Y_t + 7 = 0 \quad \text{with } Y_0 = 7$$

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Step 1: Express in standard first-order form

Rearranging to express $Y_t$ as a function of $Y_{t-1}$:

$$0.7 Y_t = Y_{t-1} + 7$$

$$Y_t = \frac{1}{0.7} Y_{t-1} + \frac{7}{0.7}$$

$$Y_t = \frac{10}{7} Y_{t-1} + 10$$

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Step 2: Find Particular Solution ($Y_p$)

Let $Y_t = Y_{t-1} = Y^*$ (equilibrium value):

$$Y^* = \frac{10}{7}Y^* + 10$$

$$Y^* - \frac{10}{7}Y^* = 10$$

$$-\frac{3}{7}Y^* = 10 \implies Y^* = -\frac{70}{3} \approx -23.333$$

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Step 3: Find Complementary Solution ($Y_c$) and General Solution

For the homogeneous part $Y_t - \frac{10}{7}Y_{t-1} = 0$:

$$Y_c = A \left(\frac{10}{7}\right)^t$$

Thus, the general solution is:

$$Y_t = A \left(\frac{10}{7}\right)^t - \frac{70}{3}$$

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Step 4: Apply the initial condition $Y_0 = 7$

At $t = 0$:

$$7 = A \left(\frac{10}{7}\right)^0 - \frac{70}{3}$$

$$7 = A - \frac{70}{3} \implies A = 7 + \frac{70}{3} = \frac{21 + 70}{3} = \frac{91}{3}$$

The specific time path is:

$$Y_t = \frac{91}{3} \left(\frac{10}{7}\right)^t - \frac{70}{3}$$

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Step 5: Calculate $Y_7$

For $t = 7$:

$$\left(\frac{10}{7}\right)^7 \approx 12.18684$$

$$Y_7 = \frac{91}{3}(12.18684) - \frac{70}{3} \approx 369.667 - 23.333 \approx 346.33$$

(Note: If the equation was printed as a standard lag $Y_t - 0.7Y_{t-1} = 7$, then $Y^* = \frac{7}{0.3} = \frac{70}{3}$, and $Y_t = \frac{70}{3} - \frac{49}{3}(0.7)^t$, yielding $Y_7 \approx 22.01$).

Final Answer:

• Time path: $Y_t = \frac{91}{3}\left(\frac{10}{7}\right)^t - \frac{70}{3}$
• $Y_7 \approx 346.33$

Step-by-Step Solution:

(a) Evaluate $\int \frac{\log x}{x} dx$:

Let $u = \log x = \ln x$. Differentiating:

$$du = \frac{1}{x} dx$$

Substitute $u$ into the integral:

$$\int \frac{\log x}{x} dx = \int u \, du = \frac{u^2}{2} + C$$

Substitute back $u = \ln x$:

$$\int \frac{\log x}{x} dx = \frac{(\ln x)^2}{2} + C$$

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(b) Evaluate $\int \frac{3x + 1}{\sqrt{5x + 2}} dx$:

Use algebraic substitution: Let $u = 5x + 2 \implies 5x = u - 2 \implies x = \frac{u - 2}{5}$. Differentiating gives:

$$dx = \frac{1}{5} du$$

Express numerator in terms of $u$:

$$3x + 1 = 3\left(\frac{u - 2}{5}\right) + 1 = \frac{3u - 6 + 5}{5} = \frac{3u - 1}{5}$$

Substitute into the integral:

$$I = \int \frac{\frac{3u - 1}{5}}{\sqrt{u}} \cdot \left(\frac{1}{5} du\right) = \frac{1}{25} \int \frac{3u - 1}{u^{1/2}} du$$

$$I = \frac{1}{25} \int \left( 3u^{1/2} - u^{-1/2} \right) du$$

Integrate term by term:

$$I = \frac{1}{25} \left[ 3 \cdot \frac{u^{3/2}}{3/2} - \frac{u^{1/2}}{1/2} \right] + C$$

$$I = \frac{1}{25} \left[ 2u^{3/2} - 2u^{1/2} \right] + C = \frac{2}{25} \sqrt{u}(u - 1) + C$$

Substitute back $u = 5x + 2$:

$$I = \frac{2}{25} \sqrt{5x + 2} \left( (5x + 2) - 1 \right) + C$$

$$I = \frac{2}{25} (5x + 1) \sqrt{5x + 2} + C$$

Final Answer:

• (a) $\frac{(\ln x)^2}{2} + C$
• (b) $\frac{2}{25}(5x + 1)\sqrt{5x + 2} + C$

Step-by-Step Solution:

Given:

• Rate of repair cost: $\frac{dC}{dt} = 5t + 12.5$ (in Rs per year)
• Age of machine: $t$ in years
• We want to find the total repair cost incurred over the first 4 years ($t = 0$ to $t = 4$).

