Business Mathematics II

Step-by-Step Solution:

Evaluate the indefinite integral:

$$I = \int \frac{x+2}{x-5} \, dx$$

Step 1: Perform algebraic division / rearrangement

Express the numerator in terms of $(x - 5)$:

$$\frac{x + 2}{x - 5} = \frac{(x - 5) + 7}{x - 5} = \frac{x - 5}{x - 5} + \frac{7}{x - 5} = 1 + \frac{7}{x - 5}$$

Step 2: Integrate term by term

$$I = \int \left( 1 + \frac{7}{x - 5} \right) dx$$

$$I = \int 1 \, dx + 7 \int \frac{1}{x - 5} \, dx$$

$$I = x + 7 \ln|x - 5| + C$$

Final Answer: $x + 7 \ln|x - 5| + C$.

Step-by-Step Solution:

Given:

• Curve: $y = \frac{3}{2}x^2$
• Bounded by the $x$-axis and ordinates $x = 0$ and $x = 2$.

Step 1: Set up the definite integral

Since $y \ge 0$ on $[0, 2]$, the area $A$ is:

$$A = \int_0^2 \frac{3}{2}x^2 \, dx$$

Step 2: Evaluate the integral

$$A = \frac{3}{2} \left[ \frac{x^3}{3} \right]_0^2 = \frac{1}{2} \left[ x^3 \right]_0^2$$

$$A = \frac{1}{2} (2^3 - 0^3) = \frac{1}{2}(8) = 4 \text{ sq. units}$$

Final Answer: The area of the region is $4$ square units.

Step-by-Step Solution:

Given function:

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

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

Treating $y$ as a constant:

$$f_x = \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(3x^2 y^2) + \frac{\partial}{\partial x}(5x^2 y)$$

$$f_x = 3y^2(2x) + 5y(2x) = 6xy^2 + 10xy$$

Step 2: Compute second-order partial derivative $f_{xx}$

Differentiating $f_x$ again with respect to $x$ (treating $y$ as constant):

$$f_{xx} = \frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}(6xy^2 + 10xy)$$

$$f_{xx} = 6y^2(1) + 10y(1) = 6y^2 + 10y$$

Final Answer: $f_{xx} = 6y^2 + 10y$.

Step-by-Step Solution:

Given matrix:

$$A = \begin{bmatrix} 1 & 2 \\ 3 & 1 \end{bmatrix}$$

Step 1: Multiply $A$ by $A$

$$A^2 = A \cdot A = \begin{bmatrix} 1 & 2 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 1 \end{bmatrix}$$

Evaluating each entry by row-column dot product:

• $c_{11} = (1)(1) + (2)(3) = 1 + 6 = 7$
• $c_{12} = (1)(2) + (2)(1) = 2 + 2 = 4$
• $c_{21} = (3)(1) + (1)(3) = 3 + 3 = 6$
• $c_{22} = (3)(2) + (1)(1) = 6 + 1 = 7$

$$A^2 = \begin{bmatrix} 7 & 4 \\ 6 & 7 \end{bmatrix}$$

Final Answer: $A^2 = \begin{bmatrix} 7 & 4 \\ 6 & 7 \end{bmatrix}$.

Step-by-Step Solution:

Evaluate the determinant:

$$\Delta = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix}$$

Method 1: Expansion along Row 1

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

Evaluate the $2 \times 2$ determinants:

• $\begin{vmatrix} 3 & 4 \\ 4 & 5 \end{vmatrix} = (3)(5) - (4)(4) = 15 - 16 = -1$
• $\begin{vmatrix} 2 & 4 \\ 3 & 5 \end{vmatrix} = (2)(5) - (4)(3) = 10 - 12 = -2$
• $\begin{vmatrix} 2 & 3 \\ 3 & 4 \end{vmatrix} = (2)(4) - (3)(3) = 8 - 9 = -1$

Substitute into the expression:

$$\Delta = 1(-1) - 2(-2) + 3(-1) = -1 + 4 - 3 = 0$$

Method 2: Elementary Row Operations

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

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

Since row 2 and row 3 are identical, the determinant is identically $0$.

