Question

Solve the following Linear Programming Problem graphically :
Maximise $Z = 600x + 400y$
subject to the constraints
$x + 2y \leq 12$
$4x + 5y \geq 20$
$2x + y \leq 12$
$x, y \geq 0$

Hint:

Always double-check the intersection points by substituting them back into the original equations. For "Maximize" problems, points further from the origin usually yield higher values if the objective coefficients are positive.

Answer 1

Concept:
• The graphical method involves plotting the constraint lines on a graph and identifying the feasible region.
• The optimal value occurs at one of the corner points (vertices) of the bounded feasible region.

Step 1: Find intercepts for each constraint line
Line $L_1: x + 2y = 12$. Intercepts: $(12, 0)$ and $(0, 6)$.
Line $L_2: 4x + 5y = 20$. Intercepts: $(5, 0)$ and $(0, 4)$.
Line $L_3: 2x + y = 12$. Intercepts: $(6, 0)$ and $(0, 12)$.

Step 2: Determine the feasible region
Test $(0,0)$ in the inequalities:
$0 + 0 \leq 12$ (True, region towards origin for $L_1$)
$0 + 0 \geq 20$ (False, region away from origin for $L_2$)
$0 + 0 \leq 12$ (True, region towards origin for $L_3$)
The overlapping region is a polygon in the first quadrant.

Step 3: Identify the corner points
Solving intersections:
$L_1 \cap L_3: x + 2y = 12$ and $2x + y = 12 \implies (4, 4)$.
$L_3 \cap \text{x-axis}: (6, 0)$.
$L_2 \cap \text{x-axis}: (5, 0)$.
$L_2 \cap \text{y-axis}: (0, 4)$.
$L_1 \cap \text{y-axis}: (0, 6)$.

Step 4: Calculate $Z$ at each vertex
$A(0, 4): Z = 600(0) + 400(4) = 1600$
$B(0, 6): Z = 600(0) + 400(6) = 2400$
$C(4, 4): Z = 600(4) + 400(4) = 2400 + 1600 = 4000$
$D(6, 0): Z = 600(6) + 400(0) = 3600$
$E(5, 0): Z = 600(5) + 400(0) = 3000$

Step 5: Conclusion
The maximum value of $Z$ is 4000, which occurs at the point $(4, 4)$.