Step 1: Set up the Definite Integral

The total accumulated cost $C$ is the definite integral of the marginal rate of repair:

$$C = \int_0^4 \frac{dC}{dt} \, dt = \int_0^4 (5t + 12.5) \, dt$$

Step 2: Evaluate the Integral

$$C = \left[ \frac{5t^2}{2} + 12.5t \right]_0^4$$

$$C = \left( \frac{5(4^2)}{2} + 12.5(4) \right) - \left( \frac{5(0)}{2} + 12.5(0) \right)$$

$$C = \left( \frac{5(16)}{2} + 50 \right) - 0$$

$$C = 40 + 50 = 90$$

Final Answer: The total cost of repair after 4 years is Rs 90.

Step-by-Step Solution:

Given system of linear equations:

$$3x + 4y + 5z = 18$$

$$2x - y + 8z = 13$$

$$5x - 2y + 7z = 20$$

Method: Cramer’s Rule (Determinant Method)

Step 1: Evaluate the Coefficient Determinant ($D$)

$$D = \begin{vmatrix} 3 & 4 & 5 \\ 2 & -1 & 8 \\ 5 & -2 & 7 \end{vmatrix}$$

Expand along Row 1:

$$D = 3 \begin{vmatrix} -1 & 8 \\ -2 & 7 \end{vmatrix} - 4 \begin{vmatrix} 2 & 8 \\ 5 & 7 \end{vmatrix} + 5 \begin{vmatrix} 2 & -1 \\ 5 & -2 \end{vmatrix}$$

• $\begin{vmatrix} -1 & 8 \\ -2 & 7 \end{vmatrix} = (-1)(7) - (8)(-2) = -7 + 16 = 9$
• $\begin{vmatrix} 2 & 8 \\ 5 & 7 \end{vmatrix} = (2)(7) - (8)(5) = 14 - 40 = -26$
• $\begin{vmatrix} 2 & -1 \\ 5 & -2 \end{vmatrix} = (2)(-2) - (-1)(5) = -4 + 5 = 1$$D = 3(9) - 4(-26) + 5(1) = 27 + 104 + 5 = 136$$

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Step 2: Evaluate $D_x$$D_x = \begin{vmatrix} 18 & 4 & 5 \\ 13 & -1 & 8 \\ 20 & -2 & 7 \end{vmatrix}$$

Expand along Row 1:

$$D_x = 18(9) - 4 \begin{vmatrix} 13 & 8 \\ 20 & 7 \end{vmatrix} + 5 \begin{vmatrix} 13 & -1 \\ 20 & -2 \end{vmatrix}$$

• $\begin{vmatrix} 13 & 8 \\ 20 & 7 \end{vmatrix} = 91 - 160 = -69$
• $\begin{vmatrix} 13 & -1 \\ 20 & -2 \end{vmatrix} = -26 - (-20) = -6$$D_x = 18(9) - 4(-69) + 5(-6) = 162 + 276 - 30 = 408$$
  $$x = \frac{D_x}{D} = \frac{408}{136} = 3$$

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Step 3: Evaluate $D_y$$D_y = \begin{vmatrix} 3 & 18 & 5 \\ 2 & 13 & 8 \\ 5 & 20 & 7 \end{vmatrix}$$

Expand along Row 1:

$$D_y = 3(91 - 160) - 18(14 - 40) + 5(40 - 65)$$

$$D_y = 3(-69) - 18(-26) + 5(-25) = -207 + 468 - 125 = 136$$

$$y = \frac{D_y}{D} = \frac{136}{136} = 1$$

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Step 4: Evaluate $D_z$$D_z = \begin{vmatrix} 3 & 4 & 18 \\ 2 & -1 & 13 \\ 5 & -2 & 20 \end{vmatrix}$$

Expand along Row 1:

$$D_z = 3(-20 - (-26)) - 4(40 - 65) + 18(-4 - (-5))$$

$$D_z = 3(6) - 4(-25) + 18(1) = 18 + 100 + 18 = 136$$

$$z = \frac{D_z}{D} = \frac{136}{136} = 1$$

Verification:

• $3(3) + 4(1) + 5(1) = 9 + 4 + 5 = 18$
• $2(3) - 1 + 8(1) = 6 - 1 + 8 = 13$
• $5(3) - 2(1) + 7(1) = 15 - 2 + 7 = 20$

Final Answer: $x = 3$, $y = 1$, $z = 1$.

Step-by-Step Solution:

Given demand function:

$$Q = 200 - 4P + 0.5P_A + Y$$

where $P = 6$, $P_A = 10$, and $Y = 500$.

Step 1: Compute baseline quantity demanded ($Q$)

$$Q = 200 - 4(6) + 0.5(10) + 500 = 200 - 24 + 5 + 500 = 681$$

Step 2: Compute Partial Elasticities of Demand

1. Price Elasticity of Demand ($\epsilon_P$):

   $$\frac{\partial Q}{\partial P} = -4$$
   $$\epsilon_P = \frac{P}{Q} \cdot \frac{\partial Q}{\partial P} = \frac{6}{681} \times (-4) = -\frac{24}{681} \approx -0.03524$$
   (In absolute value, $|\epsilon_P| \approx 0.035$, highly inelastic).

