1. Solve the following by graphical method: (i) Max \( Z=X_{1}-3 X_{2} \) Subject to, \( X_{1}+X_{2} \leq 300 \) \( X_{1}-2 X_{2} \leq 200 \) \( 2 X_{1}+X_{2} \leq 100 \) \( X_{2} \leq 200 \) \( X_{1}, X_{2} \geq 0 \)

To solve the given linear programming problem using the graphical method, we will follow these steps: 1. **Identify the objective function and constraints**: - Objective function: Maximize \( Z = X_{1} - 3X_{2} \) - Constraints: 1. \( X_{1} + X_{2} \leq 300 \) 2. \( X_{1} - 2X_{2} \leq 200 \) 3. \( 2X_{1} + X_{2} \leq 100 \) 4. \( X_{2} \leq 200 \) 5. \( X_{1}, X_{2} \geq 0 \) 2. **Graph the constraints**: - We will convert each inequality into an equation to find the boundary lines. - For \( X_{1} + X_{2} = 300 \): - When \( X_{1} = 0 \), \( X_{2} = 300 \) - When \( X_{2} = 0 \), \( X_{1} = 300 \) - For \( X_{1} - 2X_{2} = 200 \): - When \( X_{1} = 0 \), \( X_{2} = -100 \) (not valid since \( X_{2} \geq 0 \)) - When \( X_{2} = 0 \), \( X_{1} = 200 \) - For \( 2X_{1} + X_{2} = 100 \): - When \( X_{1} = 0 \), \( X_{2} = 100 \) - When \( X_{2} = 0 \), \( X_{1} = 50 \) - For \( X_{2} = 200 \): - This is a horizontal line at \( X_{2} = 200 \). 3. **Plot the lines on a graph**: - The feasible region is determined by the area that satisfies all constraints simultaneously. 4. **Identify the feasible region**: - The feasible region is the area where all the constraints overlap, and it will be bounded by the lines we plotted. 5. **Find the corner points of the feasible region**: - The corner points can be found by solving the equations of the lines where they intersect. 6. **Evaluate the objective function at each corner point**: - Calculate \( Z \) at each corner point to find the maximum value. Now, let's calculate the intersection points of the constraints to find the corner points. ### Step 1: Solve for Intersection Points 1. **Intersection of \( X_{1} + X_{2} = 300 \) and \( X_{1} - 2X_{2} = 200 \)**: \[ \begin{align*} X_{1} + X_{2} &= 300 \quad (1) \\ X_{1} - 2X_{2} &= 200 \quad (2) \end{align*} \] 2. **Intersection of \( X_{1} + X_{2} = 300 \) and \( 2X_{1} + X_{2} = 100 \)**: \[ \begin{align*} X_{1} + X_{2} &= 300 \quad (1) \\ 2X_{1} + X_{2} &= 100 \quad (3) \end{align*} \] 3. **Intersection of \( X_{1} - 2X_{2} = 200 \) and \( 2X_{1} + X_{2} = 100 \)**: \[ \begin{align*} X_{1} - 2X_{2} &= 200 \quad (2) \\ 2X_{1} + X_{2} &= 100 \quad (3) \end{align*} \] 4. **Intersection of \( X_{2} = 200 \) with other constraints**. Let's calculate these intersections. ### Step 2: Calculate Intersection Points I will now calculate the intersection points of the constraints. Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}X_{2}=200\\X_{1}+X_{2}=300\end{array}\right.\) - step1: Substitute the value of \(X_{2}:\) \(X_{1}+200=300\) - step2: Move the constant to the right side: \(X_{1}=300-200\) - step3: Subtract the numbers: \(X_{1}=100\) - step4: Calculate: \(\left\{ \begin{array}{l}X_{1}=100\\X_{2}=200\end{array}\right.\) - step5: Check the solution: \(\left\{ \begin{array}{l}X_{1}=100\\X_{2}=200\end{array}\right.