Answer :
Answer:
To solve the LPP by the graphical method:
Objective Function: Maximize
[tex]\( z = 10x + 15y \)[/tex]
Constraints:
1.
[tex]\( x + y \geq 1 \)[/tex]
2.
[tex]\( x + 2y \leq 6 \)[/tex]
3.
[tex]\( 2x + y \leq 6 \)[/tex]
4.
[tex] \( x, y \geq 0 \)[/tex]
Steps:
1. Graph the constraints:
[tex] - \( x + y = 1 \)[/tex]
intersects axes at (1, 0) and (0, 1)
[tex] - \( x + 2y = 6 \)[/tex]
intersects axes at (6, 0) and (0, 3)
[tex] - \( 2x + y = 6 \)[/tex]
intersects axes at (3, 0) and (0, 6)
2. Find intersections of lines:
[tex] - \( x + y = 1 \)[/tex]
and
[tex]\( x + 2y = 6 \):[/tex]
no valid intersection within the constraints
[tex]- \( x + y = 1 \) [/tex]
and
[tex]\( 2x + y = 6 \):[/tex]
no valid intersection within the constraints
[tex]- \( x + 2y = 6 \) [/tex]
and
[tex]\( 2x + y = 6 \): \( (2, 2) \)[/tex]
3. Identify feasible region corner points:
[tex] - \( (0, 1) \)[/tex]
[tex] - \( (0, 3) \)[/tex]
[tex] - \( (3, 0) \)[/tex]
[tex] - \( (2, 2) \)[/tex]
4. Evaluate the objective function at each corner point:
[tex] - \( (0, 1) \): \( z = 15 \)[/tex]
[tex] - \( (0, 3) \): \( z = 45 \)[/tex]
[tex]- \( (3, 0) \): \( z = 30 \)[/tex]
[tex]- \( (2, 2) \): \( z = 50 \)[/tex]
Conclusion:
The maximum value
[tex]\( z \)[/tex]
is 50 at
[tex]\( (2, 2) \).[/tex]