Answer: 0 bottles of Brand X and 10 bottles of Brand Y
(please let me know if I read the question wrong and I will do my best to fix it)
Explanation:
This problem can be solved using linear programming. Let x be the number of Brand X bottles and y be the number of Brand Y bottles. The objective is to minimize the cost, given by:
Cost = 25x + 15y
Subject to the following constraints, which represent the minimum nutrient requirements:
3x + 9y ≥ 30 (Nutrient A)
3x + 2y ≥ 16 (Nutrient B)
7x + 2y ≥ 24 (Nutrient C)
x ≥ 0, y ≥ 0 (Non-negativity constraint)
We can solve this system of inequalities graphically by finding the feasible region and the corner points.
Nutrient A constraint (3x + 9y ≥ 30):
Divide the inequality by 3:
x + 3y ≥ 10
Nutrient B constraint (3x + 2y ≥ 16):
Divide the inequality by 2:
1.5x + y ≥ 8
y ≥ 8 - 1.5x
Nutrient C constraint (7x + 2y ≥ 24):
Divide the inequality by 2:
3.5x + y ≥ 12
y ≥ 12 - 3.5x
Now we will graph the inequalities on the xy-plane:
y ≥ 10 - x/3
y ≥ 8 - 1.5x
y ≥ 12 - 3.5x
The feasible region is the area where all three inequalities are satisfied. It's a triangular region with corner points A(0,10), B(4,4), and C(8/3, 2).
Now we need to find the cost at each corner point:
Cost A (0, 10) = 25(0) + 15(10) = $150
Cost B (4, 4) = 25(4) + 15(4) = $160
Cost C (8/3, 2) = 25(8/3) + 15(2) ≈ $153.33
The minimum cost is $150, which occurs when 0 bottles of Brand X and 10 bottles of Brand Y are used. So, the public aquarium should add 0 bottles of Brand X and 10 bottles of Brand Y to satisfy the needs of the reef tank at the minimum possible cost.