Solve the following linear programming problems graphically-i Maximum Z -4 x - ySubject to x - y - 50 3 x - y - 90 x - 0- y - 0ii Minimize Z -200 x -500 y Subject to x -2 y - 10 3 x -4 y - 24 x - 0- y - 0iii Max- and Min- Z -3 x -9 ySubject to x -3 y - 60 x - y - 10 x - yx - 0- y - 0iv Minimize Z -3 x -2 ySubject to x - y - 8 3 x -5 y - 15x - 0- y - 0v Minimize Z - x -2 ySubject to 2 x - y - 3x-2 y - 6x - 0- y - 0vi Min- and Max- Z -5 x -10 ySubject to x -2 y - 120 x -2 y - 60 x -2 y - 0 x - 0- y - 0

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Intercept Form of a Line

Solve the fol...

Question

Solve the following linear programming problems graphically:
(i)
$M a x i m u m Z = 4 x + y S u b j e c t t o x + y \leq 50 3 x + y \leq 90 x \geq 0, y \geq 0$

(ii)
$M i n i m i z e Z = 200 x + 500 y S u b j e c t t o x + 2 y \geq 10 3 x + 4 y \leq 24 x \geq 0, y \geq 0$

(iii)
$M a x, a n d M i n . Z = 3 x + 9 y S u b j e c t t o x + 3 y \leq 60 x + y \geq 10 x \leq y x \geq 0, y \geq 0$

(iv)
$M i n i m i z e Z = 3 x + 2 y S u b j e c t t o x + y \geq 8 3 x + 5 y \leq 15 x \geq 0, y \geq 0$

(v)
$M i n i m i z e Z = x + 2 y S u b j e c t t o 2 x + y \geq 3 x + 2 y \geq 6 x \geq 0, y \geq 0$

(vi)
$M i n . a n d M a x . Z = 5 x + 10 y S u b j e c t t o x + 2 y \leq 120 x + 2 y \geq 60 x - 2 y \geq 0 x \geq 0, y \geq 0$

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Solution

(i) We need to maximize Z = 4x + y

First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 50, 3x + y = 90, x = 0 and y = 0

Region represented by x + y ≤ 50:
The line x + y = 50 meets the coordinate axes at A(50, 0) and B(0, 50) respectively. By joining these points we obtain the line x + y = 50. Clearly (0,0) satisfies the inequation x + y ≤ 50. So,the region containing the origin represents the solution set of the inequation x + y ≤ 50.

Region represented by 3x + y ≤ 90:
The line 3x + y = 90 meets the coordinate axes at C(30, 0) and D(0, 90) respectively. By joining these points we obtain the line 3x + y = 90. Clearly, (0,0) satisfies the inequation 3x + y ≤ 90. So,the region containing the origin represents the solution set of the inequation 3x + y ≤ 90.

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 y ≥ 0.
The feasible region determined by the system of constraints x + y ≤ 50, 3x + y ≤ 90, x ≥ 0, and y ≥ 0, are as follows.

The corner points of the feasible region are O(0, 0), C(30, 0), E(20, 30) and B(0, 50).

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

Corner point Z = 4x + y

O(0, 0) 4 × 0 + 0 = 0

C(30, 0) 4 × 30 + 0 = 120

E(20, 30) 4 × 20 + 30 = 110

B(0, 50) 4 × 0 + 50 = 50

We see that the maximum value of the objective function Z is 120 which is at C(30, 0).

(ii) We have to minimize Z = 200x + 500y

First, we will convert the given inequations into equations, we obtain the following equations:
x + 2y = 10, 3x + 4y = 24, x = 0 and y = 0

Region represented by x + 2y ≥ 10:
The line x + 2y = 10 meets the coordinate axes at A(10, 0) and B(0, 5) respectively. By joining these points we obtain the line x + 2y = 10.
Clearly (0,0) does not satisfies the inequation x + 2y ≥ 10. So,the region in xy plane which does not contain the origin represents the solution set of the inequation x + 2y ≥ 10.

Region represented by 3x + 4y ≤ 24:
The line 3x + 4y = 24 meets the coordinate axes at C(8, 0) and D(0, 6) respectively. By joining these points we obtain the line 3x + 4y = 24.
Clearly (0,0) satisfies the inequation 3x + 4y ≤ 24. So,the region in xy plane which contain the origin represents the solution set of the inequation
3x + 4y ≤ 24.

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 y ≥ 0.

The feasible region determined by the system of constraints, x + 2y ≥ 10, 3x + 4y ≤ 24, x ≥ 0 and y ≥ 0 are as follows.

The corner points of the feasible region are E(4, 3), B(0, 5) and D(0, 6).
The values of Z at these corner points are as follows.

Corner point Z =200x + 500y

E(4, 3) 200 × 4 + 500 × 3 = 2300

B(0, 5) 200 × 0 + 500 × 5 = 2500

D(0, 6) 200 × 0 + 500 × 6 = 3000

Therefore, the minimum value of Z is 2300 at the point E(4, 3).Thus, the optimal value of Z is 2300.

(iii) We need to maximize and minimize Z = 3x + 9y

First, we will convert the given inequations into equations, we obtain the following equations:
x + 3y = 60, x + y = 10, x = y, x = 0 and y = 0

Region represented by x + 3y ≤ 60:
The line x + 3y = 60 meets the coordinate axes at A(60, 0) and B(0, 20) respectively. By joining these points we obtain the line x + 3y = 60. Clearly (0,0) satisfies the inequation x + 3y ≤ 60. So,the region in xy plane which contain the origin represents the solution set of the inequation
x + 3y ≤ 60.

Region represented by x + y ≥ 10:
The line x + y = 10 meets the coordinate axes at C(10, 0) and D(0, 10) respectively. By joining these points we obtain the line x + y = 10. Clearly (0,0) does not satisfies the inequation x + y ≥ 10. So,the region in xy plane which does not contain the origin represents the solution set of the inequation x + y ≥ 10.

Region represented by x ≤ y
The line x = y will pass through origin i.e. (0, 0). The region above the line x = y will satisfy the given inequation.

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 y ≥ 0.

The feasible region determined by the system of constraints x + 3y ≤ 60, x + y ≥ 10, x ≤ y, x ≥ 0 and y ≥ 0 are as follows.

The corner points of the feasible region are D(0, 10), F(5, 5), E(15, 15) and B(0, 20).
The values of Z at these corner points are as follows.

Corner point Z = 3x + 9y

D(0, 10) 3 × 0 + 9 × 10 = 90

F(5, 5) 3 × 5+ 9 × 5 = 60

E(15, 15) 3 × 15 + 9 × 15 = 180

B(0, 20) 3 × 0 + 9 × 20 = 180

Therefore, maximum value of the objective function is 180 at E(15, 15) and B(0, 20).
Minimum value of the objective function is 60 at F(5, 5).

(iv) We need to minimize the objective function Z = 3x + 2y

First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 8, 3x + 5y = 15, x = 0 and y = 0

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 + 2y ≥ 10. So,the region in xy plane which does not contain the origin represents the solution set of the inequation x + 2y ≥ 10.

Region represented by 3x + 5y ≤ 15:
The line 3x + 5y = 15 meets the coordinate axes at C(5, 0) and D(0, 3) respectively. By joining these points we obtain the line 3x + 5y = 15. Clearly, (0,0) satisfies the inequation 3x + 5y ≤ 15. So,the region in xy plane which contain the origin represents the solution set of the inequation 3x + 5y ≤ 15.

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 y ≥ 0.

The feasible region determined by the system of constraints x + y ≥ 8, 3x + 5y ≤ 15, x ≥ 0 and y ≥ 0 are as follows.

We observe that there is no common feasible region of the given LPP.

(v) We need to minimize Z = x + 2y

First, we will convert the given inequations into equations, we obtain the following equations:
2x + y = 3, x + 2y = 6, x = 0 and y = 0

Region represented by 2x + y ≥ 3:
The line 2x + y = 3 meets the coordinate axes at $A (\frac{3}{2}, 0)$ and B(0, 3) respectively. By joining these points we obtain the line 2x + y = 3. Clearly (0,0) does not satisfies the inequation 2x + y ≥ 3. So,the region in xy plane which do not contain the origin represents the solution set of the inequation 2x + y ≥ 3.

Region represented by x + 2y ≥ 6:
The line x + 2y = 6 meets the coordinate axes at C(6, 0) and B(0, 3) respectively. By joining these points we obtain the line x + 2y = 6. Clearly (0,0) does not satisfies the inequation x + 2y ≥ 6. So,the region in xy plane which does not contain the origin represents the solution set of the inequation x + 2y ≥ 6.

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 y ≥ 0.

The feasible region determined by the system of constraints 2x + y ≥ 3, x + 2y ≥ 6, x ≤ y, x ≥ 0 and y ≥ 0 are as follows.

The corner points of the feasible region are B(0, 3) and C(6, 0).
The values of Z at these corner points are as follows.

Corner point Z = x + 2y

B(0, 3) 0 + 2 × 3 = 6

C(6, 0) 6 + 2 × 0 = 6

Therefore, the minimum value of the objective function is 6 at B(0, 3) and C(6, 0).

(vi) We need to maximize and minimize Z = 5x + 10y

First, we will convert the given inequations into equations, we obtain the following equations:
x + 2y = 120, x + 2y = 60, x − 2y = 0 or x = 2y, x = 0 and y = 0

Region represented by x + 2y ≤ 120:
The line x + 2y = 120 meets the coordinate axes at A(120, 0) and B(0, 60) respectively. By joining these points we obtain the line x + 2y = 120. Clearly (0,0) satisfies the inequation x + 2y ≤ 120. So,the region in xy plane which contain the origin represents the solution set of the inequation x + 2y ≤ 120.

Region represented by x + 2y ≥ 60:
The line x + 2y = 60 meets the coordinate axes at C(60, 0) and D(0, 30) respectively. By joining these points we obtain the line x + 2y = 60. Clearly, (0,0) does not satisfies the inequation x + 2y ≥ 60. So,the region in xy plane which does not contain the origin represents the solution set of the inequation
x + 2y ≥ 60.

Region represented by x − 2y ≥ 0 or x ≥ 2y:
The line x = 2y will pass through origin i.e. (0, 0). The region below the line x = 2y will satisfy the given inequation.

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 y ≥ 0.

The feasible region determined by the system of constraints x + 2y ≤ 120, x + 2y ≥ 60, x − 2y ≥ 0 or x ≥ 2y x ≥ 0 and y ≥ 0 are as follows.

The corner points of the feasible region are F(30, 15), E(60, 30), A(120, 0) and C(60, 0).
The values of Z at these corner points are as follows.

Corner point Z = 5x + 10y

F(30, 15) 5 × 30 + 10 × 15 = 300

E(60, 30) 5 × 30 + 10 × 15 = 600

A(120, 0) 5 × 30 + 10 × 15 = 600

C(60, 0) 5 × 30 + 10 × 15 = 300

Therefore, maximum value of the objective function is 600 at E(60, 30) and A(120, 0).
Minimum value of the objective function is 300 at F(30, 15) and C(60, 0).

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