Maximize Z = 2X₁ + 3X₂

Subject to

2X₁ + X₂ ≤ 6

X₁ – X₂ ≥ 3

X₁, X₂ ≥ 0

The solution to the above LPP is

Concept:

• In order to find the maximum value of the objective function, the constraints of the objective function are drawn and the region formed by the constraints is the feasible region.

Draw the constraints to find the feasible region:

• To draw the inequalities, first, draw the equation form of the inequalities.
• Convert all the constraints to equality and plot on the graph. Put the value of (x1, x2) obtained from the corner points of the feasible region and put it in the objective function.
• Now check the region which we have to choose depending on the sign of inequality.
• To check which region we need to choose put (0,0) in both the inequality. and check whether this inequality is satisfying or not.
• If it is satisfying the inequality then take the region containing t (0,0) else the opposite side of (0,0).

Calculation:

Given:

• Following cases are observed by the region formed by the constraints

Calculation:

Given:

• The objective function to maximize is,

Z = 2X₁ + 3X₂

• Which is subjected to the constraints,

2X₁ + X₂ ≤ 6

X₁ – X₂ ≥ 3

X₁, X₂ ≥ 0

• If the given constraints are drawn graphically then,

• The feasible region is a point here, therefore the given LPP has an optimal solution and it will occur at point (3, 0).