Question

Solve the following Linear Programming Problem graphically :
Maximise $Z = 12x + 18y$
subject to the constraints \[ x + y \le 1200, \quad x - 2y \ge 0, \quad x + 3y \ge 600, \quad x \ge 0, \, y \ge 0 \]

Hint:

When graphing constraints that pass through the origin like $x - 2y = 0$, you cannot use $(0,0)$ as a test region indicator. Choose an explicit test point resting purely on one axis line, such as $(100, 0)$, to instantly know which side to shade.

Answer 1

Concept: To find the maximum value of the objective function $Z$, we plot the boundary lines for each constraint inequality on a Cartesian plane, shade the common overlapping region (Feasible Region), identify its corner points, and evaluate $Z$ at each of those vertices.

Step 1: Plot the boundary lines and find their axis intercepts.
Convert the inequalities into linear equations to trace them on the graph:
• Line 1 ($L_1$): $x + y = 1200$
• If $x = 0$, $y = 1200 \implies (0, 1200)$
• If $y = 0$, $x = 1200 \implies (1200, 0)$
• Line 2 ($L_2$): $x - 2y = 0 \implies x = 2y$
• Passes through the origin $(0, 0)$.
• Additional guiding point: If $y = 200$, $x = 400 \implies (400, 200)$.
• Line 3 ($L_3$): $x + 3y = 600$
• If $x = 0$, $y = 200 \implies (0, 200)$
• If $y = 0$, $x = 600 \implies (600, 0)$

Step 2: Determine the direction of shading for each inequality.

• $x + y \le 1200$: Testing the origin $(0,0)$ gives $0 \le 1200$, which is True. Shade towards the origin.
• $x - 2y \ge 0$: Since it passes through $(0,0)$, test point $(400, 0) \implies 400 \ge 0$, which is True. Shade downwards/rightwards towards the $x$-axis.
• $x + 3y \ge 600$: Testing $(0,0)$ gives $0 \ge 600$, which is False. Shade away from the origin.
• $x \ge 0, y \ge 0$: Restricts the feasible region completely within the First Quadrant.

Step 3: Solve for the intersection corner points.
Let's find where the boundary lines cross each other to identify the vertices of the bounded closed polygonal feasible region:
• Point A (Intersection of $L_2$ and $L_3$): Substitute $x = 2y$ into $x + 3y = 600$: \[ 2y + 3y = 600 \implies 5y = 600 \implies y = 120, \, x = 240 \implies A(240, 120) \]
• Point B (Intersection of $L_1$ and $L_2$): Substitute $x = 2y$ into $x + y = 1200$: \[ 2y + y = 1200 \implies 3y = 1200 \implies y = 400, \, x = 800 \implies B(800, 400) \]
• Point C ($x$-intercept of $L_1$): Since $x - 2y \ge 0$ keeps the region below the line $x=2y$, the bounding lower vertices extend along the $x$-axis. Let's inspect the graph intersections! The region is bounded by $x+3y \ge 600$, $x-2y \ge 0$, and $x+y \le 1200$. Let's check the $y=0$ baseline intercepts: For $L_1$, it crosses at $(1200, 0)$. For $L_3$, it crosses at $(600, 0)$. Let's test if points on the $x$-axis between 600 and 1200 satisfy all conditions: At $(600,0)$: $600+0 \le 1200$ (True), $600-0 \ge 0$ (True), $600+0 \ge 600$ (True). So $C(600, 0)$ is a valid corner vertex. At $(1200,0)$: $1200+0 \le 1200$ (True), $1200-0 \ge 0$ (True), $1200+0 \ge 600$ (True). So $D(1200, 0)$ is a valid corner vertex. The four corner vertices of the bounded feasible region are: \[ A(240, 120), \quad B(800, 400), \quad C(1200, 0), \quad D(600, 0) \]

Step 4: Evaluate the objective function $Z = 12x + 18y$ at each corner point.
{|c|c|l|} Corner Point & Coordinates $(x, y)$ & Value of $Z = 12x + 18y$
$A$ & $(240, 120)$ & $Z = 12(240) + 18(120) = 2880 + 2160 = 5040$
$B$ & $(800, 400)$ & $Z = 12(800) + 18(400) = 9600 + 7200 = 16800$
$C$ & $(1200, 0)$ & $Z = 12(1200) + 18(0) = 14400$
$D$ & $(600, 0)$ & $Z = 12(600) + 18(0) = 7200$
Comparing the values from our optimization matrix column table, the highest calculated output numerical quantity is 16800. Conclusion: The maximum value of $Z$ is 16800, which is achieved at the point $(800, 400)$.