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)
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Step-by-Step Solution
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≥0 and y≥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≥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).