Question

Maximize \(Z = x + 2y\) by graphical method under the constraints \(x+y \le 1, -x+y \le 0, x \ge 0, y \ge 0\).

Hint:

The fundamental theorem of linear programming states that the optimal solution (maximum or minimum) for an LPP, if it exists, will always be at one of the corner points of the feasible region. Therefore, once you have identified the feasible region and its vertices, you only need to check these points.

Answer 1

Step 1: Understanding the Concept:
This is a Linear Programming Problem (LPP). The goal is to find the maximum value of a linear objective function \(Z\) subject to a set of linear inequalities (constraints). The graphical method involves plotting the constraints to find a feasible region and then testing the corner points of this region in the objective function.
Step 2: Key Formula or Approach:
1. Convert the inequalities into equations to plot the boundary lines.
2. Identify the feasible region that satisfies all constraints.
3. Determine the coordinates of the corner points (vertices) of the feasible region.
4. Evaluate the objective function \(Z\) at each corner point.
5. The maximum value of \(Z\) will be the largest value among those calculated.
Step 3: Detailed Explanation:
The constraints are:
\(x + y \le 1\)
\(-x + y \le 0 \implies y \le x\)
\(x \ge 0\)
\(y \ge 0\)
Let's plot the lines:
- \(L_1: x + y = 1\) (passes through (1,0) and (0,1))
- \(L_2: y = x\) (passes through the origin (0,0) at a 45-degree angle)
- \(L_3: x = 0\) (the y-axis)
- \(L_4: y = 0\) (the x-axis)
The feasible region is the area bounded by these lines that satisfies all the inequalities. It is a triangle with vertices at the intersection points of these lines.
Let's find the corner points:
- Point O: Intersection of \(x=0\) and \(y=0\). The coordinates are \((0, 0)\).
- Point A: Intersection of \(x+y=1\) and \(y=x\).
Substituting \(y=x\) into the first equation: \(x+x=1 \implies 2x=1 \implies x=0.5\). So, \(y=0.5\).
The coordinates are \((0.5, 0.5)\).
- Point B: Intersection of \(x+y=1\) and \(y=0\).
Substituting \(y=0\): \(x+0=1 \implies x=1\).
The coordinates are \((1, 0)\).
Now, evaluate the objective function \(Z = x + 2y\) at these corner points:
- At O(0, 0): \(Z = 0 + 2(0) = 0\)
- At A(0.5, 0.5): \(Z = 0.5 + 2(0.5) = 0.5 + 1 = 1.5\)
- At B(1, 0): \(Z = 1 + 2(0) = 1\)
Step 4: Final Answer:
Comparing the values of Z, the maximum value is 1.5, which occurs at the point \((0.5, 0.5)\).
Maximum \(Z = 1.5\).