Solve the following Linear Programming Problem (LPP) graphically : Maximize $z = 3x + 5y$ subject to the constraints : $x + 2y \le 2000$ , $x + y \le 1500$ , $y \le 600$ , $x \ge 0, y \ge 0$ .

Answer: The maximum value of $z$ is $5500$ at $x=1000$ and $y=500$ .

Step-by-step solution

Step 1: Identify the feasible region

First, we need to graph each inequality to find the feasible region. The inequalities $x \ge 0$ and $y \ge 0$ 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 $(1000, 500)$ .

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Solve the following Linear Programming Problem (LPP) graphically : Maximize z=3x+5yz = 3x + 5yz=3x+5y subject to the constraints : x+2y≤2000x + 2y \le 2000x+2y≤2000, x+y≤1500x + y \le 1500x+y≤1500, y≤600y \le 600y≤600, x≥0,y≥0x \ge 0, y \ge 0x≥0,y≥0.