All objective function and constraints are linear

Only objective function is linear

Either objective or constraints are linear

All objective function and constraints are linear

Inequation y - x ≤ 0 represents

the half that contains the positive x - axis

closed half plane above the line y = x which contains positive y - axis

half plane that contains the negative x - axis

half plane that contains the positive y - axis

The linear programming problem: Max Z = x 1 + x 2 such that 2 x 1 - x 2 ≥ - 1 , x 1 ≤ 2 , x 1 + x 2 ≤ 3 and x 1 , x 2 ≥ 0 has

An infinite numbers of solution

A manufacturing company makes two models A and B of a product. Each piece of model A requires 9 labour hours for fabricating and 1 labour hour for finishing.Each piece of model B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available are 180 and 30 respectively. The company makes a profit of Rs 8000 on each piece of model A and Rs 12000 on each piece of model B . Formulate LPP so as to maximise his profit

Suppose x pieces of model A and y pieces of model B are manufactured per week. Since each piece of model A requires 9 labour hours and each piece of model B requires 12 labour hours for fabricating, therefore, x pieces of model A and y pieces of model B require ( 9 x + 12 y ) labour hours for fabricating. But, the maximum labour hours available for fabricating are 180 . ∴ 9 x + 12 y ≤ 180 ⇒ 3 x + 4 y ≤ 60 Similarly, each piece of model A requires 1 labour hour and each piece of model B requires 3 labour hours for finishing, therefore, x pieces of model A and y pieces of model B require ( x + 3 y ) labour hours for finishing. But, the maximum labour hours available for finishing are 30 . ∴ x + 3 y ≤ 30 The profit from each piece of model A is Rs 8 , 000 and from each piece of model B is Rs 12 , 000 . Therefore, the total profit from x pieces of model A and y pieces of model B is Rs ( 8 , 000 x + 12 , 000 y ) . Thus, the given linear programming problem is Maximise Z = 8000 x + 12000 y , subject to the constraints 3 x + 4 y ≤ 60 , x + 3 y ≤ 30 and x , y ≥ 0 The feasible region determined by the given constraints can be diagrammatically represented as, The coordinates of the corner points of the feasible region are O ( 0 , 0 ) , A ( 20 , 0 ) , B ( 12 , 6 ) and C ( 0 , 10 ) . The value of the objective function at these points are given in the following table. Corner Point Z = 8000 x + 12000 y O 0 , 0 Z = 0 A 20 , 0 Z = 16000 B 12 , 6 Z = 168000 Maximum C 0 , 10 Z = 120000 Hence, the manufacturing company should produce 12 pieces of model A and 6 pieces of model B to realise maximum profit.The maximum profit is Rs 1 , 68 , 000 .

A cooperative society of farmers has 50 hectares of land to grow two crops X and Y . The profit from crops X and Y per hectare are estimated as ₹ 10500 and ₹ 9000 respectively. To control weeds, a liquid herbicide has to be used for crops X and Y at rates of 20 litres and 10 litres per hectare. Further no more than 800 litres of herbicide should be used in order to protect fish and wildlife using a pond which collects drainage from this land. How much land should be allocated to each crop so as to maximise the total profit of the society?

Let x hectare of land be allocated crop X and Y hectare to crop Y . x ≥ 0 , y ≥ 0 Profit per hectare on crop X = Rs 10500 Profit per hectare on crop Y = Rs 9000 ∴ Total profit = Rs ( 10 , 500 x + 9000 y ) Maximise Z = 10 , 500 x + 9000 y Subject to constraints x + y ≤ 50 (constraint related to land) 20 x + 10 y ≤ 800 (constraint related to use of herbicide) 2 x + y ≤ 80 x ≥ 0 , y ≥ 0 From the graph, the feasible region is O A B C . The coordinate of the corner points are O ( 0 , 0 ) , A ( 40 , 0 ) , B ( 30 , 20 ) and C ( 0 , 50 ) . Hence, the society will get maximum profit of Rs 4 , 95 , 000 by allocating 30 hectares for crop X and 20 hectares for crop Y .

Solve the following problem graphically; Minimize and Maximize z = 3 x + 9 y Subject to the constraints x + 3 y ≤ 60 x + y ≥ 10 x ≤ y x ≥ 0 , y ≥ 0 .

Given LPP as follows: Objective function: z = 3 x + 9 y Subject to the constraints: x + 3 y ≤ 60 x + y ≥ 10 x ≤ y x ≥ 0 , y ≥ 0 Table for x + 3 y = 60 x 0 60 y 20 0 Similarly, table for x + y = 10 x 0 10 y 10 0 We first draw the graph of the line x + 3 y = 60 on the X Y - plane by joining the points 0 , 20 and 60 , 0 . Now, the line divides the plane into two half. Let us identify the solution region of the inequation x + 3 y ≤ 60 . Since the coordinates of the point 0 , 0 satisfy the inequation x + 3 y ≤ 60 . 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 draw the graph for other lines also using the same process. Also, x ≥ 0 , y ≥ 0 represents the region in the first quadrant. Now, solve x + 3 y = 60 and x = y , then we get: C 15 , 15 Similarly, solve x + y = 10 and x = y , then we get: B 5 , 5 The feasible region determined by the constraints x + 3 y ≤ 60 , x + y ≥ 10 , x , y ≥ 0 as follows; The corner points of the feasible region are A ( 0 , 10 ) , B ( 5 , 5 ) , C 15 , 15 , and D ( 0 , 20 ) . The values of z at these points are as follows. Corner points z = 3 x + 9 y A ( 0 , 10 ) z = 90 B ( 5 , 5 ) z = 60 [Minimum] C ( 15 , 15 ) z = 180 [Maximum] D ( 0 , 20 ) z = 180 [Maximum] Therefore, the maximum value of z is 180 at the points C ( 15 , 15 ) & D ( 0 , 20 ) . Similarly, the minimum value of z is 60 at the point B ( 5 , 5 ) .