\) - step6: Rewrite: \(\left(X_{1},X_{2}\right) = \left(100,200\right)\) Solve the system of equations \( X_{2} = 200; X_{1} - 2X_{2} = 200 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}X_{2}=200\\X_{1}-2X_{2}=200\end{array}\right.\) - step1: Substitute the value of \(X_{2}:\) \(X_{1}-2\times 200=200\) - step2: Simplify: \(X_{1}-400=200\) - step3: Move the constant to the right side: \(X_{1}=200+400\) - step4: Add the numbers: \(X_{1}=600\) - step5: Calculate: \(\left\{ \begin{array}{l}X_{1}=600\\X_{2}=200\end{array}\right.\) - step6: Check the solution: \(\left\{ \begin{array}{l}X_{1}=600\\X_{2}=200\end{array}\right.\) - step7: Rewrite: \(\left(X_{1},X_{2}\right) = \left(600,200\right)\) Solve the system of equations \( X_{2} = 200; 2X_{1} + X_{2} = 100 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}X_{2}=200\\2X_{1}+X_{2}=100\end{array}\right.\) - step1: Substitute the value of \(X_{2}:\) \(2X_{1}+200=100\) - step2: Move the constant to the right side: \(2X_{1}=100-200\) - step3: Subtract the numbers: \(2X_{1}=-100\) - step4: Divide both sides: \(\frac{2X_{1}}{2}=\frac{-100}{2}\) - step5: Divide the numbers: \(X_{1}=-50\) - step6: Calculate: \(\left\{ \begin{array}{l}X_{1}=-50\\X_{2}=200\end{array}\right.\) - step7: Check the solution: \(\left\{ \begin{array}{l}X_{1}=-50\\X_{2}=200\end{array}\right.\) - step8: Rewrite: \(\left(X_{1},X_{2}\right) = \left(-50,200\right)\) Solve the system of equations \( X_{1} + X_{2} = 300; 2X_{1} + X_{2} = 100 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}X_{1}+X_{2}=300\\2X_{1}+X_{2}=100\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}X_{1}=300-X_{2}\\2X_{1}+X_{2}=100\end{array}\right.\) - step2: Substitute the value of \(X_{1}:\) \(2\left(300-X_{2}\right)+X_{2}=100\) - step3: Simplify: \(600-X_{2}=100\) - step4: Move the constant to the right side: \(-X_{2}=100-600\) - step5: Subtract the numbers: \(-X_{2}=-500\) - step6: Change the signs: \(X_{2}=500\) - step7: Substitute the value of \(X_{2}:\) \(X_{1}=300-500\) - step8: Simplify: \(X_{1}=-200\) - step9: Calculate: \(\left\{ \begin{array}{l}X_{1}=-200\\X_{2}=500\end{array}\right.\) - step10: Check the solution: \(\left\{ \begin{array}{l}X_{1}=-200\\X_{2}=500\end{array}\right.\) - step11: Rewrite: \(\left(X_{1},X_{2}\right) = \left(-200,500\right)\) Solve the system of equations \( X_{1} + X_{2} = 300; X_{1} - 2X_{2} = 200 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}X_{1}+X_{2}=300\\X_{1}-2X_{2}=200\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}X_{1}=300-X_{2}\\X_{1}-2X_{2}=200\end{array}\right.\) - step2: Substitute the value of \(X_{1}:\) \(300-X_{2}-2X_{2}=200\) - step3: Subtract the terms: \(300-3X_{2}=200\) - step4: Move the constant to the right side: \(-3X_{2}=200-300\) - step5: Subtract the numbers: \(-3X_{2}=-100\) - step6: Change the signs: \(3X_{2}=100\) - step7: Divide both sides: \(\frac{3X_{2}}{3}=\frac{100}{3}\) - step8: Divide the numbers: \(X_{2}=\frac{100}{3}\) - step9: Substitute the value of \(X_{2}:\) \(X_{1}=300-\frac{100}{3}\) - step10: Simplify: \(X_{1}=\frac{800}{3}\) - step11: Calculate: \(\left\{ \begin{array}{l}X_{1}=\frac{800}{3}\\X_{2}=\frac{100}{3}\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}X_{1}=\frac{800}{3}\\X_{2}=\frac{100}{3}\end{array}\right.