wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Minimize Z = 3x1 + 5x2
Subject to
x1+3x23 x1+x22 x1, x20

Open in App
Solution

First, we will convert the given inequations into equations, we obtain the following equations:
x1 + 3x2 = 3, x1 + x2 = 2, x1 = 0 and x2 = 0

Region represented by x1 + 3x2 ≥ 3 :
The line x1 + 3x2 = 3 meets the coordinate axes at A(3, 0) and B(0, 1) respectively. By joining these points we obtain the line x1 + 3x2 = 3.
Clearly (0,0) does not satisfies the inequation x1 + 3x2 ≥ 3 .So,the region in the plane which does not contain the origin represents the solution set of the inequation x1 + 3x2 ≥ 3.

Region represented by x1 + x2 ≥ 2:
The line x1 + x2 = 2 meets the coordinate axes at C(2, 0) and D(0, 2) respectively. By joining these points we obtain the line x1 + x2 = 2.Clearly (0,0) does not satisfies the inequation x1 + x2 ≥ 2. So,the region containing the origin represents the solution set of the inequation x1 + x2 ≥ 2.

Region represented by x1 ≥ 0 and x2 ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x1 ≥ 0 and x2 ≥ 0.

The feasible region determined by the system of constraints, x1 + 3x2 ≥ 3 , x1 + x2 ≥ 2, x1 ≥ 0, and x2 ≥ 0, are as follows.

The corner points of the feasible region are O(0, 0), B(0, 1), E32, 12 and C(2, 0).

The values of Z at these corner points are as follows.

Corner point Z = 3x1 + 5x2
O(0, 0) 3 × 0 + 5 × 0 = 0
B(0, 1) 3 × 0 + 5 × 1 = 5
E32, 12 3 × 32 + 5 × 12 = 7
C(2, 0) 3 × 2 + 5 × 0 = 6

Therefore, the minimum value of Z is 0 at the point O(0, 0). Hence, x1 = 0 and x2 = 0 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 0.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Foot of the Perpendicular from a Point on a Plane
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon