Step 1 :Let x be the number of adults and y be the number of students.
Step 2 :Set up the constraint equations: \(x+y \le 210\), \(2y \le x\).
Step 3 :Set up the objective function, which represents the total proceeds: \(P(x,y) = 12x + 6y\).
Step 4 :Solve the inequality \(2y \le x\) to find the relationship between x and y: \(y \ge \frac{1}{2}x\).
Step 5 :Since the number of participants cannot exceed 210, we have \(x + y = 210\). Substitute the previous relationship: \(x + \frac{1}{2}x = 210\).
Step 6 :Solve the equation for x: \(\frac{3}{2}x = 210\) => \(x = 140\).
Step 7 :Find y using the equation \(x + y = 210\) => \(y = 70\).
Step 8 :The maximum proceeds can be obtained with 140 adults and 70 students attending.