Linear Programming Test

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Linear Programming Test - 5

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Weekly Quiz Competition

Question 1

1 / -0

Maximise Z = 3x + 4y subject to the constraints: x + y ≤ 4, x ≥ 0, y ≥ 0.

Maximum Z = 17 at (0, 5)

Maximum Z = 16 at (0, 4)

Maximum Z = 19 at (1, 5)

Maximum Z = 18 at (1, 4)

Solution

Objective function is Z = 3x + 4 y ……(1).
The given constraints are : x + y ≤ 4, x ≥ 0, y ≥ 0.

The corner points obtained by constructing the line x+ y= 4, are (0,0),(0,4) and (4,0).

Corner points	Z = 3x +4y
O ( 0 ,0 )	Z = 3(0)+4(0) = 0
A ( 4 , 0 )	Z = 3(4) + 4 (0) = 12
B ( 0 , 4 )	Z = 3(0) + 4 ( 4) = 16 …( Max. )

therefore Z = 16 is maximum at ( 0 , 4 ) .

Question 2

1 / -0

Minimise Z = – 3x + 4 y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.

Minimum Z = – 14 at (5, 0)

Minimum Z = – 15 at (5, 1)

Minimum Z = – 13 at (4, 1)

Minimum Z = – 12 at (4, 0)

Solution

Objective function is Z = - 3x + 4 y ……………………(1).
The given constraints are : x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.

The corner points obtained by constructing the line x+2y=8 and 3x +2y = 12 are (0,0),(0,2),(3,0) and (20/19,45/19)

Corner points	Z = 5x + 3 y
O(0 , 0 )	0
B ( 2 , 0 )	10
C( 0 , 3 )	9
D ( 20/19 , 45/19 )	235/19 ……………….(Max.)

Here , Z = -12 is minimum at C ( 4 , 0 )

Question 3

1 / -0

Minimize Z = 3x + 5y such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.

Minimum Z = 9 at (3/2,1/2) 110 V 60 Hz

Minimum Z = 10 at (3/2,1/2)

Minimum Z = 8 at (3/2,1/2)

Minimum Z = 7 at (3/2,1/2)

Solution

Objective function is Z = 3x + 5 y ……………………(1).
The given constraints are : x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0 .

The corner points obtained by drawing the lines x+3y=3 and x+y=2 are (0,0), (3,0),(0,2) and (3/2,1/2)

Corner points	Z = 3x + 5 y
A(3 , 0 )	9
B ( 0 ,2 )	10
C( 3/2 , 1/2 )	7………………..( Min. )

Here Z = 7 is minimum at (3/2 , 1/2 )

Question 4

1 / -0

Maximise Z = 5x + 3y subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.

Maximum Z = 235/19 at (20/19,45/19)

Maximum Z = 275/19 at (20/19,45/19)

Maximum Z = 245/19 at (20/19,45/19)

Maximum Z = 255/19 at (20/19,45/19)

Solution

Objective function is Z = 5x + 3 y ……………………(1).
The given constraints are : 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.

The corner points obtained by drawing the lines 3x+5y=15 and 5x+2y=10 are (0,0),(0,3), (2,0) and (20/19,45/19)

Corner points	Z = 5x + 3 y
O(0 , 0 )	0
B ( 2 , 0 )	10
C( 0 , 3 )	9
D ( 20/19 , 45/19 )	235/19 ……………….(Max.)

Here , Z = 235/19 is maximum at ( 20/19 , 45/19 ) .

Question 5

1 / -0

Maximize Z = 3x + 2y subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.

Maximum Z = 22 at (4, 3)

Maximum Z = 20 at (4, 3)

Maximum Z = 24 at (4, 3)

Maximum Z = 18 at (4, 3)

Solution

Objective function is Z = 3x + 2 y ……………………(1).
The given constraints are : x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0. The corner points obtained by drawing the lines 3x+y=15 and x+2y=10 graphically are (0,0),(0,5), (5,0) and (4,3).

