To solve the Linear Programming (LP) problem graphically, we follow these steps:

Step 1: Define the Problem

We aim to maximize the objective function:

$Z = 5x_1 + 3x_2$

Subject to constraints:

1. $3x_1 + 5x_2 \leq 15$
2. $5x_1 + 2x_2 \leq 10$
3. $x_1, x_2 \geq 0$

Step 2: Graph the Constraints

1. Rewrite each inequality constraint as an equation:

   • $3x_1 + 5x_2 = 15$
   • $5x_1 + 2x_2 = 10$

2. Determine points of intersection for each equation with the axes:

   • For $3x_1 + 5x_2 = 15$:
     • When $x_1 = 0$, $x_2 = 3$ ($(0, 3)$)
     • When $x_2 = 0$, $x_1 = 5$ ($(5, 0)$)
   • For $5x_1 + 2x_2 = 10$:
     • When $x_1 = 0$, $x_2 = 5$ ($(0, 5)$)
     • When $x_2 = 0$, $x_1 = 2$ ($(2, 0)$)

3. Plot these lines on a graph, shading the feasible region that satisfies all constraints, including $x_1, x_2 \geq 0$.

Step 3: Identify Corner Points

The feasible region will be a polygon. Locate the vertices (corner points) of this region:

• Intersection of $3x_1 + 5x_2 = 15$ and $5x_1 + 2x_2 = 10$: Solve the system of equations: $3x_1 + 5x_2 = 15$ $5x_1 + 2x_2 = 10$ Multiply the first equation by 2 and the second by 5 to align coefficients of $x_2$: $6x_1 + 10x_2 = 30$ $25x_1 + 10x_2 = 50$ Subtract: $19x_1 = 20 \quad \Rightarrow \quad x_1 = \frac{20}{19}$ Substitute $x_1$ back into $3x_1 + 5x_2 = 15$: $3\left(\frac{20}{19}\right) + 5x_2 = 15 \quad \Rightarrow \quad \frac{60}{19} + 5x_2 = 15$ $5x_2 = 15 - \frac{60}{19} \quad \Rightarrow \quad 5x_2 = \frac{225}{19} \quad \Rightarrow \quad x_2 = \frac{45}{19}$ The intersection point is $\left(\frac{20}{19}, \frac{45}{19}\right)$.

Step 4: Evaluate the Objective Function

Calculate $Z = 5x_1 + 3x_2$ at each vertex of the feasible region:

1. At $(0, 0)$: $Z = 5(0) + 3(0) = 0$
2. At $(5, 0)$: $Z = 5(5) + 3(0) = 25$
3. At $(0, 3)$: $Z = 5(0) + 3(3) = 9$
4. At $\left(\frac{20}{19}, \frac{45}{19}\right)$: $Z = 5\left(\frac{20}{19}\right) + 3\left(\frac{45}{19}\right) = \frac{100}{19} + \frac{135}{19} = \frac{235}{19} \approx 12.37$

Step 5: Determine the Optimal Solution

The maximum value of $Z$ is $25$, occurring at $(5, 0)$.

Solution

The optimal solution is: $x_1 = 5, \, x_2 = 0, \, \text{with } Z = 25.$

Would you like a detailed explanation of any step or a graph of the solution?

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Related Questions

1. How is the graphical method used for more than two variables?
2. What are the limitations of the graphical method?
3. Can the Simplex method solve this LP problem, and how does it compare?
4. How can we verify the feasibility of a solution in LP problems?
5. What are some real-world applications of linear programming?

Tip

Always check that the feasible region is bounded to ensure the existence of a maximum or minimum value for the objective function.