ECZ 2020 · O-Level Paper 2 · Question 12 — Linear Programming

A carpenter intends to manufacture at least 10 tables and at least 20 chairs. Each table requires 4 hours of assembling and 2 hours of varnishing. Each chair requires 3 hours of assembling and 1 hour of varnishing. There are 240 hours available for assembling and 100 hours for varnishing. Let x represent the number of tables and y the number of chairs.

(a) Given that x represents the number of tables and y the number of chairs, write four inequalities which represent these conditions.[4]

Verified working — step 1

At least 10 tables: x >= 10.

(b) Using a scale of 2 cm to represent 10 pieces of furniture on each axis, draw the x and y axes for 0 <= x <= 70 and 0 <= y <= 100 respectively and shade the unwanted region to show clearly the region where the solution of the inequalities lie.[4]

Verified working — step 1

Draw x = 10 (vertical), y = 20 (horizontal), 4x + 3y = 240 through (60, 0) and (0, 80), and 2x + y = …

(c) Each table sold yields a profit of K300.00 while each chair sold yields a profit of K250.00. Find the best combination of the number of tables and chairs to gain the maximum profit.[2]

Verified working — step 1

Profit P = 300x + 250y. Because tables and chairs are whole items, test the lattice points on the binding line 4x + 3y = 240:

(d) Calculate this estimate of the maximum profit.[2]

Verified working — step 1

P = 300(12) + 250(64) = 3600 + 16000 = …

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Original examination question © Examinations Council of Zambia. Worked solution and commentary © G12 Titan — not to be reproduced without permission.

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