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Maximize Z = 7x + 10y Subject to X + Y ≤ 30000 Y ≤ 12000 X ≥ 6000 X ≥ Y X , Y ≥ 0 - Mathematics

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Maximize Z = 7x + 10y
Subject to 

\[x + y \leq 30000\]
\[ y \leq 12000\]
\[ x \geq 6000\]
\[ x \geq y\]
\[ x, y \geq 0\]

 

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Solution

We have to maximize Z = 7x + 10y
First, we will convert the given inequations into equations, we obtain the following equations:
x y = 30000,y = 12000, = 6000, x = yx = 0 and y = 0.

Region represented by x y ≤ 30000:
The line x y = 30000 meets the coordinate axes at \[A\left( 30000, 0 \right)\] and \[B\left( 0, 30000 \right)\]  respectively. By joining these points we obtain the line x y = 30000.
Clearly (0,0) satisfies the inequation x y ≤ 30000. So,the region containing the origin represents the solution set of the inequation x y ≤ 30000.

The line y = 12000 is the line that passes through C(0,12000) and parallel to x axis.

The line x = 6000 is the line that passes through (6000, 0) and parallel to y axis.

Region represented by x ≥ y
The line x = y is the line that passes through origin.The points to the right of the line x= y satisfy the inequation x ≥ y.
Like by taking the point (−12000, 6000).Here, 6000 > −12000 which implies y > x. Hence, the points to the left of the line x = y will not satisfy the given inequation x ≥ y.

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, x y ≤ 30000, y ≤ 12000, x ≥ 6000, x   y , x ≥ 0 and y ≥ 0 are as follows:

The corner points of the feasible region are D(6000, 0), \[A\left( 3000, 0 \right)\] , \[F\left( 18000, 12000 \right)\] and  \[E\left( 12000, 12000 \right)\] .

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

Corner point  Z = 7x + 10y
D(6000, 0) 7 × 6000 + 10 × 0 = 42000
\[A\left( 3000, 0 \right)\]
7× 3000 + 10 × 0 = 21000
\[F\left( 18000, 12000 \right)\]
7 × 18000 + 10 × 12000 = 246000
\[E\left( 12000, 12000 \right)\] 
7 × 12000 + 10 ×12000 = 204000

We see that the maximum value of the objective function Z is 246000 which is at  \[F\left( 18000, 12000 \right)\]  that means at = 18000 and y = 12000.
Thus, the optimal value of Z is 246000.

 
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 9 | Page 32

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