Maximize Z = 3 x + 4 y subject to the constraints, x + y ≤ 10 , x , y ≥ 0 .

Given LPP as follows: Objective function: Z = 3 x + 4 y Subject to x + y ≤ 10 , x , y ≥ 0 We first draw the graph of the line x + y = 10 on the X Y - plane by joining the points 0 , 10 and 10 , 0 . Now, the line divides the plane into two half. Let us identify the solution region of the inequation x + y ≤ 10 . Since the coordinates of the point 0 , 0 satisfy the inequation x + y ≤ 10 . So, the solution region for this inequation is that half-plane which contains the point 0 , 0 . Also, since the inequality is slack, so all the points on the line are also included in the solution region. Also, x ≥ 0 , y ≥ 0 represents the region in the first quadrant. The feasible region determined by the constraints x + y ≤ 10 , x ≥ 0 , y ≥ 0 as follows; The corner points of the feasible region are O ( 0 , 0 ) , A ( 10 , 0 ) , and B ( 0 , 10 ) . The values of Z at these points are as follows. Corner Points Z = 3 x + 4 y O 0 , 0 Z = 0 A 10 , 0 Z = 30 B 0 , 10 Z = 40 [maximum] Therefore, the maximum value of Z is 40 at the point B 0 , 10 .

Maximize Z = 3 x + 4 y subject to the constraints, x + y ≤ 10 , x , y ≥ 0 .

Given LPP as follows: Objective function: Z = 3 x + 4 y Subject to x + y ≤ 10 , x , y ≥ 0 We first draw the graph of the line x + y = 10 on the X Y - plane by joining the points 0 , 10 and 10 , 0 . Now, the line divides the plane into two half. Let us identify the solution region of the inequation x + y ≤ 10 . Since the coordinates of the point 0 , 0 satisfy the inequation x + y ≤ 10 . So, the solution region for this inequation is that half-plane which contains the point 0 , 0 . Also, since the inequality is slack, so all the points on the line are also included in the solution region. Also, x ≥ 0 , y ≥ 0 represents the region in the first quadrant. The feasible region determined by the constraints x + y ≤ 10 , x ≥ 0 , y ≥ 0 as follows; The corner points of the feasible region are O ( 0 , 0 ) , A ( 10 , 0 ) , and B ( 0 , 10 ) . The values of Z at these points are as follows. Corner Points Z = 3 x + 4 y O 0 , 0 Z = 0 A 10 , 0 Z = 30 B 0 , 10 Z = 40 [maximum] Therefore, the maximum value of Z is 40 at the point B 0 , 10 .

Minimize Z = 6 x - 5 y subject to the constraints, 2 x + 5 y ≤ 20 , x , y ≥ 0 .

Given LPP as follows: Objective function: Z = 6 x - 5 y Subject to 2 x + 5 y ≤ 20 , x , y ≥ 0 We first draw the graph of the line 2 x + 5 y = 20 on the X Y - plane by joining the points 0 , 4 and 10 , 0 . Now, the line divides the plane into two half. Let us identify the solution region of the inequation 2 x + 5 y ≤ 20 . Since the coordinates of the point 0 , 0 satisfy the inequation 2 x + 5 y ≤ 20 . So, the solution region for this inequation is that half-plane which contains the point 0 , 0 . Also, since the inequality is slack, so all the points on the line are also included in the solution region. Also, x ≥ 0 , y ≥ 0 represents the region in the first quadrant. The feasible region determined by the constraints 2 x + 5 y ≤ 20 , x ≥ 0 , y ≥ 0 as follows; The corner points of the feasible region are O ( 0 , 0 ) , A ( 10 , 0 ) , and B ( 0 , 4 ) . The values of Z at these points are as follows. Corner Points Z = 6 x - 5 y O 0 , 0 Z = 0 A 10 , 0 Z = 60 B 0 , 4 Z = - 20 [minimum] Therefore, the minimum value of Z is - 20 at the point B 0 , 4 .

Minimize Z = 7 x - 5 y subject to the constraints, 4 x + 5 y ≤ 40 , x , y ≥ 0 .

