Advertisement Remove all ads

Find the Maximum and Minimum Value of 2x + Y Subject to the Constraints: X + 3y ≥ 6, X − 3y ≤ 3, 3x + 4y ≤ 24, − 3x + 2y ≤ 6, 5x + Y ≥ 5, X, Y ≥ 0. - Mathematics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Sum

Find the maximum and minimum value of 2x + y subject to the constraints:
x + 3y ≥ 6, x − 3y ≤ 3, 3x + 4y ≤ 24, − 3x + 2y ≤ 6, 5x + y ≥ 5, xy ≥ 0.

Advertisement Remove all ads

Solution

First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 4, x + y = 3, x  2y = 2, x = 0 and y = 0.
The line x + 3y = 6 meets the coordinate axis at \[A\left( 6, 0 \right)\]  and B(0, 2). Join these points to obtain the line x + 3y = 6.
Clearly, (0, 0) does not satisfies the inequation x + 3y ≥ 6. So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.

The line x  3y = 3 meets the coordinate axis at C(3, 0) and D(0, 1). Join these points to obtain the line x  3y = 3.
Clearly, (0, 0) satisfies the inequation x  3y ≤ 3. So, the region in xy-plane that contains the origin represents the solution set of the given equation.

The line 3x + 4y = 24 meets the coordinate axis at E(8, 0) and F(0, 6). Join these points to obtain the line 3x + 4y = 24.
Clearly, (0, 0) satisfies the inequation 3x + 4y ≤ 24. So, the region in xy-plane that contains the origin represents the solution set of the given equation.

The line 3x + 2y = 6 meets the coordinate axis at G(−2, 0) and H(0, 3). Join these points to obtain the line 3x + 2y = 6.
Clearly, (0, 0) satisfies the inequation 3x + 2y ≤ 6. So, the region in xy-plane that contains the origin represents the solution set of the given equation.

The line 5x + y = 5 meets the coordinate axis at \[I\left( 1, 0 \right)\] and J(0, 5). Join these points to obtain the line 5x + y = 5.

Clearly, (0, 0) does not satisfies the inequation 5x + ≥ 5. So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.

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.
These lines are drawn using a suitable scale.

The corner points of the feasible region are \[\left( \frac{4}{13}, \frac{45}{13} \right)\] k \[\left( \frac{4}{3}, 5 \right)\] ,L  \[\left( \frac{84}{13}, \frac{15}{13} \right)\], M \[\left( \frac{9}{2}, \frac{1}{2} \right)\] , N \[\left( \frac{9}{14}, \frac{25}{14} \right)\] The values of Z at these corner points are as follows.

Corner point Z = 2x y
\[\left( \frac{4}{13}, \frac{45}{13} \right)\]
 
 
2 x \[\frac{4}{13}\] + \[\frac{45}{13}\] = \[\frac{53}{13}\]
K \[\left( \frac{4}{3}, 5 \right)\] 2 x \[\frac{4}{3}\] + 5 = \[\frac{23}{3}\]
L \[\left( \frac{84}{13}, \frac{15}{13} \right)\] 2 × \[\frac{84}{13}\] + \[\frac{15}{13}\] = \[\frac{183}{13}\]
M \[\left( \frac{9}{2}, \frac{1}{2} \right)\] 2 × \[\frac{9}{2}\] + \[\frac{1}{2}\] = \[\frac{19}{2}\]
N \[\left( \frac{9}{14}, \frac{25}{14} \right)\] 2 × \[\frac{9}{14}\] + \[\frac{25}{14}\] = \[\frac{43}{14}\]
We see that the minimum value of the objective function Z is \[\frac{43}{14}\]  which is at \[\left( \frac{9}{14}, \frac{25}{14} \right)\] and maximum value of the objective function is \[\frac{183}{13}\]  which is at \[\left( \frac{84}{13}, \frac{15}{13} \right)\] . 
Concept: Graphical Method of Solving Linear Programming Problems
  Is there an error in this question or solution?

APPEARS IN

RD Sharma Class 12 Maths
Chapter 30 Linear programming
Exercise 30.2 | Q 23 | Page 33

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×