# Topic: Linear Programming (Test 1)

Topic: Linear Programming
Q.1
A linear programming problem is one that is concerned with
A. finding the upper limits of a linear function of several variables
B. finding the lower limit of a linear function of several variables
C. finding the limiting values of a linear function of several variables
D. finding the optimal value (maximum or minimum) of a linear function of several variables
Explaination / Solution:

A linear programming problem is one that is concerned with finding the optimal value (maximum or minimum) of a linear function of several variables .

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Q.2
Let R be the feasible region (convex polygon) for a linear programming problem and let Z = ax + by be the objective function. When Z has an optimal value (maximum or minimum), where the variables x and y are subject to constraints described by linear inequalities,
A. None of these
B. optimal value must occur at the midpoints of the corner points (vertices) of the feasible region.
C. optimal value must occur at the centroid of the feasible region.
D. optimal value must occur at a corner point (vertex) of the feasible region.
Explaination / Solution:

Let R be the feasible region (convex polygon) for a linear programming problem and let Z = ax + by be the objective function. When Z has an optimal value (maximum or minimum), where the variables x and y are subject to constraints described by linear inequalities then , optimal value must occur at a corner point (vertex) of the feasible region.

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Q.3
In Corner point method for solving a linear programming problem the second step after finding the feasible region of the linear programming problem and determining its corner points is
A. Evaluate the objective function Z = ax + by at the mid points
B. None of these
C. Evaluate the objective function Z = ax + by at each corner point.
D. Evaluate the objective function Z = ax + by at the center point
Explaination / Solution:

In Corner point method for solving a linear programming problem the second step after finding the feasible region of the linear programming problem and determining its corner points is : To evaluate the objective function Z = ax + by at each corner point.

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Q.4
In a LPP, the objective function is always
B. Linear
C. constant
D. cubic
Explaination / Solution:

In a LPP, the objective function is always linear. this is because these problems are always subjected to linear inequalities, where we maximise or minimise the linear functions.

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Q.5
Maximise Z = 3x + 4y subject to the constraints: x + y ≤ 4, x ≥ 0, y ≥ 0.
A. Maximum Z = 16 at (0, 4)
B. Maximum Z = 19 at (1, 5)
C. Maximum Z = 17 at (0, 5)
D. Maximum Z = 18 at (1, 4)
Explaination / 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 ) .

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Q.6
Minimize Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0.
A. Minimum Z = 6 at all the points on the line segment joining the points (6, 0) and (0, 3).
B. Minimum Z = 8 at all the points on the line segment joining the points (6, 0) and (0, 3).
C. Minimum Z = 7 at all the points on the line segment joining the points (6, 0) and (0, 3).
D. Minimum Z = 9 at all the points (6, 0) and (0, 3).
Explaination / 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).

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Q.7
Maximize Z = – x + 2y, subject to the constraints: x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.
A. Maximum Z = 12 at (2, 6)
B. Maximum Z = 14 at (2, 6)
C. Z has no maximum value
D. Maximum Z = 10 at (2, 6)
Explaination / Solution:

Objective function is Z = - x + 2 y ……………………(1).
The given constraints are : x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.

 Corner points Z =  - x + 2y D(6,0 ) -6 A(4,1) -2 B(3,2) 1

Here , the open half plane has points in common with the feasible region .
Therefore , Z has no maximum value.

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Q.8
A factory manufactures two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines tomanufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of Rs 7 and screws B at a profit of Rs 10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum profit.
A. 32 packages of screws A and 22 packages of screws B; Maximum profit = Rs 414
B. 30 packages of screws A and 20 packages of screws B; Maximum profit = Rs 410
C. 32 packages of screws A and 20 packages of screws B; Maximum profit = Rs 412
D. 30 packages of screws A and 22 packages of screws B; Maximum profit = Rs 412
Explaination / Solution:

Let number of packages of screws A produced = x
And number of packages of screws B produced = y
Therefore , the above L.P.P. is given as :
Maximise , Z = 7x +10y , subject to the constraints : 4x +6y ≤ 240 and. 6x +3y ≤ 240 i.e. 2x +3y ≤ 120 and 2x +y ≤ 80 , x, y ≥ 0.

 Corner points Z =7 x +10 y O( 0 , 0 ) 0 D(40,0 ) 280 A(0,40) 400 B(30,20) 410…………………(Max.)

Here Z = 410 is maximum.
i.e 30 packages of screws A and 20 packages of screws B; Maximum profit = Rs 410.

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Q.9
There are two types of fertilizers F1 and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F1 costs Rs 6/kg and F2 costsRs 5/kg, determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?
A. 130 kg of fertilizer F1 and 80 kg of fertilizer F2; Minimum cost = Rs 1300
B. 120 kg of fertilizer F1 and 80 kg of fertilizer F2; Minimum cost = Rs 1200
C. 100 kg of fertilizer F1 and 80 kg of fertilizer F2; Minimum cost = Rs 1000
D. 110 kg of fertilizer F1 and 80 kg of fertilizer F2; Minimum cost = Rs 1100
Explaination / Solution:

Let number of kgs. of fertilizer F1 = x
And number of kgs. of fertilizer F2 = y
Therefore , the above L.P.P. is given as :
Minimise , Z = 6x +5y , subject to the constraints : 10/100 x + 5/100y ≥ 14 and 6/100x + 10/100y ≥ 14, i.e. 2 x + y ≥ 280 and 3x + 5y ≥ 700, x,y ≥ 0.,

 Corner points Z =6x +5 y A ( 0 , 280 ) 1400 D(700/3,0 ) 1400 B(100,80) 1000………….(Min.)

Corner points Z =6x +5 y A ( 0 , 280 ) 1400 D(700/3,0 ) 1400 B(100,80) 1000………….(Min.) Here Z = 1000 is minimum.
i.e. 100 kg of fertilizer F1 and 80 kg of fertilizer F2; Minimum cost = Rs 1000.

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Q.10
Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5).Let F = 4x + 6y be the objective function. Maximum of F – Minimum of F =
A. 60
B. 18
C. 48
D. 42