Final Answer: The value is $0$.

Step-by-Step Solution:

Given input-output coefficient matrix:

$$A = \begin{bmatrix} 0.8 & 0.2 \\ 0.9 & 0.7 \end{bmatrix}$$

Step 1: Form the Leontief matrix $(I - A)$

$$I - A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 0.8 & 0.2 \\ 0.9 & 0.7 \end{bmatrix} = \begin{bmatrix} 1 - 0.8 & -0.2 \\ -0.9 & 1 - 0.7 \end{bmatrix} = \begin{bmatrix} 0.2 & -0.2 \\ -0.9 & 0.3 \end{bmatrix}$$

Step 2: Check Hawkins-Simon Condition 1

The principal diagonal elements must be positive:

$$(1 - a_{11}) = 0.2 > 0 \quad (\text{Satisfied})$$

$$(1 - a_{22}) = 0.3 > 0 \quad (\text{Satisfied})$$

Step 3: Check Hawkins-Simon Condition 2

The determinant $|I - A|$ must be strictly positive:

$$|I - A| = \begin{vmatrix} 0.2 & -0.2 \\ -0.9 & 0.3 \end{vmatrix} = (0.2)(0.3) - (-0.2)(-0.9)$$

$$|I - A| = 0.06 - 0.18 = -0.12$$

Since $|I - A| = -0.12 < 0$, the second Hawkins-Simon condition fails.

Economic Interpretation:

The economy requires more inputs than it can produce as final outputs, meaning the system is incapable of sustaining positive net production.

Final Answer: The system is NOT viable as per Hawkins-Simon conditions because $|I - A| = -0.12 < 0$.

Step-by-Step Solution:

Given differential equation:

$$\left( \frac{dy}{dx} \right)^2 - \left( \frac{dy}{dx} \right)^5 + 7xy = 0$$

1. Determine the Order:

• The order of a differential equation is the order of the highest derivative present.
• The only derivative present is the first derivative $\frac{dy}{dx}$.
• Therefore:
  $$\mathbf{\text{Order} = 1}$$

2. Determine the Degree:

• The degree is the exponent of the highest power of the highest-order derivative occurring in polynomial form.
• The highest-order derivative is $\frac{dy}{dx}$, and its highest power is $5$.
• Therefore:
  $$\mathbf{\text{Degree} = 5}$$

Final Answer:

• Order = 1
• Degree = 5

Step-by-Step Solution:

Given differential equation:

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

Step 1: Separate variables

$$dy = \frac{1}{x} \, dx$$

Step 2: Integrate both sides

$$\int dy = \int \frac{1}{x} \, dx$$

$$y(x) = \ln|x| + C$$

Final Answer: $y = \ln|x| + C$ (where $C$ is an arbitrary constant of integration).

Step-by-Step Solution:

Given first-order homogeneous difference equation:

$$y_{t+1} - 2y_t = 0 \quad \text{with } y_0 = 3$$

Step 1: Rewrite in standard recursive form

$$y_{t+1} = 2y_t$$

or with index shifted:

$$y_t = 2y_{t-1}$$

Step 2: General solution

The general solution of a homogeneous first-order difference equation $y_t = b y_{t-1}$ is:

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

Here $b = 2$, so:

$$y_t = A \cdot 2^t$$

Step 3: Apply the initial condition $y_0 = 3$

At $t = 0$:

$$y_0 = A \cdot 2^0 = A = 3$$

Thus, the particular solution is:

$$y_t = 3 \cdot 2^t$$

Final Answer: $y_t = 3 \cdot 2^t$.

