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Minimize Z = 2x + 4y Subject to X + Y ≥ 8 X + 4 Y ≥ 12 X ≥ 3 , Y ≥ 2

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प्रश्न

Minimize Z = 2x + 4y
Subject to 

\[x + y \geq 8\]
\[x + 4y \geq 12\]
\[x \geq 3, y \geq 2\]

 

योग
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उत्तर

First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 8, x + 4y = 12, x = 3, y = 2

Region represented by x + y ≥ 8:
The line x + y = 8 meets the coordinate axes at A(8, 0) and B(0, 8) respectively. By joining these points we obtain the line x + y = 8.
Clearly (0,0) does not satisfies the inequation x + y ≥ 8. So,the region in xy plane which does not contain the origin represents the solution set of the inequation x + y ≥ 8.

Region represented by x + 4y ≥ 12:
The line x + 4y = 12 meets the coordinate axes at C(12, 0) and D(0, 3) respectively. By joining these points we obtain the line x + 4y = 12.
Clearly (0,0) satisfies the inequation x + 4y ≥ 12. So,the region in xy plane which contain the origin represents the solution set of the inequation x + 4y ≥ 12.

The line x = 3 is the line that passes through the point (3, 0) and is parallel to Y axis.x ≥ 3 is the region to the right of the line x = 3.

The line y = 2 is the line that passes through the point (0, 12) and is parallel to X axis.y ≥ 2 is the region above the line y = 2.

The corner points of the feasible region are E(3, 5) and F(6, 2).

The values of Z at these corner points are as follows.

Corner point Z = 2x + 4y
E(3, 5) 2 × 3 + 4 × 5 = 26
F(6, 2) 2 × 6 + 4 × 2 = 20

Therefore, the minimum value of Z is 20 at the point F(6, 2). Hence, x = 6 and y =2 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 20.

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अध्याय 29: Linear programming - Exercise 30.2 [पृष्ठ ३२]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 29 Linear programming
Exercise 30.2 | Q 10 | पृष्ठ ३२

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