Question

Minimize Z = 200x + 500y by graphical method subject to the following constraints:
x + 2y \(\ge\) 10, 3x + 4y \(\le\) 24, x \(\ge\) 0, y \(\ge\) 0.

Hint:

Always shade the feasible region clearly on the graph to avoid confusion. To find which side of a line satisfies an inequality, pick a test point (like the origin (0,0), if the line doesn't pass through it) and check if it satisfies the inequality.

Answer 1

Step 1: Understanding the Concept:
This is a Linear Programming Problem (LPP). The goal is to find the minimum value of a linear objective function Z, subject to a set of linear inequalities called constraints. The graphical method involves plotting these constraints to find the feasible region and then evaluating the objective function at the corner points (vertices) of this region. The optimal solution (minimum or maximum) will occur at one of these vertices.
Step 2: Key Formula or Approach:
1. Convert the inequalities into equations to plot the lines.
2. Graph the lines on a 2D plane.
3. Identify the feasible region, which is the area that satisfies all the given constraints simultaneously.
4. Determine the coordinates of the corner points of the feasible region.
5. Evaluate the objective function Z = 200x + 500y at each corner point.
6. The smallest value of Z will be the minimum value.
Step 3: Detailed Explanation:
The constraints are: \[ x + 2y \ge 10 \] \[ 3x + 4y \le 24 \] \[ x \ge 0, y \ge 0 \] 1. Graph the lines:
Line 1: \(x + 2y = 10\). It passes through (10, 0) and (0, 5). The region \(x + 2y \ge 10\) is the area on and above this line.
Line 2: \(3x + 4y = 24\). It passes through (8, 0) and (0, 6). The region \(3x + 4y \le 24\) is the area on and below this line.
The constraints \(x \ge 0, y \ge 0\) restrict the feasible region to the first quadrant.
2. Find the feasible region and corner points:
The feasible region is the area in the first quadrant that is above the line \(x+2y=10\) and below the line \(3x+4y=24\). The vertices (corner points) of this region are:
Point A: Intersection of \(x=0\) and \(3x+4y=24\). \(3(0) + 4y = 24 \implies y=6\). So, A = (0, 6).
Point B: Intersection of \(x=0\) and \(x+2y=10\). \(0 + 2y = 10 \implies y=5\). So, B = (0, 5).
Point C: Intersection of \(x+2y=10\) and \(3x+4y=24\).
From \(x+2y=10\), we have \(x = 10-2y\). Substitute this into the second equation: \[ 3(10-2y) + 4y = 24 \] \[ 30 - 6y + 4y = 24 \] \[ 30 - 2y = 24 \implies 2y = 6 \implies y = 3 \] Substitute \(y=3\) back into \(x=10-2y\): \[ x = 10 - 2(3) = 4 \] So, C = (4, 3).
3. Evaluate Z at corner points:
The objective function is Z = 200x + 500y.
At A(0, 6): \(Z = 200(0) + 500(6) = 3000\).
At B(0, 5): \(Z = 200(0) + 500(5) = 2500\).
At C(4, 3): \(Z = 200(4) + 500(3) = 800 + 1500 = 2300\).
4. Find the minimum value:
Comparing the values of Z, the minimum value is 2300.
Step 4: Final Answer:
The minimum value of Z is 2300, which occurs at the point (x, y) = (4, 3).