Step 1 :First, we need to set up the constraints for the problem. The total number of ads placed in all three newspapers cannot exceed 9, so we have the constraint \(x_{1} + x_{2} + x_{3} \le 9\).
Step 2 :Next, we need to ensure that at least 11,000 people read the ads. This gives us the constraint \(2000x_{1} + 500x_{2} + 1000x_{3} \ge 11000\).
Step 3 :Since the number of ads cannot be negative, we also have the constraints \(x_{1} \ge 0\), \(x_{2} \ge 0\), and \(x_{3} \ge 0\).
Step 4 :The objective function to minimize is the total cost of the ads, which is \(200x_{1} + 100x_{2} + 50x_{3}\).
Step 5 :Now, we can solve this linear programming problem using a method such as the simplex method or graphical method. However, since this problem only involves three variables, we can also solve it by inspection.
Step 6 :By inspection, we can see that placing all 9 ads in the Tribune will reach 9,000 people, which is not enough. Placing 8 ads in the Tribune and 1 ad in the Sentinel will reach 10,000 people, which is still not enough. Placing 7 ads in the Tribune and 2 ads in the Sentinel will reach 11,000 people, which meets the requirement.
Step 7 :The cost of placing 7 ads in the Tribune and 2 ads in the Sentinel is \(7*50 + 2*200 = 350 + 400 = 750\) dollars.
Step 8 :By checking other combinations, we can see that no other combination can reach at least 11,000 people with a lower cost. Therefore, the company should place 2 ads in the Sentinel and 7 ads in the Tribune to minimize the advertising costs.
Step 9 :The minimum cost is \(\boxed{750}\) dollars.