Solve the following Linear Programming Problem (LPP) graphically : Maximize z=3x+5y subject to the constraints : x+2y≤2000, x+y≤1500, y≤600, x≥0,y≥0.
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Step-by-Step Solution
Step 1: Identify the feasible region
First, we need to graph each inequality to find the feasible region. The inequalities x≥0 and y≥0 restrict our solution to the first quadrant. We will plot the boundary lines for the other inequalities and shade the region that satisfies all of them.
Step 2: Find the vertices of the feasible region
The vertices of the feasible region are the points where the boundary lines intersect. We find these intersection points by solving the equations of the lines. For example, the intersection of x+y=1500 and y=600 gives x=900. The intersection of x+2y=2000 and y=600 gives x=800. The intersection of x+y=1500 and x+2y=2000 gives y=500 and x=1000. The vertices are O(0,0), A(1500,0), B(1000,500), C(800,600), and D(0,600).
Step 3: Evaluate the objective function at each vertex
To find the maximum value of the objective function z=3x+5y, we evaluate z at each of the vertices of the feasible region.
Step 4: Calculate z for each vertex
We substitute the coordinates of each vertex into the objective function z=3x+5y to find the corresponding z value. The values obtained are 0, 4500, 5500, 5400, and 3000.
Step 5: Determine the maximum value
Comparing the values of z calculated at each vertex, the maximum value is 5500, which occurs at the point (1000,500).