Maximize Z = 4x + y Subject to the constraints, x + y ≤ 50, 3x + y ≤ 90 and x ≥ 0, y ≥ 0.

Question

Solve the linear programming problem by graphical method:

Maximize Z = 4x + y Subject to the constraints, x + y ≤ 50, 3x + y ≤ 90 and x ≥ 0, y ≥ 0.

Answer 1

Converting the given in equations into equations 

x + y = 50 …..(1) 

3x + y = 90 …..(2) 

Region represented by x + y ≤ 50 : 

The line x + y = 50 meets the coordinate axis at A(50, 0) and B(0, 50). x + y = 50

x	50	0
y	0	50

A(50, 0); B(0, 50) 

Join the points A and 5 to obtain the line. 

Clearly (0, 0) satisfies the in equation x + y ≤ 50. 

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

Region represented by 3x + y ≤ 90: 

The line 3x + y = 90 meets the coordinate axis at the points C(30, 0) and D(0, 90). 

3x + y = 90

x	30	0
y	0	90

C(30, 0); D(0, 90) 

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 equation x ≥ 0 and y ≥ 0. The shaded region OCEB represents the common region of the above in equations. This region is the feasible region of the given LPP. 

E(20, 30) is the point of intersection of the lines x + y = 50 and lines 3x + y = 90. 

The coordinates of this region of solution are O(0, 0), C(30, 0), E(20, 30) and B(0, 50).

The value of objective function on these points are given in this table :

Point	x-coordinate	y-coordinate	Objective function Z = 4x + y
O	0	0	ZO = 4(0) + (0) = 0
C	30	0	ZC = 4(30) + 0 = 120
E	20	30	ZE = 4(20) + 30 = 110
B	0	50	ZB = 4(0) + 50 = 50

It is clear from the table that objective function is maximum on point C(30, 0). Hence maximum value of Z = 120.