Minimize and maximize Z = x + 2y Subject to the constraints, x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200 and x ≥ 0,y ≥ 0

Question

Solve the linear programming problem by graphical method:

Minimize and maximize Z = x + 2y Subject to the constraints,
x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200 and x ≥ 0,y ≥ 0

Answer 1

Converting the given in equations into equations

x + 2y = 100 …..(1)

2x – y = 0 …..(2)

2x + y = 200 …..(3)

x = 0 …..(4)

y = 0 …..(5)

Region represented by x + 2y ≥ 100 : 

The line x + 2y = 100 meets the coordinate axis at A(10,0) and 5(0, 5).

Table for x + 2y = 100

x	100	0
y	0	50

A(100, 0); B(0, 50)

Join the points A and B to obtain the line. We find that the point (0, 0) does not satisfy the in equation x + 2y ≥ 100. 

So, the region opposite to the origin represents the solution set of the in equation.

Region represented by 2x – y ≤ 0: 

The line 2x – y = 0 meets the coordinate points at C(0, 0) and D( 100, 200).

2x – y = 0

x	0	100
y	0	200

C(0, 0), D(100, 200)

Join the points C(0,0) to D(100, 200) to obtain the line. 

Clearly (0, 0) satisfies. The in equation 2(0) – 0 = 0 ≤ 0. 

So the region containing origin represents the solution set of the in equation.

Region represent by 2x + y ≤ 200 : 

The line 2x + y = 200 meets the coordinate axis at ,A(100,0) and B(0, 200).

2x + y = 200

x	100	0
y	0	200

E(100, 0); F(0, 200)

Join points E and F to obtain the line. 

Clearly (0, 0) satisfies the in equation 2x + y ≤ 200. 

So the region containing the origin represents the solution set of this in equation.

Region represented by x ≥ 0 and y ≥ 0 : 

Since every point in the first quadrant satisfies these in equations. 

So the first quadrant is the region represented by the in equations x ≥ 0 and y ≥ 0.

The shaded region BEFD represents the common region of the above in equations. This region is the feasible region of the given LPP.

where point E(20,40) is the point of intersection of lines x + 2y = 100 and 2x – y = 0.

Point F(50, 100) is the point of intersection of lines 2x + y = 200 and 2x – y = 0.

The value of objective function on these comer points of feasible region are E(20, 40), B(0, 50), D(0, 200) and F(50, 100).

Point	x-coordinate	y-coordinate	Objective function Z = x + 2y
E	20	40	ZE = 20 + 2 x 40 = 100
B	0	50	ZB = 0 + 2 x 50 = 100
F	50	100	ZF = 50 + 2 x 100 = 250
D	0	200	ZD = 0 + 2 x 200 = 400

It is clear from the table that minimum value of objective function is = 100 at points E(20, 40) and B(0, 50) which is minimum on every point on line AEB and maximum value of objective function is on the point D(0, 200) where Z = 400.

Hence Minimum value of Z = 100

and Maximum value of Z = 400.