Step-by-Step Solution:

Given:

• Marginal Cost function: $MC = 3x^2 - 10x + 3$
• Total cost of producing $1$ unit: $TC(1) = \text{Rs } 7$

Step 1: Find the Total Cost function $TC(x)$ by integration

$$TC(x) = \int MC \, dx = \int (3x^2 - 10x + 3) \, dx$$

$$TC(x) = 3\left(\frac{x^3}{3}\right) - 10\left(\frac{x^2}{2}\right) + 3x + C$$

$$TC(x) = x^3 - 5x^2 + 3x + C$$

Step 2: Determine the integration constant ($C$) using $TC(1) = 7$

$$7 = 1^3 - 5(1^2) + 3(1) + C$$

$$7 = 1 - 5 + 3 + C$$

$$7 = -1 + C \implies C = 7 + 1 = 8$$

Thus, the Total Cost function is:

$$TC(x) = x^3 - 5x^2 + 3x + 8$$

Step 3: Find the Average Cost function $AC(x)$

$$AC(x) = \frac{TC(x)}{x} = \frac{x^3 - 5x^2 + 3x + 8}{x}$$

$$AC(x) = x^2 - 5x + 3 + \frac{8}{x}$$

Final Answer: The Average Cost function is $AC(x) = x^2 - 5x + 3 + \frac{8}{x}$.

Step-by-Step Solution:

(a) Evaluate $\int \frac{\log x + 5}{x} \, dx$:

Let $u = \log x + 5 = \ln x + 5$. Differentiating with respect to $x$:

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

Substitute into the integral:

$$I = \int u \, du = \frac{u^2}{2} + C$$

Substitute back $u = \log x + 5$:

$$I = \frac{(\log x + 5)^2}{2} + C$$

(Alternatively: $\int \frac{\ln x}{x} dx + 5 \int \frac{1}{x} dx = \frac{(\ln x)^2}{2} + 5\ln x + C_1$, which is algebraically equivalent up to a constant).

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

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

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

Express the term $(2x + 1)$ in terms of $u$:

$$2x + 1 = 2\left(\frac{u - 2}{5}\right) + 1 = \frac{2u - 4 + 5}{5} = \frac{2u + 1}{5}$$

Substitute into the integral:

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

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

Integrate term by term:

$$I = \frac{1}{25} \left[ 2 \cdot \frac{u^{5/2}}{5/2} + \frac{u^{3/2}}{3/2} \right] + C$$

$$I = \frac{1}{25} \left[ \frac{4}{5} u^{5/2} + \frac{2}{3} u^{3/2} \right] + C$$

Factor out common terms $\frac{2}{15} u^{3/2}$:

$$I = \frac{1}{25} \cdot \frac{2}{15} u^{3/2} (6u + 5) + C = \frac{2}{375} u^{3/2} (6u + 5) + C$$

Substitute back $u = 5x + 2$:

$$6u + 5 = 6(5x + 2) + 5 = 30x + 12 + 5 = 30x + 17$$

$$I = \frac{2}{375} (30x + 17)(5x + 2)^{3/2} + C$$

Final Answer:

• (a) $\frac{(\log x + 5)^2}{2} + C$
• (b) $\frac{2}{375}(30x + 17)(5x + 2)^{3/2} + C$

Step-by-Step Solution:

Given market functions under pure competition:

• Demand function: $P_d = 8 - 2x$
• Supply function: $P_s = 2x$ where $x$ denotes quantity and $P$ denotes price.

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Step 1: Find Market Equilibrium Quantity ($x^*$) and Price ($P^*$)

At equilibrium, $P_d = P_s$:

$$8 - 2x = 2x$$

$$8 = 4x \implies x^* = 2 \text{ units}$$

Substitute $x^* = 2$ into either equation:

$$P^* = 2(2) = 4$$

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Step 2: Compute Consumer’s Surplus ($CS$)

Consumer’s surplus is the integral of $(P_d - P^*)$ from $0$ to $x^*$:

$$CS = \int_0^{x^*} (P_d - P^*) \, dx = \int_0^2 (8 - 2x - 4) \, dx = \int_0^2 (4 - 2x) \, dx$$

$$CS = \left[ 4x - x^2 \right]_0^2 = 4(2) - (2)^2 = 8 - 4 = 4$$

(Geometrically: Triangle area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times (8 - 4) = 4$).