Given LPP as follows: Objective function: Z = 7 x - 5 y Subject to 4 x + 5 y ≤ 40 , x , y ≥ 0 We first draw the graph of the line 4 x + 5 y = 40 on the X Y - plane by joining the points 0 , 8 and 10 , 0 . Now, the line divides the plane into two half. Let us identify the solution region of the inequation 4 x + 5 y ≤ 40 . Since the coordinates of the point 0 , 0 satisfy the inequation 4 x + 5 y ≤ 40 . So, the solution region for this inequation is that half-plane which contains the point 0 , 0 . Also, since the inequality is slack, so all the points on the line are also included in the solution region. Also, x ≥ 0 , y ≥ 0 represents the region in the first quadrant. The feasible region determined by the constraints 4 x + 5 y ≤ 40 , x ≥ 0 , y ≥ 0 as follows; The corner points of the feasible region are O ( 0 , 0 ) , A ( 10 , 0 ) , and B ( 0 , 8 ) . The values of Z at these points are as follows. Corner Points Z = 7 x - 5 y O 0 , 0 Z = 0 A 10 , 0 Z = 70 B 0 , 8 Z = - 40 [minimum] Therefore, the minimum value of Z is - 40 at the point B 0 , 8 .

Minimize Z = 10 x - 6 y subject to the constraints, 6 x + 5 y ≤ 30 , x , y ≥ 0 .

Given LPP as follows: Objective function: Z = 10 x - 6 y Subject to 6 x + 5 y ≤ 30 , x , y ≥ 0 We first draw the graph of the line 6 x + 5 y = 30 on the X Y - plane by joining the points 0 , 6 and 5 , 0 . Now, the line divides the plane into two half. Let us identify the solution region of the inequation 6 x + 5 y ≤ 30 . Since the coordinates of the point 0 , 0 satisfy the inequation 6 x + 5 y ≤ 30 . So, the solution region for this inequation is that half-plane which contains the point 0 , 0 . Also, since the inequality is slack, so all the points on the line are also included in the solution region. Also, x ≥ 0 , y ≥ 0 represents the region in the first quadrant. The feasible region determined by the constraints 6 x + 5 y ≤ 30 , x ≥ 0 , y ≥ 0 as follows; The corner points of the feasible region are O ( 0 , 0 ) , A ( 5 , 0 ) , and B ( 0 , 6 ) . The values of Z at these points are as follows. Corner Points Z = 10 x - 6 y O 0 , 0 Z = 0 A 5 , 0 Z = 50 B 0 , 6 Z = - 36 [minimum] Therefore, the minimum value of Z is - 36 at the point B 0 , 6 .

Maximize Z = - 3 x + 4 y Subject to the constraints, x + 2 y ≤ 8 , 3 x + 2 y ≤ 12 , x ≥ 0 , y ≥ 0 .

Given LPP as follows: Objective function: Z = - 3 x + 4 y Subject to x + 2 y ≤ 8 , 3 x + 2 y ≤ 12 , x ≥ 0 , y ≥ 0 We first draw the graph of the line x + 2 y = 8 on the X Y - plane by joining the points 0 , 4 and 8 , 0 . Now, the line divides the plane into two half. Let us identify the solution region of the inequation x + 2 y ≤ 8 . Since the coordinates of the point 0 , 0 satisfy the inequation x + 2 y ≤ 8 . So, the solution region for this inequation is that half-plane which contains the point 0 , 0 . Also, since the inequality is slack, so all the points on the line are also included in the solution region. Similarly, we can find the solution region for 3 x + 2 y ≤ 12 Also, x ≥ 0 , y ≥ 0 represents the region in the first quadrant. The feasible region determined by the constraints x + 2 y ≤ 8 , 3 x + 2 y ≤ 12 , x ≥ 0 and y ≥ 0 as follows; The corner points of the feasible region are O ( 0 , 0 ) , A ( 4 , 0 ) , B ( 2 , 3 ) , and C ( 0 , 4 ) . The values of Z at these points are as follows. Corner Points Z = - 3 x + 4 y O 0 , 0 Z = 0 A 4 , 0 Z = - 12 B 2 , 3 Z = 6 C 0 , 4 Z = 16 → maximum Therefore, the maximum value of Z is 16 at the point C 0 , 4 .

A factory manufactures two types of screws A and B . Each type of screw requires the use of two machines automatic and a hand operated. It takes 4 minutes on automatic and 6 minutes on hand operated machines to manufacture a package of screws A while it takes 6 minutes on automatic and 3 minutes on hand operated machine 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 to 70 paise and screw B at a profit of ₹ 1 . Assuming that he sells 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.

