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

**Answer : ****Option D**

**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 .

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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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.

**Answer : ****Option D**

**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.

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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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

**Answer : ****Option C**

**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.

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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In a LPP, the objective function is always

**A. ** quadratic

**B. ** Linear

**C. ** constant

**D. ** cubic

**Answer : ****Option B**

**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.

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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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)

**Answer : ****Option A**

**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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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).

**Answer : ****Option A**

**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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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)

**Answer : ****Option C**

**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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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

**Answer : ****Option B**

**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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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

**Answer : ****Option C**

**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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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

**Answer : ****Option A**

**Explaination / Solution: **

Here the objective function is given by : F = 4x +6y .

Corner points | Z = 4x +6 y |

(0, 2 ) | 12………………..(Min.) |

(3,0) | 12………………….(Min.) |

(6,0 ) | 24 |

(6 , 8 ) | 72 |

(0 , 5 ) | 30 |

Maximum of F – Minimum of F = 72 – 12 = 30 .

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