\) - step13: Rewrite: \(\left(X_{1},X_{2}\right) = \left(\frac{800}{3},\frac{100}{3}\right)\) Solve the system of equations \( X_{1} - 2X_{2} = 200; 2X_{1} + X_{2} = 100 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}X_{1}-2X_{2}=200\\2X_{1}+X_{2}=100\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}X_{1}=200+2X_{2}\\2X_{1}+X_{2}=100\end{array}\right.\) - step2: Substitute the value of \(X_{1}:\) \(2\left(200+2X_{2}\right)+X_{2}=100\) - step3: Simplify: \(400+5X_{2}=100\) - step4: Move the constant to the right side: \(5X_{2}=100-400\) - step5: Subtract the numbers: \(5X_{2}=-300\) - step6: Divide both sides: \(\frac{5X_{2}}{5}=\frac{-300}{5}\) - step7: Divide the numbers: \(X_{2}=-60\) - step8: Substitute the value of \(X_{2}:\) \(X_{1}=200+2\left(-60\right)\) - step9: Simplify: \(X_{1}=80\) - step10: Calculate: \(\left\{ \begin{array}{l}X_{1}=80\\X_{2}=-60\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}X_{1}=80\\X_{2}=-60\end{array}\right.\) - step12: Rewrite: \(\left(X_{1},X_{2}\right) = \left(80,-60\right)\) ### Step 2: Intersection Points From the calculations, we have the following intersection points: 1. **Intersection of \( X_{1} + X_{2} = 300 \) and \( X_{1} - 2X_{2} = 200 \)**: - \( (X_{1}, X_{2}) = \left(\frac{800}{3}, \frac{100}{3}\right) \) or approximately \( (266.67, 33.33) \) 2. **Intersection of \( X_{1} + X_{2} = 300 \) and \( 2X_{1} + X_{2} = 100 \)**: - No valid intersection since \( X_{1} \) would be negative. 3. **Intersection of \( X_{1} - 2X_{2} = 200 \) and \( 2X_{1} + X_{2} = 100 \)**: - No valid intersection since \( X_{2} \) would be negative. 4. **Intersection of \( X_{2} = 200 \) with \( X_{1} + X_{2} = 300 \)**: - \( (X_{1}, X_{2}) = (100, 200) \) 5. **Intersection of \( X_{2} = 200 \) with \( X_{1} - 2X_{2} = 200 \)**: - No valid intersection since \( X_{1} \) would be negative. 6. **Intersection of \( X_{2} = 200 \) with \( 2X_{1} + X_{2} = 100 \)**: - No valid intersection since \( X_{1} \) would be negative. ### Step 3: Valid Corner Points The valid corner points of the feasible region are: 1. \( \left(\frac{800}{3}, \frac{100}{3}\right) \) or approximately \( (266.67, 33.33) \) 2. \( (100, 200) \) ### Step 4: Evaluate the Objective Function Now, we will evaluate the objective function \( Z = X_{1} - 3X_{2} \) at these corner points. 1. For \( \left(\frac{800}{3}, \frac{100}{3}\right) \): \[ Z = \frac{800}{3} - 3 \cdot \frac{100}{3} = \frac{800}{3} - \frac{300}{3} = \frac{500}{3} \approx 166.67 \] 2. For \( (100, 200) \): \[ Z = 100 - 3 \cdot 200 = 100 - 600 = -500 \] ### Step 5: Conclusion The maximum value of \( Z \) occurs at the point \( \left(\frac{800}{3}, \frac{100}{3}\right) \) with a maximum value of approximately \( 166.67 \). Thus, the solution to the linear programming problem is: - **Maximum \( Z \approx 166.67 \) at \( (X_{1}, X_{2}) \approx (266.67, 33.33) \)**.