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Chapter 5

Linear Programming

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1 question set from this chapter. Most repeated first.

  • 2081 · Mathematics · 4+6 marks

Define linear programming with appropriate terminologies. Maximize: P=x+2y+zP = x + 2y + z subject to x+y+2z≤10,2x+3y+z≤12,x≥0,y≥0,z≥0x + y + 2z \le 10, 2x + 3y + z \le 12, x \ge 0, y \ge 0, z \ge 0 by simplex method.

Answer

Linear programming (LP) is a mathematical technique for finding the maximum or minimum value of a linear objective function of several variables, subject to linear constraints (equations or inequalities) and non-negativity conditions. It is used in production planning, resource allocation, diet and transportation problems.

Terminologies

  • Decision variables: the unknowns (x,y,zx, y, z) to be determined.
  • Objective function: the linear function to be optimised, here P=x+2y+zP = x + 2y + z.
  • Constraints: linear restrictions on the variables, here x+y+2z≤10x+y+2z\le 10 and 2x+3y+z≤122x+3y+z\le 12.
  • Non-negativity restrictions: x,y,z≥0x,y,z\ge 0.
  • Feasible solution / feasible region: a solution satisfying all constraints / the set of all such solutions.
  • Optimal solution: the feasible solution giving the best value of the objective function.
  • Slack variable: a non-negative variable added to a "≤\le" constraint to turn it into an equation.
  • Basic variables: the variables that are non-zero in a basic feasible solution (their columns form an identity matrix); non-basic variables are set to zero.

Solution by simplex method

Add slack variables S1,S2≥0S_1, S_2\ge 0:

x+y+2z+S1=10,2x+3y+z+S2=12,P−x−2y−z=0x+y+2z+S_1=10,\qquad 2x+3y+z+S_2=12,\qquad P-x-2y-z=0

Initial table

BasisxxyyzzS1S_1S2S_2bbRatio
S1S_1112101010/1=1010/1=10
S2S_2231011212/3=412/3=4
Cj−ZjC_j-Z_j−1-1−2-2−1-1000

Most negative Cj−Zj=−2C_j-Z_j=-2, so yy enters; minimum ratio 44, so S2S_2 leaves (pivot =3=3).

Iteration 1 (R2→R2/3R_2\to R_2/3, R1→R1−R2R_1\to R_1-R_2, objective row →\to row +2R2+2R_2)

BasisxxyyzzS1S_1S2S_2bbRatio
S1S_11/31/305/35/31−1/3-1/366÷53=1856\div\frac53=\frac{18}{5}
yy2/32/311/31/301/31/344÷13=124\div\frac13=12
Cj−ZjC_j-Z_j1/31/30−1/3-1/302/32/38

The only negative entry is −1/3-1/3 under zz, so zz enters. The minimum ratio is 18/518/5, so S1S_1 leaves (pivot =5/3=5/3).

Iteration 2 (R1→35R1R_1\to \tfrac35R_1, R2→R2−13R1R_2\to R_2-\tfrac13R_1, objective row →\to row +13R1+\tfrac13R_1)

BasisxxyyzzS1S_1S2S_2bb
zz1/51/5013/53/5−1/5-1/518/518/5
yy3/53/510−1/5-1/52/52/514/514/5
Cj−ZjC_j-Z_j2/52/5001/51/53/53/546/546/5

All entries of the objective row are ≥0\ge 0, so the solution is optimal.

Result

x=0,y=145,z=185,Pmax⁡=2⋅145+185=465=9.2x=0,\quad y=\frac{14}{5},\quad z=\frac{18}{5},\quad P_{\max}=2\cdot\frac{14}{5}+\frac{18}{5}=\frac{46}{5}=9.2

Check: x+y+2z=145+365=10x+y+2z=\frac{14}{5}+\frac{36}{5}=10 and 2x+3y+z=425+185=122x+3y+z=\frac{42}{5}+\frac{18}{5}=12, so both constraints are satisfied (both resources fully used).

Answer checked by Claude Sonnet · Oct 21, 2026

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Questions from TSC Lower Secondary Level Teacher Exam Questions 2082 - General (Lower Secondary Level Teacher · General Examination · 2082), TSC Lower Secondary Level Math Questions (Lower Secondary · Mathematics · 2081), TSC Lower Secondary Level Sample Questions (Lower Secondary · English · 2078) and TSC Lower Secondary Level Mathematics Exam Question 2082 (Lower Secondary Level (3rd Class) Teacher · Mathematics · 2082). Answers are written for this site; check them against your class notes.