A company plans x radio advertisements and y newspaper advertisements. A radio advertisement costs Rs. 4000 and reaches 5000 people, and a newspaper advertisement costs Rs. 6000 and reaches 8000 people. The budget is Rs. 60000, at most 12 advertisements can be placed in all, and at least 3 of them must be on the radio. If each advertisement reaches different people, then the greatest number of people that can be reached is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 79000
The linear programming problem is
Maximise \(Z = 5000x + 8000y\)
subject to
\(4000x + 6000y \le 60000\), that is, \(2x + 3y \le 30\)
\(x + y \le 12\)
\(x \ge 3, \; y \ge 0\)
Meeting points of the boundary lines:
\(2x + 3y = 30\) and \(x + y = 12\): \(y = 6, \; x = 6\)
\(2x + 3y = 30\) and \(x = 3\): \(y = 8\)
The corner points are (3, 0), (12, 0), (6, 6) and (3, 8). Values of Z:
\(Z(3, 0) = 15000\)
\(Z(12, 0) = 60000\)
\(Z(6, 6) = 30000 + 48000 = 78000\)
\(Z(3, 8) = 15000 + 64000 = 79000\)
So the greatest reach is 79000 people, with 3 radio and 8 newspaper advertisements.
Why the other options are incorrect
Option (1) 60000 — Incorrect. This is the reach with 12 radio advertisements only, at (12, 0).
Option (2) 78000 — Incorrect. This is the value at (6, 6), which is 1000 less than at (3, 8).
Option (3) 80000 — Incorrect. This needs 10 newspaper advertisements and no radio one, which breaks \(x \ge 3\).