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Maximize Z = 15x + 10y Subject to 3 X + 2 Y ≤ 80 2 X + 3 Y ≤ 70 X , Y ≥ 0 - Mathematics

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Sum

Maximize Z = 15x + 10y
Subject to 

\[3x + 2y \leq 80\]
\[2x + 3y \leq 70\]
\[ x, y \geq 0\]

 

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Solution

First, we will convert the given inequations into equations, we obtain the following equations:
3x + 2y = 80, 2x + 3y = 70, x = 0 and y = 0
Region represented by 3x + 2y ≤ 80 :
The line 3x + 2y = 80 meets the coordinate axes at \[A\left( \frac{80}{3}, 0 \right)\] and  \[B\left( 0, 40 \right)\]respectively. By joining these points we obtain the line 3x + 2y = 80.
Clearly (0,0) satisfies the inequation 3x + 2y ≤ 80 . So,the region containing the origin represents the solution set of the inequation 3x + 2y ≤ 80 .
Region represented by 2x + 3y ≤ 70:
The line 2x + 3y = 70 meets the coordinate axes at \[C\left( 35, 0 \right)\] and  \[D\left( 0, \frac{70}{3} \right)\] respectively. By joining these points we obtain the line 2x + 3y ≤ 70.
Clearly (0,0) satisfies the inequation 2x + 3y ≤ 70. So,the region containing the origin represents the solution set of the inequation 2x + 3y ≤ 70.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and ≥ 0.
 The feasible region determined by the system of constraints, 3x + 2y ≤ 80, 2x + 3y ≤ 70, x ≥ 0, and y ≥ 0 are as follows.

The corner points of the feasible region are O(0, 0), \[A\left( \frac{80}{3}, 0 \right)\] \[E\left( 20, 10 \right)\] and \[D\left( 0, \frac{70}{3} \right)\]

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

Corner point Z = 15x + 10y
O(0, 0) 15 × 0 + 10 × 0 = 0
\[A\left( \frac{80}{3}, 0 \right)\]
15 × \[\frac{80}{3}\]+ 10 × 0 = 400
 
 
\[E\left( 20, 10 \right)\]
15 × 20 + 10 × 10 = 400
\[D\left( 0, \frac{70}{3} \right)\] 
15 × 0 + 10 × \[\frac{70}{3}\] =\[\frac{700}{3}\]
 

We see that the maximum value of the objective function Z is 400 which is at \[A\left( \frac{80}{3}, 0 \right)\] and \[E\left( 20, 10 \right)\] Thus, the optimal value of Z is 400.

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 6 | Page 32

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