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Maximize Z = 3x + 4y Subject to 2 X + 2 Y ≤ 80 2 X + 4 Y ≤ 120 - Mathematics

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Maximize Z = 3x + 4y
Subject to

\[2x + 2y \leq 80\]
\[2x + 4y \leq 120\]

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Solution

We have to maximize Z = 3x + 4y
First, we will convert the given inequations into equations, we obtain the following equations:
2x + 2y = 80, 2x + 4y = 120
Region represented by 2x + 2y ≤ 80:
The line 2x + 2y = 80 meets the coordinate axes at \[A\left( 40, 0 \right)\] and  \[B\left( 0, 40 \right)\] respectively. By joining these points we obtain the line 2x + 2y = 80.
Clearly (0,0) satisfies the inequation 2x + 2y ≤ 80. So,the region containing the origin represents the solution set of the inequation 2x + 2y ≤ 80.

Region represented by 2x + 4y ≤ 120:
The line 2x + 4y = 120 meets the coordinate axes at

\[C\left( 60, 0 \right)\] and  \[D\left( 0, 30 \right)\] respectively. By joining these points we obtain the line 2x + 4y ≤ 120.
Clearly (0,0) satisfies the inequation 2x + 4y ≤ 120. So,the region containing the origin represents the solution set of the inequation 2x + 4y ≤ 120.

The feasible region determined by the system of constraints, 2x + 2y ≤ 80, 2x + 4y ≤ 120 are as follows:
The corner points of the feasible region are O(0, 0), \[A\left( 40, 0 \right)\] \[E\left( 20, 20 \right)\] and  \[D\left( 0, 30 \right)\]
The values of Z at these corner points are as follows:
Corner point Z = 3x + 4y
O(0, 0) 3 × 0 + 4 × 0 = 0
\[A\left( 40, 0 \right)\]
3× 40 + 4 × 0 = 120
\[E\left( 20, 20 \right)\] 
3 × 20 + 4 × 20 = 140
\[D\left( 0, 30 \right)\] 
10 × 0 + 4 ×30 = 120

We see that the maximum value of the objective function Z is 140 which is at \[E\left( 20, 20 \right)\]  that means at = 20 and y = 20.
Thus, the optimal value of Z is 140.
 

 
Concept: Graphical Method of Solving Linear Programming Problems
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APPEARS IN

RD Sharma Class 12 Maths
Chapter 30 Linear programming
Exercise 30.2 | Q 8 | Page 32

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