⋆
]]
⋆
DSC2602: Rational Decision Making
May/June Examination 2026 — Comprehensive Revision Guide
(Based on May/June 2025 & May/June 2023 Past Papers)
⋆ ⋄ ⋆ ⋄ ⋆ ⋄ ⋆ ⋄ ⋆
[ Decision Sciences / Quantitative Management [
_ Exam Revision Guide
DSC2602
Module Code:
Rational Decision Making
Module Name:
May/June 2025 & May/June 2023
Paper / Exam:
2026
Year:
100
Total Marks:
Economic & Management Sciences
Faculty:
Study this guide carefully. Focus on understanding methods, not just memorising
formulas — examiners test application.
Exam Revision Notes | DSC2602 | 2026
,DSC2602 | Exam Revision Rational Decision Making
Question 1 [25 marks]
(a) [10 marks]
Question: A furniture company produces two products: chairs and tables. Each chair
requires 3 hours of carpentry and 2 hours of finishing. Each table requires 5 hours of
carpentry and 1.5 hours of finishing. The company has 150 hours of carpentry and 60
hours of finishing available per week. The profit contribution is R180 per chair and R250
per table. Define the decision variables and formulate a Linear Programming (LP) model
to maximise weekly profit.
Answer: Decision Variables:
• Let x1 = number of chairs produced per week
• Let x2 = number of tables produced per week
Objective Function (Maximise Profit):
Maximise Z = 180x1 + 250x2
Subject to the constraints:
3x1 + 5x2 ≤ 150 (carpentry hours)
2x1 + 1.5x2 ≤ 60 (finishing hours)
x1 , x2 ≥ 0 (non-negativity)
Key Concept
An LP model always has three parts: decision variables (what we control), an
objective function (what we optimise), and constraints (resource or demand
limits). Non-negativity constraints are always required.
(b) [10 marks]
Question: Solve the LP model in (a) using the graphical method. Show the feasible
region and identify the optimal solution.
Page 2 of 23
, DSC2602 | Exam Revision Rational Decision Making
Answer: Step 1: Find boundary line intersections with axes.
For 3x1 + 5x2 = 150: when x1 = 0, x2 = 30; when x2 = 0, x1 = 50.
For 2x1 + 1.5x2 = 60: when x1 = 0, x2 = 40; when x2 = 0, x1 = 30.
Step 2: Find intersection of the two constraint lines.
From constraint 2: x1 = 60−1.5x2
2 = 30 − 0.75x2
Substitute into constraint 1:
3(30 − 0.75x2 ) + 5x2 = 150
90 − 2.25x2 + 5x2 = 150 ⇒ 2.75x2 = 60 ⇒ x2 = 21.82
x1 = 30 − 0.75(21.82) = 30 − 16.36 = 13.64
Step 3: Evaluate the objective function at each corner of the feasible region.
Corner Point x1 x2 Z =
180x1 +
250x2
Origin 0 0 R0
Carpentry axis inter- 0 30 R7 500
cept
Finishing axis inter- 30 0 R5 400
cept
Intersection 13.64 21.82 R7 910
Optimal solution: Produce 13.64 chairs and 21.82 tables per week for a maximum
profit of R7 910 (rounded to whole units: 14 chairs, 22 tables).
⋆ Exam Tip
At the exam, always check all corner points of the feasible region. The optimal
is never in the interior — it always lies on a corner (vertex). If results are frac-
tional and the problem deals with whole units, note the rounding but accept the
continuous LP answer unless asked for integer programming.
Page 3 of 23
]]
⋆
DSC2602: Rational Decision Making
May/June Examination 2026 — Comprehensive Revision Guide
(Based on May/June 2025 & May/June 2023 Past Papers)
⋆ ⋄ ⋆ ⋄ ⋆ ⋄ ⋆ ⋄ ⋆
[ Decision Sciences / Quantitative Management [
_ Exam Revision Guide
DSC2602
Module Code:
Rational Decision Making
Module Name:
May/June 2025 & May/June 2023
Paper / Exam:
2026
Year:
100
Total Marks:
Economic & Management Sciences
Faculty:
Study this guide carefully. Focus on understanding methods, not just memorising
formulas — examiners test application.
Exam Revision Notes | DSC2602 | 2026
,DSC2602 | Exam Revision Rational Decision Making
Question 1 [25 marks]
(a) [10 marks]
Question: A furniture company produces two products: chairs and tables. Each chair
requires 3 hours of carpentry and 2 hours of finishing. Each table requires 5 hours of
carpentry and 1.5 hours of finishing. The company has 150 hours of carpentry and 60
hours of finishing available per week. The profit contribution is R180 per chair and R250
per table. Define the decision variables and formulate a Linear Programming (LP) model
to maximise weekly profit.
Answer: Decision Variables:
• Let x1 = number of chairs produced per week
• Let x2 = number of tables produced per week
Objective Function (Maximise Profit):
Maximise Z = 180x1 + 250x2
Subject to the constraints:
3x1 + 5x2 ≤ 150 (carpentry hours)
2x1 + 1.5x2 ≤ 60 (finishing hours)
x1 , x2 ≥ 0 (non-negativity)
Key Concept
An LP model always has three parts: decision variables (what we control), an
objective function (what we optimise), and constraints (resource or demand
limits). Non-negativity constraints are always required.
(b) [10 marks]
Question: Solve the LP model in (a) using the graphical method. Show the feasible
region and identify the optimal solution.
Page 2 of 23
, DSC2602 | Exam Revision Rational Decision Making
Answer: Step 1: Find boundary line intersections with axes.
For 3x1 + 5x2 = 150: when x1 = 0, x2 = 30; when x2 = 0, x1 = 50.
For 2x1 + 1.5x2 = 60: when x1 = 0, x2 = 40; when x2 = 0, x1 = 30.
Step 2: Find intersection of the two constraint lines.
From constraint 2: x1 = 60−1.5x2
2 = 30 − 0.75x2
Substitute into constraint 1:
3(30 − 0.75x2 ) + 5x2 = 150
90 − 2.25x2 + 5x2 = 150 ⇒ 2.75x2 = 60 ⇒ x2 = 21.82
x1 = 30 − 0.75(21.82) = 30 − 16.36 = 13.64
Step 3: Evaluate the objective function at each corner of the feasible region.
Corner Point x1 x2 Z =
180x1 +
250x2
Origin 0 0 R0
Carpentry axis inter- 0 30 R7 500
cept
Finishing axis inter- 30 0 R5 400
cept
Intersection 13.64 21.82 R7 910
Optimal solution: Produce 13.64 chairs and 21.82 tables per week for a maximum
profit of R7 910 (rounded to whole units: 14 chairs, 22 tables).
⋆ Exam Tip
At the exam, always check all corner points of the feasible region. The optimal
is never in the interior — it always lies on a corner (vertex). If results are frac-
tional and the problem deals with whole units, note the rounding but accept the
continuous LP answer unless asked for integer programming.
Page 3 of 23