Corner points	Z = 3x + 2y
O(0 ,0 )	0
A(5,0)	15
B(0,5)	10
C(4,3)	18……………………..(Max.)

Here , Z = 18 is maximum at ( 4, 3 ).

Question 6

1 / -0

Minimize Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0.

Minimum Z = 9 at all the points (6, 0) and (0, 3)

Minimum Z = 6 at all the points on the line segment joining the points (6, 0) and (0, 3)

Minimum Z = 7 at all the points on the line segment joining the points (6, 0) and (0, 3)

Minimum Z = 8 at all the points on the line segment joining the points (6, 0) and (0, 3)

Solution

Objective function is Z = x + 2 y ……………………(1).
The given constraints are : 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0 .

Corner points	Z = x + 2y
A(0 ,3 )	6…………………..(Minimum)
B(6,0)	6………………………(Minimum)

Here , Z = 18 is minimum at ( 0, 3 ) and ( 6 , 0 ) .
Minimum Z = 6 at all the points on the line segment joining the points (6, 0) and (0, 3).

Question 7

1 / -0

Minimize Z = 5x + 10 y subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0

Minimum Z = 300 at (60, 0)

Minimum Z = 320 at (60, 0)

Minimum Z = 310 at (60, 0)

Minimum Z = 330 at (60, 0)

Solution

Objective function is Z = 5x + 10 y ……………………(1).
The given constraints are : x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0 .

The corner points are obtained by drawing the lines x+2y =120, x+y = 60 and x-2y = 0. The points so obtained are (60,30),(120,0), (60,0) and (40,20)

Corner points	Z = 5x + 10y
D(60 ,30 )	600
A(120,0)	600
B(60,0)	300……………………..(Min.)
C(40,20)	400

Here , Z = 300 is minimum at ( 60, 0 ).

Question 8

1 / -0

Minimize Z = x + 2y subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0.

Minimum Z = 140 at all the points on the line segment joining the points (0, 50) and (20, 40)

Minimum Z = 150 at all the points on the line segment joining the points (0, 50) and (20, 40)

Minimum Z = 100 at all the points on the line segment joining the points (0, 50) and (20, 40)

Minimum Z = 120 at all the points on the line segment joining the points (0, 50) and (20, 40)

Solution

Objective function is Z = x + 2 y ……………………(1).
The given constraints are : x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0.

The corner points are obtained by drawing the lines x+2y=100, 2x-y=0 and 2x+y=200.

The points so obtained are (0,50),(20,40), (50,100) and (0,200)

Corner points	Z = x + 2y
D(0 ,50 )	100……………..(Min.)
A(20,40)	100……………………..(Min.)
B(50,100)	250
C(0,200)	400

Here , Z = 100 is minimum at ( 0, 50) and ( 20 ,40).
Minimum Z = 100 at all the points on the line segment joining the points (0, 50) and (20, 40).

Question 9
1 / -0

Maximize Z = x + y, subject to x – y ≤ –1, –x + y ≤ 0, x, y ≥ 0.

A
Maximum Z = 14 at (2, 6)

B
Maximum Z = 12 at (2, 6)

C
Z has no maximum value

D
Maximum Z = 8 at (2, 6)a

Solution

Objective function is Z = x + y ……………………(1).
The given constraints are : x – y ≤ –1, –x + y ≤ 0, x, y ≥ 0.
Here , there is no common feasible region between the lines x – y = - 1 and - x + y = 0 .
Therefore , it has no solution. Thus , Z has no maximum value .

Question 10

1 / -0

Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs Rs 60/kg and Food Q costs Rs 80/kg. Food P contains 3 units/kg of Vitamin A and 5 units / kg of Vitamin B while food Q contains 4 units/kg of Vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture.