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Step 3: Compute Producer’s Surplus ($PS$)

Producer’s surplus is the integral of $(P^* - P_s)$ from $0$ to $x^*$:

$$PS = \int_0^{x^*} (P^* - P_s) \, dx = \int_0^2 (4 - 2x) \, dx$$

$$PS = \left[ 4x - x^2 \right]_0^2 = 4(2) - (2)^2 = 8 - 4 = 4$$

(Geometrically: Triangle area $= \frac{1}{2} \times 2 \times (4 - 0) = 4$).

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Step 4: Total Surplus

$$TS = CS + PS = 4 + 4 = 8$$

Final Answer:

• Equilibrium: $x^* = 2 \text{ units}$, $P^* = 4$
• Consumer’s Surplus: $4$
• Producer’s Surplus: $4$

Step-by-Step Solution:

Given system of linear equations:

$$4x + 0y + 6z = 100 \quad \text{--- (1)}$$

$$3x + 6y + z = 100 \quad \text{--- (2)}$$

$$3x + 4y + 3z = 100 \quad \text{--- (3)}$$

Method: Cramer’s Rule

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

$$D = \begin{vmatrix} 4 & 0 & 6 \\ 3 & 6 & 1 \\ 3 & 4 & 3 \end{vmatrix}$$

Expand along Row 1:

$$D = 4 \begin{vmatrix} 6 & 1 \\ 4 & 3 \end{vmatrix} - 0 + 6 \begin{vmatrix} 3 & 6 \\ 3 & 4 \end{vmatrix}$$

$$D = 4(18 - 4) + 6(12 - 18) = 4(14) + 6(-6) = 56 - 36 = 20$$

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Step 2: Calculate $D_x$

Replace Column 1 with constant terms:

$$D_x = \begin{vmatrix} 100 & 0 & 6 \\ 100 & 6 & 1 \\ 100 & 4 & 3 \end{vmatrix} = 100 \begin{vmatrix} 1 & 0 & 6 \\ 1 & 6 & 1 \\ 1 & 4 & 3 \end{vmatrix}$$

Expand along Row 1:

$$D_x = 100 \left[ 1(18 - 4) - 0 + 6(4 - 6) \right] = 100 [ 14 + 6(-2) ] = 100 [ 14 - 12 ] = 100(2) = 200$$

$$x = \frac{D_x}{D} = \frac{200}{20} = 10$$

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Step 3: Calculate $D_y$

Replace Column 2 with constant terms:

$$D_y = \begin{vmatrix} 4 & 100 & 6 \\ 3 & 100 & 1 \\ 3 & 100 & 3 \end{vmatrix} = 100 \begin{vmatrix} 4 & 1 & 6 \\ 3 & 1 & 1 \\ 3 & 1 & 3 \end{vmatrix}$$

Expand along Row 1:

$$D_y = 100 \left[ 4(3 - 1) - 1(9 - 3) + 6(3 - 3) \right] = 100 [ 4(2) - 6 + 0 ] = 100 [ 8 - 6 ] = 100(2) = 200$$

$$y = \frac{D_y}{D} = \frac{200}{20} = 10$$

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Step 4: Calculate $D_z$

Replace Column 3 with constant terms:

$$D_z = \begin{vmatrix} 4 & 0 & 100 \\ 3 & 6 & 100 \\ 3 & 4 & 100 \end{vmatrix} = 100 \begin{vmatrix} 4 & 0 & 1 \\ 3 & 6 & 1 \\ 3 & 4 & 1 \end{vmatrix}$$

Expand along Row 1:

$$D_z = 100 \left[ 4(6 - 4) - 0 + 1(12 - 18) \right] = 100 [ 4(2) + (-6) ] = 100 [ 8 - 6 ] = 100(2) = 200$$

$$z = \frac{D_z}{D} = \frac{200}{20} = 10$$

Verification:

• $4(10) + 6(10) = 40 + 60 = 100$ (Matches)
• $3(10) + 6(10) + 10 = 30 + 60 + 10 = 100$ (Matches)
• $3(10) + 4(10) + 3(10) = 30 + 40 + 30 = 100$ (Matches)

Final Answer: $x = 10$, $y = 10$, $z = 10$.