2. Cross-Price Elasticity of Demand ($\epsilon_{PA}$):

   $$\frac{\partial Q}{\partial P_A} = 0.5$$
   $$\epsilon_{PA} = \frac{P_A}{Q} \cdot \frac{\partial Q}{\partial P_A} = \frac{10}{681} \times 0.5 = \frac{5}{681} \approx +0.00734$$
   (Positive value confirms goods are substitutes).

3. Income Elasticity of Demand ($\epsilon_Y$):

   $$\frac{\partial Q}{\partial Y} = 1$$
   $$\epsilon_Y = \frac{Y}{Q} \cdot \frac{\partial Q}{\partial Y} = \frac{500}{681} \times 1 \approx +0.73421$$
   ($0 < \epsilon_Y < 1$ confirms this is a normal/necessity good).

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Step 3: Analyze Impacts on Demand

• (i) When $P$ decreases by $10\%$ ($\% \Delta P = -10\%$):

  $$\% \Delta Q \approx \epsilon_P \times (\% \Delta P) = (-0.03524) \times (-10\%) = +0.3524\%$$
  Absolute change: $\Delta Q = -4 \times \Delta P = -4(-0.6) = +2.4 \text{ units}$. Demand increases by approximately $0.35\%$ (from $681$ to $683.4$).

• (ii) When $P_A$ increases by $10\%$ ($\% \Delta P_A = +10\%$):

  $$\% \Delta Q \approx \epsilon_{PA} \times (\% \Delta P_A) = (+0.00734) \times (+10\%) = +0.0734\%$$
  Absolute change: $\Delta Q = 0.5 \times 1 = +0.5 \text{ units}$. Demand increases by approximately $0.073\%$ (from $681$ to $681.5$).

• (iii) When $Y$ increases by $10\%$ ($\% \Delta Y = +10\%$):

  $$\% \Delta Q \approx \epsilon_Y \times (\% \Delta Y) = (+0.73421) \times (+10\%) = +7.342\%$$
  Absolute change: $\Delta Q = 1 \times 50 = +50 \text{ units}$. Demand increases by approximately $7.34\%$ (from $681$ to $731$).

Final Answer:

• Price Elasticity: $\epsilon_P \approx -0.0352$
• Cross-Price Elasticity: $\epsilon_{PA} \approx +0.0073$
• Income Elasticity: $\epsilon_Y \approx +0.7342$
• Demand Changes: (i) increases by $+0.35\%$ (+2.4 units), (ii) increases by $+0.073\%$ (+0.5 units), (iii) increases by $+7.34\%$ (+50 units).

Step-by-Step Solution:

Given production function:

$$Q(L, K) = L^2 - 2KL + 2K^2$$

where $L$ = labour units and $K$ = capital units.

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(i) Find Marginal Products of Labour ($MP_L$) and Capital ($MP_K$) at $L = 30, K = 20$:

• Marginal Product of Labour ($MP_L$):

  $$MP_L = \frac{\partial Q}{\partial L} = 2L - 2K$$
  At $L = 30, K = 20$:
  $$MP_L = 2(30) - 2(20) = 60 - 40 = 20$$

• Marginal Product of Capital ($MP_K$):

  $$MP_K = \frac{\partial Q}{\partial K} = -2L + 4K$$
  At $L = 30, K = 20$:
  $$MP_K = -2(30) + 4(20) = -60 + 80 = 20$$

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(ii) Find Marginal Rate of Technical Substitution ($MRTS$):

The Marginal Rate of Technical Substitution of labour for capital along an isoquant ($dQ = 0$) is:

$$MRTS_{LK} = -\frac{dK}{dL} = \frac{MP_L}{MP_K} = \frac{2L - 2K}{-2L + 4K} = \frac{L - K}{2K - L}$$

At $L = 30, K = 20$:

$$MRTS_{LK} = \frac{30 - 20}{2(20) - 30} = \frac{10}{40 - 30} = \frac{10}{10} = 1$$

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(iii) Estimate the increase in capital needed when labour decreases by 0.08:

Using the total differential of output:

$$dQ = MP_L \cdot dL + MP_K \cdot dK$$

To maintain the current level of output, $dQ = 0$:

$$0 = MP_L \cdot dL + MP_K \cdot dK$$

$$dK = -\frac{MP_L}{MP_K} dL = -MRTS_{LK} \cdot dL$$

Given $dL = -0.08$ (a decrease):

$$dK = -(1)(-0.08) = +0.08 \text{ units of capital}$$

Final Answer:

• (i) $MP_L = 20$, $MP_K = 20$
• (ii) $MRTS_{LK} = 1$
• (iii) Capital must be increased by $0.08$ units.