मराठी

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 - Mathematics

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प्रश्न

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   

बेरीज
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उत्तर

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,

2x + 4y = 8 and 3x + y = 6 gives (8/5, 6/5)

The coordinates of the corner points of the feasible region are O(0, 0), A(0, 2), B \[\left(\frac{8}{5}, \frac{6}{5} \right)\] 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
\[\left( \frac{8}{5}, \frac{6}{5} \right)\]
\[2 \times \frac{8}{5} + 5 \times \frac{6}{5} = \frac{46}{5}\] = 9.2

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

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पाठ 30: Linear programming - Exercise 30.2 [पृष्ठ ३३]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 30 Linear programming
Exercise 30.2 | Q 26 | पृष्ठ ३३

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