Step-by-Step Solution:

Given production function:

$$Q = 100 L^{0.5} K^{0.5}$$

where $L$ = labor and $K$ = capital.

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Step 1: Compute the partial derivatives

• With respect to labor ($L$):
  $$\frac{\partial Q}{\partial L} = 100(0.5 L^{-0.5} K^{0.5}) = 50 L^{-0.5} K^{0.5} = 0.5 \frac{Q}{L}$$
• With respect to capital ($K$):
  $$\frac{\partial Q}{\partial K} = 100(0.5 L^{0.5} K^{-0.5}) = 50 L^{0.5} K^{-0.5} = 0.5 \frac{Q}{K}$$

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Step 2: Use the Total Differential formula

The total differential $dQ$ approximates the change in output:

$$dQ = \frac{\partial Q}{\partial L} dL + \frac{\partial Q}{\partial K} dK$$

$$dQ = \left( 0.5 \frac{Q}{L} \right) dL + \left( 0.5 \frac{Q}{K} \right) dK$$

Divide both sides by $Q$ to express in terms of percentage/proportional changes:

$$\frac{dQ}{Q} = 0.5 \left(\frac{dL}{L}\right) + 0.5 \left(\frac{dK}{K}\right)$$

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Step 3: Substitute the given percentage changes

• Labor decreases by $3\%$:
  $$\frac{dL}{L} = -3\% = -0.03$$
• Capital increases by $5\%$:
  $$\frac{dK}{K} = +5\% = +0.05$$

Substitute these values:

$$\frac{dQ}{Q} = 0.5(-0.03) + 0.5(+0.05)$$

$$\frac{dQ}{Q} = -0.015 + 0.025 = +0.010 = +1.0\%$$

Final Answer: The approximate change in output $Q$ is an increase of $1\%$.

Step-by-Step Solution:

Given first-order linear differential equation:

$$\frac{dy}{dx} + \frac{1}{x} \cdot y = x^3$$

Step 1: Identify standard form and integrating factor

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

$$P(x) = \frac{1}{x}, \quad Q(x) = x^3$$

The integrating factor $I(x)$ is:

$$I(x) = e^{\int P(x) \, dx} = e^{\int \frac{1}{x} \, dx} = e^{\ln x} = x$$

Step 2: Multiply the differential equation by $I(x)$

$$x \left( \frac{dy}{dx} + \frac{1}{x} y \right) = x(x^3)$$

$$\frac{d}{dx}[y \cdot x] = x^4$$

Step 3: Integrate both sides with respect to $x$

$$\int \frac{d}{dx}[y \cdot x] \, dx = \int x^4 \, dx$$

$$y \cdot x = \frac{x^5}{5} + C$$

Step 4: Solve explicitly for $y$

Divide both sides by $x$:

$$y(x) = \frac{x^4}{5} + \frac{C}{x}$$

Final Answer: $y(x) = \frac{x^4}{5} + \frac{C}{x}$ (where $C$ is an arbitrary constant of integration).

Step-by-Step Solution:

Given Bertrand Duopoly Reaction Functions:

$$P_t^X = 45 + 0.8 P_{t-1}^Y \quad \text{--- (1)}$$

$$P_t^Y = 45 + 0.8 P_{t-1}^X \quad \text{--- (2)}$$

with initial price set by Firm X in period 0: $P_0^X = 300$.

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Step 1: Derive a single difference equation for $P_t^X$

Lag equation (2) by one time period to express $P_{t-1}^Y$:

$$P_{t-1}^Y = 45 + 0.8 P_{t-2}^X$$

Substitute this into equation (1):

$$P_t^X = 45 + 0.8 (45 + 0.8 P_{t-2}^X)$$

$$P_t^X = 45 + 36 + 0.64 P_{t-2}^X$$

$$P_t^X = 81 + 0.64 P_{t-2}^X$$

$$P_t^X - 0.64 P_{t-2}^X = 81$$

This is a linear second-order difference equation with two-period step dynamics.