Let manufacture should produce x packets of screw of type A and y packets of screw of type B. ∴ Profit on x packets = ₹ 0 . 70 x and profit on y packets = ₹ 1 y ∴ objective function to earn maximum profit z = 0 . 70 x + y Time to produce x screw of type A = 4 x minute and time to produce y screws of type B = 6 y minute But automatic machine is available for 4 hours only. Hence, 4 x + 6 y ⩽ 4 hour ⇒ 4 x + 6 y ⩽ 240 minutes Similarly, time taken to prepare screw of type A on handmade machine = 6 x minute nd time taken to prepare screws of type B on handmade machine = 3 y minutes But handmade machine is available for 4 hours only a day. ∴ 6 x + 3 y ⩽ 4 hours ⇒ 6 x + 3 y ⩽ 240 minute ∵ x and y are the number of screws. ∴ x ⩾ 0 , y ⩾ 0 Mathematically formulation of this Linear programming problem is as the following: Maximum z = 0 . 70 x + y Subject to the constraints 4 x + 6 y ⩽ 240 6 x + 3 y ⩽ 240 x ⩾ 0 , y ⩾ 0 Converting the given inequation into equations. 4 x + 6 y = 240 . . . . . ( 1 ) 6 x + 3 y = 240 . . . . ( 2 ) Region represented by 4 x + 6 y ⩽ 240 : the line 4 x + 6 y = 240 meets the coordinate axis at A 60 , 0 and B 0 , 40 . 4 x + 6 y = 240 x 60 0 y 0 40 A 60 , 0 ; B 0 , 40 join the points A and B to obtain the line, Clearly 0 , 0 satisfies the inequation 4 × 0 + 6 × 0 = 0 ⩽ 240 . So, the region containing the origin represents the solution set of the inequation. Region represented by 6 x + 3 y ⩽ 240 : the line 6 x + 3 y = 240 meets the coordinate axis at points C 40 , 0 and D 0 , 80 . 6 x + 3 y = 240 x 40 0 y 0 80 C 40 , 0 ; D 0 , 80 join C and D to obtain the line clearly 0 , 0 satisfies the given inequation 6 × 0 + 3 × 0 = 0 ⩽ 240 . So, the region containing the origin represents the solution set of the inequation. Region represented x ⩾ 0 and y ⩾ 0 : Since every point in the first quadrant satisfies these inequations. So the first quadrant in the region represented by the inequation x ⩾ 0 and y ⩾ 0 . The coordinate of point of intersection of lines 4 x + 6 y = 240 and 6 x + 3 y = 240 are x = 30 and y = 20 . The shaded region OAED represent the common region of the inequation. This region is the feasible solution region of the inequations. The comer points of this solution region are O 0 , 0 , A 40 , 0 , E 30 , 20 and D 0 , 40 . The value of objective function on these points is as following table: Points X-coordinate Y-coordinate objective function z = 0 . 70 x + y O 0 0 z o = 0 . 70 × 0 + 0 = 0 A 40 0 z A = 0 . 70 × 40 + 0 = 28 E 30 20 z E = 0 . 70 × 30 + 20 = 41 D 0 40 z D = 0 . 70 × 0 + 40 = 40 From the table the value of objective function is maximum at point E 30 , 20 = ₹ 41 Hence, the manufacturer should produce 30 packets of screw of type A and 20 packets of screw of type B to get the maximum profit ₹ 41

A manufacturer has employed 5 skilled men and 10 semi-skilled men and makes two models A and B of an article.The making of one item of model A requires 2 hours work by a skilled man and 2 hours work by a semi-skilled man. One item of model B requires 1 hour by a skilled man and 3 hours by a semi-skilled man.No man is expected to work more than 8 hours per day. The manufacturer's profit on an item of model A is ₹ 15 and on an item of model B is ₹ 10 . How many of items of each model should be made per day in order to maximize daily profit ? Formulate the above L P P and solve it graphically and find the maximum profit.

