solve the following LPP using graphic method?

Question

max z=2x1 +3x2 subject to:
4x1 +3x2 <\=18
5x1+2x2</=19
x1,x2>/=0

Answer 1

To solve the given linear programming problem (LPP) using the graphical method, we'll follow these steps:

1. Formulate the Problem:

   • Objective Function: Maximize \( Z = 2x_1 + 3x_2 \)
   • Subject to constraints:
     1. \( 4x_1 + 3x_2 \leq 18 \)
     2. \( 5x_1 + 2x_2 \leq 19 \)
     3. \( x_1 \geq 0 \)
     4. \( x_2 \geq 0 \)

2. Graph the Constraints: We'll convert the inequalities into equations and find their intersections with the axes.

   • For \( 4x_1 + 3x_2 = 18 \):

     • If \( x_1 = 0 \): \[ 3x_2 = 18 \implies x_2 = 6 \] (Point: (0,6))
     • If \( x_2 = 0 \): \[ 4x_1 = 18 \implies x_1 = 4.5 \] (Point: (4.5,0))
     • This line will be drawn between points (0,6) and (4.5,0).

   • For \( 5x_1 + 2x_2 = 19 \):

     • If \( x_1 = 0 \): \[ 2x_2 = 19 \implies x_2 = 9.5 \] (Point: (0,9.5))
     • If \( x_2 = 0 \): \[ 5x_1 = 19 \implies x_1 = 3.8 \] (Point: (3.8,0))
     • This line will be drawn between points (0,9.5) and (3.8,0).

3. Determine the Feasible Region:

   • The feasible region is where all constraints overlap in the first quadrant (since \( x_1 \) and \( x_2 \) must be non-negative).
   • Use the lines from the equations and shade the region that satisfies all the inequalities. Since both of our constraints are "<=" inequalities, we shade below the lines for both constraints.

4. Find Vertices of the Feasible Region: We need to find the intersection of the lines:

   • \( 4x_1 + 3x_2 = 18 \) and \( 5x_1 + 2x_2 = 19 \):

   To solve this system, we can manipulate the equations: \[ \begin{align*}

   1. & \quad 4x_1 + 3x_2 = 18 \
   2. & \quad 5x_1 + 2x_2 = 19 \end{align*} \]

   From equation (1), express \( x_2 \) in terms of \( x_1 \): \[ 3x_2 = 18 - 4x_1 \implies x_2 = 6 - \frac{4}{3}x_1 \]

   Substitute this expression for \( x_2 \) into equation (2): \[ 5x_1 + 2\left(6 - \frac{4}{3}x_1\right) = 19 \] \[ 5x_1 + 12 - \frac{8}{3}x_1 = 19 \] \[ \left(5 - \frac{8}{3}\right)x_1 + 12 = 19 \] \[ \frac{15}{3}x_1 - \frac{8}{3}x_1 = 7 \implies \frac{7}{3}x_1 = 7 \implies x_1 = 3 \]

   Substitute \( x_1 = 3 \) back into the expression for \( x_2 \): \[ x_2 = 6 - \frac{4}{3}(3) = 6 - 4 = 2 \] Thus, one vertex of the feasible region is \( (3, 2) \).

5. Analyze the Other Vertices: The other vertices come from the intersections with the axes:

   • (0, 6) from the first constraint.
   • (4.5, 0) from the first constraint.
   • (3.8, 0) from the second constraint.
   • (0, 9.5) from the second constraint.

   However, we will only consider points that are inside the feasible region defined by the constraints.

6. Evaluation of the Objective Function: Now, we evaluate \( Z = 2x_1 + 3x_2 \) at each vertex inside the feasible region:

   • At \( (0, 6) \): \[ Z = 2(0) + 3(6) = 18 \]
   • At \( (3, 2) \): \[ Z = 2(3) + 3(2) = 6 + 6 = 12 \]
   • At \( (4.5, 0) \): \[ Z = 2(4.5) + 3(0) = 9 \]
   • (3.8, 0) is outside the feasible region defined by the constraints so we don't include it.

7. Conclusion: The maximum value of \( Z \) is at \( (0, 6) \): \[ Z_{max} = 18 \] Thus, the optimal solution to the LPP is:

   • \( x_1 = 0, x_2 = 6 \) with maximum \( Z = 18 \).