Find graphically, the maximum value of Z = 2x + 5y, subject to constraints given below:
2x + 4y ≤ 8

3x + y ≤ 6

x + y ≤ 4

x ≥ 0, y ≥ 0 [CBSE 2015]

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Solution

The given linear programming problem is Maximize Z = 2x + 5y

subject to the constraints

2x + 4y ≤ 8

3x + y ≤ 6

x + y ≤ 4

x ≥ 0, y ≥ 0

Converting the inequations into equations, we obtain the following equations of straight lines:

2x + 4y = 8, 3x + y = 6, x + y = 4

The line 2x + 4y = 8 meets the coordinate axes at (4, 0) and (0, 2).

The line 3x + y = 6 meets the coordinate axes at (2, 0) and (0, 6).

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

The feasible region determined by the given constraints can be diagrammatically represented as,

The coordinates of the corner points of the feasible region are O(0, 0), A(0, 2), B $(\frac{8}{5}, \frac{6}{5})$ and C(2, 0).

The value of the objective function at these points are given in the following table.

Corner Point Z = 2x + 5y

(0, 0) 2 × 0 + 5 × 0 = 0

(2, 0) 2 × 2 + 5 × 0 = 4

(0, 2) 2 × 0 + 5 × 2 = 10 → Maximum

$(\frac{8}{5}, \frac{6}{5})$ $2 \times \frac{8}{5} + 5 \times \frac{6}{5} = \frac{46}{5}$

Thus, the maximum value of Z is 10 at x = 0, y = 2.

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