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MHT-CET Maths · Linear Programming

Formulation and Special Cases — Word Problems and Infinitely Many Optima

Turn a word problem into variables, an objective and inequalities in consistent units; and recognise the tie — when two adjacent corners give the same optimum, every point of the edge between them is optimal.

Why this matters

7 PYQs, none HARD. Two are formulation stems (a machine-time problem and a minimum-cost alloy), and five ask what the solution set of an optimisation LOOKS like — and the answer, every time, has been 'infinitely many points', because the objective was parallel to an edge of the region. That pattern has been set in 2023 (twice) and 2025 (three times); it is the one thing on this page worth memorising.

Concept 1 of 2

Formulating an LPP: Variables, Objective, Constraints in One Unit

Intuition

Name the decision variables (how many of each item), write profit or cost as the objective, and turn each resource limit into an inequality — after converting every quantity to the same unit. 'At most' is ≤\le, 'at least' is ≥\ge, and x,y≥0x, y \ge 0 is always there.

Definition

  • Machine I: 1010 h 4040 min =640= 640 min; item A takes 2020, B takes 1515: 20x+15y≤64020x + 15y \le 640. Machine II: 88 h 2020 min =500= 500 min; 5x+8y≤5005x + 8y \le 500. Profit 25x+18y25x + 18y, maximise.
  • The option with 5x+8y≥5005x + 8y \ge 500 reverses a capacity limit; the option with 20x+5y≤820x + 5y \le 8 mixed hours and minutes. Both are built from the two standard slips.
  • Minimum cost: copper xx g at ₹8, brass yy g at ₹5; chip weight at least 55 g (x+y≥5x + y \ge 5), brass at most 44 (y≤4y \le 4), copper at least 22 (x≥2x \ge 2). Minimise 8x+5y8x + 5y: corners (2,3)(2,3), (2,4)(2,4), (5,0)(5,0) give 31,36,4031, 36, 40; minimum ₹31.
  • Formulation stems stop at the model; the optimisation stems continue into the corner-point method.

Standard form

Maximise/Minimise Z=px+qysubject to aix+biy≤ci (or ≥), x,y≥0\text{Maximise/Minimise } Z = px + qy \quad \text{subject to } a_ix + b_iy \le c_i \ (\text{or } \ge),\ x, y \ge 0

Worked example

A factory makes chairs (xx) and tables (yy). A chair needs 22 h of carpentry and 11 h of finishing; a table needs 33 h and 22 h. Carpentry is limited to 3636 h a week and finishing to 2020 h. Profits are ₹500 and ₹800. Formulate.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Linear ProgrammingMODERATE
A manufacturing company produces two items, A and B. Each toy should be processed by two machines, I and II. Machine I can be operated for maximum 10 hours 40 minutes. It takes 20 minutes for an item of A and 15 minutes for B. Machine II can be operated for a total time at 8 hours 20 minutes. It takes 5 minutes for an item A and 8 minutes for B . The profit per item of AA is ₹ 25 and per item of BB is ₹ 18 . The formulation of an L.P.P. to maximize the profit (where xx is number of items A and yy is the number of item BB ) is

[Q146 · 26 April Shift II · 2025]

Mixing hours and minutes

20x+15y≤102320x + 15y \le 10\frac23 is the same constraint in hours and would be fine — but 20x+5y≤820x + 5y \le 8 pairs minute coefficients with an hour limit. Convert everything to one unit before writing the inequality.

Concept 2 of 2

Infinitely Many Optima: The Objective Parallel to an Edge

Intuition

If the objective line Z=ax+byZ = ax + by has the same slope as an edge of the region, the sliding line touches that whole edge last. Two adjacent corners then give the same optimal value, and every point between them is optimal.

Definition

  • Z=x+yZ = x + y, x+y≤10x + y \le 10, 5x+3y≥155x + 3y \ge 15, x≤6x \le 6: corners (0,10)(0,10) and (6,4)(6,4) both give Z=10Z = 10 because the edge x+y=10x + y = 10 is parallel to the objective. Maximum at infinitely many points.
  • Minimise Z=30x+20yZ = 30x + 20y, x+y≤8x + y \le 8, x+2y≥4x + 2y \ge 4, 6x+4y≥126x + 4y \ge 12: (0,3)(0,3) and (1,32)\left(1, \frac32\right) both give 6060 — the edge 6x+4y=126x + 4y = 12 has slope −32-\frac32, the same as 30x+20y30x + 20y.
  • Minimise Z=x+yZ = x + y, x+y≥2x + y \ge 2, x+2y≤8x + 2y \le 8, y≤3y \le 3: (2,0)(2,0) and (0,2)(0,2) tie at 22; the region is bounded, so 'infinitely many points, bounded set'.
  • Test: compare the ratio a:ba : b of the objective with the ratio of the coefficients of each binding constraint; a match means a tie. x+yx + y vs x+y≤10x + y \le 10: match.
  • 'Unique', 'two distinct points' and 'does not exist' are the distractors; a tie is never at exactly two points.

Tie condition

Z=ax+by ∥ edge ax+by=c ⇒ optimum on the whole edgeZ = ax + by \ \parallel\ \text{edge } ax + by = c \ \Rightarrow\ \text{optimum on the whole edge}

Worked example

Maximise Z=2x+4yZ = 2x + 4y subject to x+2y≤8x + 2y \le 8, x≤6x \le 6, x,y≥0x, y \ge 0. Describe the solution set.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Linear ProgrammingMODERATE
The maximum value of z=x+yz=x+y, subjected to x+y≤10x+y\leq10, 5x+3y≥155x+3y\geq15, x≤6x\leq6, x,y≥0x,y\geq0

[Q126 · 9th May Shift 1 · 2023]

Answering 'two distinct points'

Two corners tie only because the whole edge between them ties. The optimum is the segment, so the honest count is infinite, never two.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Formulating an LPP: Variables, Objective, Constraints in One Unit

    Standard form

    Maximise/Minimise Z=px+qysubject to aix+biy≤ci (or ≥), x,y≥0\text{Maximise/Minimise } Z = px + qy \quad \text{subject to } a_ix + b_iy \le c_i \ (\text{or } \ge),\ x, y \ge 0
  • Infinitely Many Optima: The Objective Parallel to an Edge

    Tie condition

    Z=ax+by ∥ edge ax+by=c ⇒ optimum on the whole edgeZ = ax + by \ \parallel\ \text{edge } ax + by = c \ \Rightarrow\ \text{optimum on the whole edge}

Watch out for (2)

Drill every past-year question on this subtopic

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