Solve the following Linear Programming Problem graphically: Maximise $Z = 600x + 400y$ subject to the constraints $x + 2y \le 12$ $4x + 5y \ge 20$ $2x + y \le 12$ $x, y \ge 0$ (linear programming, maximization, constraints, graphical solution, objective function)

Step 1: Define the objective function and constraints

We are given the objective function $Z = 600x + 400y$ which we need to maximize. The problem also specifies four inequality constraints that define the feasible region for $x$ and $y$. The conditions $x \ge 0$ and $y \ge 0$ imply that the solution must lie in the first quadrant.

Step 2: Convert inequalities to equations and find intercepts

To graph the inequalities, we first treat them as equalities to find the boundary lines. For each line, we find its $x$ and $y$ intercepts by setting $x=0$ and $y=0$ respectively. These points allow us to draw each line on the coordinate plane.

Step 3: Plot lines and determine feasible region

We plot the lines corresponding to the constraints. Then, we determine the feasible region by testing a point (like the origin (0,0) if it's not on the line) for each inequality. For $x+2y \le 12$, the region is towards the origin. For $4x+5y \ge 20$, the region is away from the origin. For $2x+y \le 12$, the region is towards the origin. Combined with $x, y \ge 0$, this gives us the feasible region.

Step 4: Find the corner points of the feasible region

The corner points of the feasible region are critical for finding the maximum or minimum of the objective function. These points occur at the intersections of the boundary lines. We solve pairs of linear equations to find these intersection points. We must ensure that the intersection points also satisfy all other constraints, ensuring they are valid vertices of the feasible region. After correctly identifying the region, the points from the previous step are found to be: $(5{,}0)$, $(6{,}0)$, $(4{,}4)$, $(0{,}6)$ and $(0{,}4)$. For consistency, we label these as $A=(5{,}0)$, $B=(6{,}0)$, $C=(4{,}4)$, $D=(0{,}6)$, $E=(0{,}4)$. Note that the question uses 'P, Q, R', let's use those instead: $P=(5{,}0)$, $Q=(6{,}0)$, $R=(4{,}4)$, $S=(0{,}6)$, $T=(0{,}4)$. The points $(0{,}4)$, $(5{,}0)$, $(6{,}0)$, $(4{,}4)$ and $(0{,}6)$ form a polygon.

Step 5: Evaluate objective function at corner points

According to the Corner Point Theorem, the optimal (maximum or minimum) value of an objective function, if it exists, occurs at one of the corner points of the feasible region. We substitute the coordinates of each corner point into the objective function $Z$ to find the corresponding value of $Z$ at each point.

Step 6: Determine the maximum value

By comparing the values of $Z$ calculated at each corner point, we identify the maximum value. In this case, the maximum value of $Z$ is 4000, which occurs at the point $(4{,}4)$. This point lies within the feasible region and satisfies all given constraints.

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