6 min readConcept sheet1-mark MCQ or assertion-reason; 1 or 2-mark definitions

Constraints, objective function and optimisation

An LPP has an objective function to optimise and constraints that limit the variables. Optimisation means finding the maximum or minimum value of Z.

Objective functionZ = ax + by (a, b constants), the linear function to be maximised or minimised

ConstraintsThe linear inequalities (or equations) the variables must satisfy, e.g. 2x + y ≤ 12

Non-negative restrictionsx ≥ 0, y ≥ 0. The syllabus calls every other constraint non-trivial and allows up to three of them

Optimisation problemA problem of finding the maximum or minimum of a function; an LPP is a special type

The parts of an LPP

Term	Meaning	In the tailor example
Decision variables	Quantities we choose	x shirts, y kurtas
Objective function	Linear function to maximise or minimise	Z = 40x + 60y (maximise)
Constraints	Linear conditions from limited resources	2x + y ≤ 12, x + 2y ≤ 12
Non-negative restrictions	Variables cannot be negative	x ≥ 0, y ≥ 0
Optimal value	The best (maximum or minimum) value of Z	₹ 400
Optimal solution	The point that gives the optimal value	(4, 4): 4 shirts and 4 kurtas

Turning words into inequalities

Words in the question	Inequality
at most 40, not more than 40, up to 40	≤ 40
at least 40, not less than 40, a minimum of 40	≥ 40
exactly 40	= 40
x is at least twice y	x ≥ 2y
y exceeds x by at least 5	y − x ≥ 5

Objective function vs constraint

Objective functionOne linear expression, Z = ax + by. We want its best value. It has no inequality sign.

ConstraintA linear condition with ≤, ≥ or =. There can be several. Together they decide which points are allowed.

The syllabus limits problems to three non-trivial constraints (constraints other than x ≥ 0, y ≥ 0).

Example 1: name the parts and test points

Problem

For: Minimise Z = 4x + 6y subject to x + 2y ≥ 8, 3x + y ≥ 9, x ≥ 0, y ≥ 0, name the objective function and the non-trivial constraints. Is (2, 3) allowed? Is (1, 5)?

Working

Objective function: Z = 4x + 6y (to be minimised).
Non-trivial constraints: x + 2y ≥ 8 and 3x + y ≥ 9.
(2, 3): 2 + 6 = 8 ≥ 8 is true; 6 + 3 = 9 ≥ 9 is true. Allowed, and Z = 8 + 18 = 26.
(1, 5): 1 + 10 = 11 ≥ 8 is true; 3 + 5 = 8 ≥ 9 is false. Not allowed.

Answer

(2, 3) satisfies every constraint (Z = 26); (1, 5) breaks 3x + y ≥ 9.

Example 2: write the constraints

Problem

A shop sells x tables and y chairs. The number of chairs must be at least three times the number of tables, and the shop can hold at most 40 pieces. Write the constraints.

Working

'Chairs at least three times tables': y ≥ 3x
'At most 40 pieces': x + y ≤ 40
Counts cannot be negative: x ≥ 0, y ≥ 0

Answer

y ≥ 3x (that is, 3x − y ≤ 0), x + y ≤ 40, x ≥ 0, y ≥ 0.

In real life

A dietitian planning a hospital meal minimises cost (the objective function) while meeting the minimum energy and protein each patient needs (the constraints).

Exam angle

1 marks: “In 'Maximise Z = 3x + 2y subject to x + y ≤ 4, x ≥ 0, y ≥ 0', name the objective function and the non-trivial constraint.” Objective function Z = 3x + 2y; non-trivial constraint x + y ≤ 4.

1 marks: “Write as an inequality: the number of cups y is at least twice the number of plates x.” y ≥ 2x.

2 marks: “Is (3, 2) a feasible point for 2x + y ≤ 10, x + 3y ≤ 15, x ≥ 0, y ≥ 0? Find Z = 4x + 5y there.” 2(3) + 2 = 8 ≤ 10 and 3 + 6 = 9 ≤ 15, and both values are non-negative, so yes. Z = 12 + 10 = 22.

Remember it

Z is the goal; the constraints are the rules.