linear programming problem notes and assignment of XII Maths
Chapter- Linear Programming
12.1.1 An Optimisation Problem - A problem which seeks to maximise or minimise a
function is called an optimisation problem. An optimisation problem may
involve maximisation of profit, production etc or minimisation of cost, from available
resources etc.
12.1.2 A Linnear Programming Problem (LPP)- A linear programming problem deals with the optimisation (maximisation/
minimisation) of a linear function of two variables (say x and y) known as objective
function subject to the conditions that the variables are non-negative and satisfy a set
of linear inequalities (called linear constraints). A linear programming problem is a
special type of optimisation problem.
12.1.3 Objective Function - Linear function Z = ax + by, where a and b are constants, which has to be maximised or minimised is called a linear objective function.
12.1.4 Decision Variables In the objective function Z = ax + by, x and y are called decision variables.
12.1.5 Constraints The linear inequalities or restrictions on the variables of an LPP are called constraints. The conditions x ≥ 0, y ≥ 0 are called non-negative constraints.
12.1.6 Feasible Region The common region determined by all the constraints including non-negative constraints x ≥ 0, y ≥ 0 of an LPP is called the feasible region for the problem.
12.1.7 Feasible Solutions Points within and on the boundary of the feasible region for an LPP represent feasible solutions.
12.1.8 Infeasible Solutions Any Point outside feasible region is called an infeasible solution.
12.1.9 Optimal (feasible) Solution Any point in the feasible region that gives the optimal value (maximum or minimum) of the objective function is called an optimal solution. Following theorems are fundamental in solving LPPs.
Solve each of the following linear programming problems by graphical method:
1.Minimize and maximize Z = 18x + 10y
Subject to : 4x + y ≥ 20; 2x + 3y ≥ 30 x, y ≥ 0
[ Ans: Z at (3, 8) = 18 × 3 + 8 × 10 = 134
Z at (0,20) = 18 × 0 + 10 × 20 = 200
Z at (15,0) = 18 × (15) + 10 × 0 = 270]
2. Maximize Z = 10x + 6y
Subject to : 2x + 5y ≤ 34; 3x + y ≤ 12; x, y ≥ 0
[ Ans : At (2,16) , maximum value is 56]
3.Maximize and minimize Z = 3x + 5y
Subject to : x + 2y ≤ 20; x + y ≤ 15; y ≤ 5,
x, y ≥ 0. [Ans : At (10,5), maximum value is 55, minimum value is 25 at (0,5)]
4. Maximize Z = 20x + 10y
Subject to the following constraints
x + 2y ≤ 28; 3x + y ≤ 24; x ≥ 2
[ Ans : maximum is 200 at (4,12) , minimum is 40 at (2,0)]
5. Minimize and maximize z = 6x + 3y
Subject to the constraint
4x + y ≥ 80; x + 5y ≥ 115; 3x + 2y ≤ 150
x ≥ 0, y ≥ 0
[ Ans : minimum 150 at (15,20), maximum 732 at (2,72) ]
CBSE Corner
6. Solve the following Linear Programming Problem graphically :
Minimise z = 3x + 8y
subject to the constraints
3x + 4y≥ 8 ;5x + 2y≥11 ; x≥0, y≥0. A.I.C.B.S.E. Compt. Sep 2023
[ Ans : minimum value is 8 at ( 8/3,0)
7. A.I.C.B.S.E. Sample paper 2023
[ Ans : The minimum value of Z is 100 at (0,50)
and (20,40) .
OR
[Z has no maximum value as at z =1,it intersect the feasible region ]
8. Maximise z = 6x + 3y,
subject to the constraints 4x + y ≥ 80, ; 3x + 2y ≤ 150, ;x + 5y ≥115,
x ≥ 0, y ≥ 0. A.I.C.B.S.E. March 2023 set -01
[ Ans: At (40,15), maximum value is 285]
9.
A.I.C.B.S.E. March 2023 set - 02
[ Ans : Minimum 300 at (60,0) ]
10.Minimize z = 500x + 400y subject to constraints
x + y ≤ 200, x ≥ 20, y ≤ 4x, x, y ≥ 0.
A.I.C.B.S.E. March 2023 set - 03
[ Ans: minimum value is 42000 at (20,80) ]
11. Solve the following linear programming problem graphically :
Maximise z = 5x + 3y subject to the constraints
3x + 5y ≤ 15, 5x + 2y ≤ 10, x, y ≥ 0.
A.I.C.B.S.E. March 2023 set - 04
[ Ans maximum is 235/19 at ( 20/19, 45/19 ) ]
12. Minimise: Z = 60x + 80y
subject to constraints: 3x + 4y ≥ 8; 5x + 2y ≥ 11
A.I.C.B.S.E. March 2023 set - 05
[ Ans : minimum value is 160 at (2, ½ ) & (8/3, 0)
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