33. Solve the following Linear Programming Problem graphically : Maximise $Z = 600x + 400y$ subject to the constraints x + 2y ≤ 12 $4x + 5y$ ≥ 20 $2x + y$ ≤ 12 x, y ≥ 0 (Linear Programming, Graphical Method, Optimization, Constraints, Objective Function)

Step 1: Convert inequalities to equations and find intercepts

To graph the inequalities, we first convert them into linear equations. For each equation, we find the $x$ -intercept (by setting $y=0$) and the $y$ -intercept (by setting $x=0$). These two points allow us to draw the line corresponding to each equation.

Step 2: Graph the lines and identify the feasible region

We plot each line on the coordinate plane. Then, we test a point (like the origin (0,0)) for each inequality to determine which side of the line represents the feasible region. The intersection of all feasible regions, including $x \ge 0$ and $y \ge 0$, forms the bounded feasible region. In this case, the feasible region is a polygon.

Step 3: Find the corner points of the feasible region

The corner points of the feasible region are the vertices of the polygon formed by the intersection of the inequalities. These points are found by solving pairs of linear equations. We need to find the intersection points that lie within the feasible region. The corner points are (5,0), (6,0), (4,4), and (0,4) for this problem. The calculated intersections that result in negative x or y values are outside the first quadrant and thus not part of the feasible region $x, y \ge 0$.

Step 4: Evaluate the objective function at each corner point

The maximum or minimum value of the objective function (Z) occurs at one of the corner points of the feasible region. We substitute the coordinates of each corner point into the objective function $Z = 600x + 400y$ to find the corresponding value of Z.

Step 5: Determine the maximum value of Z

By comparing the values of Z obtained 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).

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