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Step 2: Solve the difference equation for even periods ($t = 2k$)

Let $k = t / 2$. Then:

$$P_{2k}^X = 81 + 0.64 P_{2(k-1)}^X$$

1. Equilibrium price ($P^*$):

   $$P^* = 81 + 0.64 P^* \implies 0.36 P^* = 81 \implies P^* = \frac{81}{0.36} = 225$$

2. General solution for even periods:

   $$P_{2k}^X = P^* + (P_0^X - P^*)(0.64)^k$$
   $$P_{2k}^X = 225 + (300 - 225)(0.64)^k = 225 + 75(0.64)^k$$

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Step 3: Calculate $P_{10}^X$ (Period 10, so $k = 10/2 = 5$)

$$P_{10}^X = 225 + 75(0.64)^5$$

Compute $(0.64)^5$:

$$(0.64)^5 \approx 0.107374$$

$$P_{10}^X = 225 + 75(0.107374) = 225 + 8.053 = 233.05$$

(Sequential check: $P_0^X=300 \implies P_1^Y=285 \implies P_2^X=273 \implies P_4^X=255.72 \implies P_6^X=244.66 \implies P_8^X=237.58 \implies P_{10}^X=233.05$).

Final Answer:

• Difference Equation: $P_t^X = 81 + 0.64 P_{t-2}^X$
• Price in period 10 ($P_{10}^X$): $233.05$

Step-by-Step Solution:

Given dynamic macroeconomic data:

• National income: $y_t = C_t + I_t$
• Consumption function: $C_t = 200 + 0.75y_{t-1}$
• Investment function: $I_t = 50 + 0.15y_{t-1}$
• Initial national income: $y_0 = 3,000$

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Step 1: Formulate the First-Order Difference Equation

Substitute $C_t$ and $I_t$ into $y_t$:

$$y_t = (200 + 0.75y_{t-1}) + (50 + 0.15y_{t-1})$$

$$y_t = (200 + 50) + (0.75 + 0.15)y_{t-1}$$

$$y_t = 250 + 0.90 y_{t-1}$$

$$y_t - 0.90 y_{t-1} = 250$$

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Step 2: Determine Particular Solution ($y^*$ / Equilibrium Level)

Set $y_t = y_{t-1} = y^*$:

$$y^* - 0.90 y^* = 250$$

$$0.10 y^* = 250 \implies y^* = \frac{250}{0.10} = 2,500$$

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Step 3: Determine Complementary Function ($y_c$) and General Time Path

Homogeneous equation: $y_t - 0.90 y_{t-1} = 0 \implies y_c = A(0.90)^t$. General solution:

$$y_t = y^* + A(0.90)^t = 2,500 + A(0.90)^t$$

Apply the initial condition $y_0 = 3,000$:

$$3,000 = 2,500 + A(0.90)^0 \implies A = 3,000 - 2,500 = 500$$

Thus, the exact time path is:

$$y_t = 2,500 + 500(0.90)^t$$

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Step 4: Comment on Stability of the Time Path

In the solution $y_t = y^* + A(b)^t$, the base is $b = 0.90$:

1. Since $|b| = 0.90 < 1$, the term $(0.90)^t \to 0$ as $t \to \infty$.
2. Since $b > 0$, the deviation converges without oscillation (non-oscillatory convergence).
3. Therefore, the time path is dynamically stable and converges monotonically to the intertemporal equilibrium income of $2,500$.

Final Answer:

• Time path: $y_t = 2,500 + 500(0.90)^t$
• Stability: Dynamically stable (monotonically converges to $2,500$ as $t \to \infty$).

