ITPEC FE Morning October 2021 Question 79
Company A has two media plans. One plan is for a series of 30-second radio advertisements (ads) and the other is for 60-second TV ads. Exposure is rated in points. The company is considering an exposure of at least 60 points in the city and at most 40 points outside the city. The table below shows the cost and exposure of both TV and radio ads. Which of the following is the set of expressions that can determine the number of media ads to minimize the total cost?
Explanation
City points (8 per TV ad, 10 per radio ad) at least 60, outside points (6 and 3) at most 40 — each constraint is built from one column of the table, with the direction taken from "at least" and "at most".
Let x be the number of TV ads and y the number of radio ads, as in the objective 2000x + 800y.
- •In the city: TV gives 8 points and radio 10, at least 60 points: 8x + 10y ≥ 60
- •Outside the city: TV gives 6 points and radio 3, at most 40 points: 6x + 3y ≤ 40
- •Counts cannot be negative: x ≥ 0, y ≥ 0
- •Total cost 2000x + 800y is minimized
Answer (d)
Why not others:
- •(a) — builds each constraint from a row of the table (TV's 8 and 6, radio's 10 and 3) and turns "at least 60" into ≤
- •(b) — directions are right, but the coefficients again come from the rows, so one constraint mixes TV's city and outside points
- •(c) — coefficients are right, but the city requirement is written as ≤ 60 although it says at least 60
Key rule: In a linear programming set-up, one constraint per resource or requirement, summing each variable's contribution to it; "at least" gives ≥ and "at most" gives ≤.
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