Question

Show the solution zone of the following inequalities on a graph paper:
$$\begin{gathered} 5x+y\ge 10 \\ x+y\ge 6 \\ x+4y\ge 12 \\ x\ge 0,y\ge 0 \end{gathered}$$
Find x and y for which 3x + 2y is minimum subject to these inequalities. Use a graphical method.

Answer 1

First, we will convert the given inequations into equations, we obtain the following equations:
5x + y = 10, x +y = 6, x + 4y = 12, x = 0 and y = 0

Region represented by 5x + y ≥ 10:
The line 5x + y = 10 meets the coordinate axes at A(2, 0) and B(0, 10) respectively. By joining these points we obtain the line 5x + y = 10.
Clearly (0,0) does not satisfies the inequation 5x + y ≥ 10. So,the region in xy plane which does not contain the origin represents the solution set of the inequation 5x + y ≥ 10.

Region represented by x +y ≥ 6:
The line x +y = 6 meets the coordinate axes at C(6,0) and D(0, 6) respectively. By joining these points we obtain the line
2x +3y = 30.Clearly (0,0) does not satisfies the inequation x +y ≥ 6. So,the region which does not contain the origin represents the solution set of the inequation 2x +3y ≥ 30.

Region represented by x + 4y ≥ 12
The line x + 4y = 12 meets the coordinate axes at E(12, 0) and F(0, 3) respectively. By joining these points we obtain the line
x + 4y = 12.Clearly (0,0) does not satisfies the inequation x + 4y ≥ 12. So,the region which does not contain the origin represents the solution set of the inequation x + 4y ≥ 12.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 5x + y ≥ 10, x +y ≥ 6,x + 4y ≥ 12, x ≥ 0, and y ≥ 0, are as follows.

The corner points of the feasible region are B(0, 10), G(1,5), H(4,2) and E(12,0).

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

Corner point	Z = 3x + 2y
B(0, 10)	3 × 0 + 3 × 10 = 30
G(1,5)	3 × 1 + 2 × 5 = 13
H(4,2)	3 × 4 + 2 × 2 = 16
E(12,0)	3 × 12 + 2 × 0 = 36

Therefore, the minimum value of Z is 13 at the point G(1,5). Hence, x = 1 and y = 5 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 13.