To determine the optimal number of radio and television ads to maximize reach under the given constraints, let's set up the problem step-by-step.

Let:

• $x$ = the number of radio ads.
• $y$ = the number of television ads.

Objective Function:

We want to maximize the number of people reached, which can be represented as: $\text{Maximize } Z = 3000x + 7000y$

Constraints:

1. Budget Constraint: Each radio ad costs $200, and each television ad costs $500. The total budget is $40,000. $200x + 500y \leq 40000$

2. Minimum Ads Requirement: At least 10 ads of each type must be used. $x \geq 10$ $y \geq 10$

3. Radio Ads Requirement: The number of radio ads must be at least as great as the number of television ads. $x \geq y$

Now, we can solve this as a linear programming problem to find the values of $x$ and $y$ that maximize $Z$.

Step-by-Step Solution:

1. Rewrite the budget constraint: $200x + 500y \leq 40000$ Simplify by dividing each term by 100: $2x + 5y \leq 400$

2. Constraints Summary:

   • $2x + 5y \leq 400$
   • $x \geq 10$
   • $y \geq 10$
   • $x \geq y$

Solving the Constraints

Let's solve the system to determine feasible values for $x$ and $y$ that maximize $Z$.

Step-by-Step Calculation:

1. Substitute values for boundary cases under these constraints and solve for maximum reach $Z$:
   • If $x = 50$ and $y = 10$, both budget and constraint are satisfied with