Minimum cost = Rs 180 at all points lying on segment joining (8/3,0)and(2,1/2)

Minimum cost = Rs 160 at all points lying on segment joining (8/3,0)and(2,1/2)

Minimum cost = Rs 170 at all points lying on segment joining (8/3,0)and(2,1/2)

Minimum cost = Rs 180 at all points lying on segment joining (8/3,0)and(2,1/2)

Solution

Let number of food type P = x
And number of units of food type Q = y
Therefore , the above L.P.P. is given as :
Minimise , Z = 60x +80y , subject to the constraints : 3 x + 4y ≥ 8, 5x + 2y ≥ 11, x,y ≥ 0.

The corner points can be obtained by drawing the lines 3x+4y=8 and 5x+2y=11 graphically.

The points so obtained are (8/3,0), (2,1/2), (0,11/2)

Corner points	Z = - x + 2y
D(8/3,0 )	160………………….(Min.)
A(2,1/2)	160………………(Min.)
B(0,11/2)	440

Here Z = 160 is minimum. i.e. Minimum cost = Rs 160 at all points lying on segment joining (8/3,0)and(2,1/2) .

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Linear Programming Test - 5

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Linear Programming Test - 5

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Question 1

Maximum Z = 17 at (0, 5)

Maximum Z = 16 at (0, 4)

Maximum Z = 19 at (1, 5)

Maximum Z = 18 at (1, 4)

Solution

Question 2

Minimum Z = – 14 at (5, 0)

Minimum Z = – 15 at (5, 1)

Minimum Z = – 13 at (4, 1)

Minimum Z = – 12 at (4, 0)

Solution

Question 3

Minimum Z = 9 at (3/2,1/2) 110 V 60 Hz

Minimum Z = 10 at (3/2,1/2)

Minimum Z = 8 at (3/2,1/2)

Minimum Z = 7 at (3/2,1/2)

Solution

Question 4

Maximum Z = 235/19 at (20/19,45/19)

Maximum Z = 275/19 at (20/19,45/19)

Maximum Z = 245/19 at (20/19,45/19)

Maximum Z = 255/19 at (20/19,45/19)

Solution

Question 5

Maximum Z = 22 at (4, 3)

Maximum Z = 20 at (4, 3)

Maximum Z = 24 at (4, 3)

Maximum Z = 18 at (4, 3)

Solution

Question 6

Minimum Z = 9 at all the points (6, 0) and (0, 3)

Minimum Z = 6 at all the points on the line segment joining the points (6, 0) and (0, 3)

Minimum Z = 7 at all the points on the line segment joining the points (6, 0) and (0, 3)

Minimum Z = 8 at all the points on the line segment joining the points (6, 0) and (0, 3)

Solution

Question 7

Minimum Z = 300 at (60, 0)

Minimum Z = 320 at (60, 0)

Minimum Z = 310 at (60, 0)

Minimum Z = 330 at (60, 0)

Solution

Question 8

Minimum Z = 140 at all the points on the line segment joining the points (0, 50) and (20, 40)

Minimum Z = 150 at all the points on the line segment joining the points (0, 50) and (20, 40)

Minimum Z = 100 at all the points on the line segment joining the points (0, 50) and (20, 40)

Minimum Z = 120 at all the points on the line segment joining the points (0, 50) and (20, 40)

Solution

Question 9

Maximum Z = 14 at (2, 6)

Maximum Z = 12 at (2, 6)

Z has no maximum value

Maximum Z = 8 at (2, 6)a

Solution

Question 10

Minimum cost = Rs 180 at all points lying on segment joining (8/3,0)and(2,1/2)

Minimum cost = Rs 160 at all points lying on segment joining (8/3,0)and(2,1/2)

Minimum cost = Rs 170 at all points lying on segment joining (8/3,0)and(2,1/2)

Minimum cost = Rs 180 at all points lying on segment joining (8/3,0)and(2,1/2)

Solution

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