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+1 vote
9.3k views
in Linear Programming by (48.0k points)

Solve the linear programming problem by graphical method :

Maximize Z = 3x + 4y subject to the x + y ≤ 4 constraints x ≥ 0, y ≥ 0

1 Answer

+2 votes
by (46.3k points)
selected by
 
Best answer

Converting the given in equations into equation x + y = 4 

Region represented by x + y ≤ 4 :

The line x + y = 4 meets the coordinate axis at A(4, 0) and B(0, 4).

x 4 0
y 0 4

A(4, 0); B(0,4) Join the points A and B to obtain a line. 

Clearly (0, 0) satisfies the in equation x + 2y ≤ 4. 

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

Region represented by x ≥ 0,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 OAB represents the common region of the above in equations. This region is the feasible region of the given Linear Programming Problem.

The coordinates of the corner points of the shaded feasible region are O(0, 0), A(4.0) and B(0, 4). 

The values of the objective function of these points are given in the following table :

Point x-coordinate y-coordinate Objective function Z = 3x + 4y
O 0 0 Z= 3(0) + 4(0) = 0
A 4 0 ZA = 3(4) + 4(0) = 12
B 0 4 ZB = 3(0) + 4(4) = 16

It is clear from the table, the objective function has maximum value at point B(0, 4). 

So, the required solution of the given LPP, is x = 0, y = 4 and the maximum value is Z = 16.

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