Step-by-Step Solution:

Problem Formulation:

$$\text{Maximize } Z = 5x_1 + 3x_2$$

$$\text{Subject to the constraints:}$$

$$2x_1 + x_2 \le 5 \quad \text{--- (1)}$$

$$x_1 + x_2 \le 4 \quad \text{--- (2)}$$

$$x_1 \ge 0, \quad x_2 \ge 0$$

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Method 1: Graphical Method

Step 1: Plot the constraint boundary lines

1. Line 1 ($2x_1 + x_2 = 5$):
   • When $x_1 = 0 \implies x_2 = 5 \implies (0, 5)$
   • When $x_2 = 0 \implies 2x_1 = 5 \implies x_1 = 2.5 \implies (2.5, 0)$
2. Line 2 ($x_1 + x_2 = 4$):
   • When $x_1 = 0 \implies x_2 = 4 \implies (0, 4)$
   • When $x_2 = 0 \implies x_1 = 4 \implies (4, 0)$

Step 2: Find intersection point of the two constraint lines

Subtract equation (2) from equation (1):

$$(2x_1 + x_2) - (x_1 + x_2) = 5 - 4 \implies x_1 = 1$$

Substitute $x_1 = 1$ into equation (2):

$$1 + x_2 = 4 \implies x_2 = 3$$

The intersection point is $(1, 3)$.

Step 3: Evaluate the objective function $Z = 5x_1 + 3x_2$ at all corner points

The feasible region is bounded by the convex polygon with vertices:

Corner Point $(x_1, x_2)$	$Z = 5x_1 + 3x_2$	Value
$(0, 0)$	$5(0) + 3(0)$	$0$
$(2.5, 0)$	$5(2.5) + 3(0)$	$12.5$
$(1, 3)$	$5(1) + 3(3) = 5 + 9$	$14$ (Optimal Maximum)
$(0, 4)$	$5(0) + 3(4)$	$12$

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Method 2: Simplex Method

Step 1: Standard Canonical Form

Introduce non-negative slack variables $s_1 \ge 0, s_2 \ge 0$:

$$2x_1 + x_2 + s_1 = 5$$

$$x_1 + x_2 + s_2 = 4$$

$$Z - 5x_1 - 3x_2 + 0s_1 + 0s_2 = 0$$

Tableau 1 (Initial):

Basic	$x_1$	$x_2$	$s_1$	$s_2$	RHS	Ratio
$s_1$	2	1	1	0	5	$5/2 = 2.5$ (Pivot)
$s_2$	1	1	0	1	4	$4/1 = 4$
$Z$	-5	-3	0	0	0

• Entering variable: $x_1$ (most negative indicator $-5$).
• Leaving variable: $s_1$ (minimum ratio $2.5$).
• Pivot element: $2$.
• Perform row operations:
  • $R_1 \leftarrow R_1 / 2$: $[1, 0.5, 0.5, 0 \mid 2.5]$
  • $R_2 \leftarrow R_2 - R_1$: $[0, 0.5, -0.5, 1 \mid 1.5]$
  • $R_Z \leftarrow R_Z + 5R_1$: $[0, -0.5, 2.5, 0 \mid 12.5]$

Tableau 2:

Basic	$x_1$	$x_2$	$s_1$	$s_2$	RHS	Ratio
$x_1$	1	0.5	0.5	0	2.5	$2.5 / 0.5 = 5$
$s_2$	0	0.5	-0.5	1	1.5	$1.5 / 0.5 = 3$ (Pivot)
$Z$	0	-0.5	2.5	0	12.5

• Entering variable: $x_2$ (indicator $-0.5$).
• Leaving variable: $s_2$ (minimum ratio $3$).
• Pivot element: $0.5$.
• Perform row operations:
  • $R_2 \leftarrow R_2 / 0.5$: $[0, 1, -1, 2 \mid 3]$
  • $R_1 \leftarrow R_1 - 0.5 R_2$: $[1, 0, 1, -1 \mid 1]$
  • $R_Z \leftarrow R_Z + 0.5 R_2$: $[0, 0, 2, 1 \mid 14]$

Tableau 3 (Optimal):

Basic	$x_1$	$x_2$	$s_1$	$s_2$	RHS
$x_1$	1	0	1	-1	1
$x_2$	0	1	-1	2	3
$Z$	0	0	2	1	14

All objective row coefficients are $\ge 0$. The optimal solution is reached:

• $x_1 = 1$
• $x_2 = 3$
• $s_1 = 0, s_2 = 0$
• $Z_{\text{max}} = 14$

Final Answer: The optimal solution is $x_1 = 1$, $x_2 = 3$, with maximum value $Z = 14$.