Q.An advertising agency wants to reach at least 60 lakh people using two media, television and newspaper. Each television advertisement reaches 4 lakh people and costs Rs 50,000; each newspaper advertisement reaches 2 lakh people and costs Rs 20,000. Because of a budget rule, the agency must place at least 5 television advertisements. Find the number of advertisements of each type that minimizes the total cost while meeting the reach requirement.

Let $x=$ number of television ads and $y=$ number of newspaper ads.

Reach requirement. TV reaches 4 lakh/ad, newspaper 2 lakh/ad, total needed at least 60 lakh: $4x+2y\ge60$, i.e. $2x+y\ge30$.

Budget rule: $x\ge5$.

At $x=5$: $2(5)+y\ge30\Rightarrow y\ge20$, giving vertex $(5,20)$.

Where the reach line meets $y=0$: $2x\ge30\Rightarrow x\ge15$, giving vertex $(15,0)$ — check $x\ge5$: $15\ge5$ ✓. …

Simplify the reach inequality by the common factor 2, use the given lower bound on TV ads directly as one boundary, find the two …

Writing the reach constraint as ≤ (a cap) instead of ≥ (a minimum requirement); forgetting that $x\ge5$ repl …