Given that total available hours for skilled men = 8 × 5 = 40 and total available hours for semi-skilled men = 8 × 10 = 80 Let x items of model A and y items of model B be made. ∴ x , y ≥ 0 (number of items can not be negative) We have, The making of model A requires 2 hours work by a semi skilled man and model B requires 1 hour by skilled man ∴ 2 x + y ≤ 40 And, the making of model A requires 2 hours work by a semi skilled man and model B requires 3 hours work by semi-skilled man. ∴ 2 x + 3 y ≤ 80 And, the profit, Z = 15 x + 10 y Changing the inequalities to equations, we have 2 x + y = 40 , 2 x + 3 y = 80 Solving both equations we get 10 , 20 . The point of intersections with coordinate axes are 0 , 40 , 20 , 0 and 0 , 80 3 , 40 , 0 The feasible region determined by the system of constraints is O A B C The corner points are A 0 , 80 3 , B 10 , 20 , C 20 , 0 Corner points Z = 15 x + 10 y A 0 , 80 3 800 3 B 10 , 20 350 C 20 , 0 300 The maximum value of Z = 350 which is attained at B 10 , 20 . Hence, the maximum profit is ₹ 350 when 10 units of model A and 20 units of model B are produced.

MEDIUM

12th Karnataka Board

IMPORTANT

Earn 100

Maximize $Z = 3 x + 4 y$

subject to the constraints, $x + y \leq 10$ , $x, y \geq 0$ .

Important Questions on Linear Programming

MEDIUM

12th Karnataka Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 12 - Linear Programming>Exercise>Q 9

Minimize $Z = 6 x - 5 y$

subject to the constraints, $2 x + 5 y \leq 20$ , $x, y \geq 0$ .

MEDIUM

12th Karnataka Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 12 - Linear Programming>Exercise>Q 10

Minimize $Z = 7 x - 5 y$

subject to the constraints, $4 x + 5 y \leq 40$ , $x, y \geq 0$ .

MEDIUM

12th Karnataka Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 12 - Linear Programming>Exercise>Q 11

Minimize $Z = 10 x - 6 y$

subject to the constraints, $6 x + 5 y \leq 30$ , $x, y \geq 0$ .

MEDIUM

12th Karnataka Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 12 - Linear Programming>Exercise>Q 12

Maximize $Z = - 3 x + 4 y$

Subject to the constraints, $x + 2 y \leq 8, 3 x + 2 y \leq 12, x \geq 0, y \geq 0$ .

HARD

12th Karnataka Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 12 - Linear Programming>Exercise>Q 13

A company produces two types of goods, $A$ and $B$ , that require gold and silver. Each unit of type $A$ requires $3 g$ of silver and $1 g$ of gold, while that of type $B$ requires $1 g$ of silver and $2 g$ of gold. The company can use at the most $9 g$ of silver and $8 g$ of gold. If each unit of type $A$ brings a profit of $₹ 120$ and that of type $B$ $₹ 150$ , then find the number of units of each type that the company should produce to maximise profit.

Formulate the above LPP and solve it graphically. Also, find the maximum profit.

HARD

12th Karnataka Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 12 - Linear Programming>Exercise>Q 14

A furniture trader deals in only two items - chairs and tables. He has $50, 000$ rupees to invest and a space to store at most $35$ items. A chair costs him $1000$ rupees and a table costs him $2000$ rupees . The trader earns a profit of $150$ rupees and $250$ rupees on a chair and table, respectively. Formulate the above problem as an LPP to maximise the profit and solve it graphically.

EASY

12th Karnataka Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 12 - Linear Programming>Exercise>Q 15

Solve the linear programming problem graphically:

Maximize $Z = 4 x + y,$ subject to the constraints $x + y \leq 50, 3 x + y \leq 90, x \geq 0, y \geq 0 .$

HARD

12th Karnataka Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 12 - Linear Programming>Exercise>Q 16

A company manufactures and sells two models of lamps

L_{1}

and

L_{2}

, the profit being

₹ 15

and

₹ 10

respectively. The process involves two workers

W_{1}

and

W_{2}

who are available for this kind of work

100

hours and

80

hours per month respectively,

W_{1}

assembles

L_{1}

20

and

L_{2}

30

minutes.

W_{2}

paints

L_{1}

20

and

L_{2}

10

minutes. Assuming that all lamps made can be sold, formulate LPP for determining the productions figures for maximum profit.

Maximize Z=3x+4y subject to the constraints, x+y≤10, x, y≥0.

Important Questions on Linear Programming

Learn and Prepare for any exam you want !

Maximize $Z = 3 x + 4 y$

subject to the constraints, $x + y \leq 10$ , $x, y \geq 0$ .