### Instruction:
A shipping company is trying to optimize its container loading process. Each ship can carry up to 2,000 containers, with a maximum weight capacity of 50,000 tons. The company has three types of containers: Type A (20 tons), Type B (15 tons), and Type C (10 tons). Determine the optimal number of each container type to maximize the ship's loading capacity without exceeding the weight limit. Explain your reasoning.


### Response:

To maximize the ship's loading capacity, we should try to fill as much of the 50,000-ton weight limit as possible while also trying to reach the maximum number of containers (2,000). 

Let x be the number of Type A containers, y be the number of Type B containers, and z be the number of Type C containers. We have two constraints:

1) Weight constraint: 20x + 15y + 10z ≤ 50,000
2) Container constraint: x + y + z ≤ 2,000

We want to maximize the total weight, so our objective function is W = 20x + 15y + 10z.

First, let's simplify the weight constraint by dividing both sides by 5:
4x + 3y + 2z ≤ 10,000

Now, let's analyze the problem using some logic. Since Type A containers are the heaviest, we should prioritize filling the ship with them first. However, if we only load Type A containers, we would need 2,500 of them to reach the weight limit, which exceeds the container limit. So, we must include lighter containers as well.

Suppose we start with 1,000 Type A containers. This takes up half of the container space and contributes 20,000 tons to the weight. Now we have:

4(1000) + 3y + 2z ≤ 10,000
1,000 + y + z ≤ 2,000

Solving for y and z in these equations, we get:

3y + 2z ≤ 6,000
y + z ≤ 1,000

From the second equation, we can see that we still have room for 1,000 more containers. To maximize the weight, we should now prioritize filling the remaining space with Type B containers. Suppose we fill the ship with 1,000 Type B containers:

3(1000) + 2z ≤ 6,000
z ≤ 1,000

Now, we have reached the container limit and only need to consider the weight constraint:

3,000 + 2z ≤ 6,000
2z ≤ 3,000
z ≤ 1,500

Since z must be a whole number, the maximum value for z is 1,500. However, this would mean that we are not using any Type C containers, which contradicts our assumption of including all three types of containers.

So, let's try another approach: instead of maximizing the number of Type A containers first, let's maximize the number of Type B containers while still having some Type A containers on board.

Suppose we start with 500 Type A containers. This contributes 10,000 tons to the weight and leaves room for 1,500 more containers:

4(500) + 3y + 2z ≤ 10,000
500 + y + z ≤ 2,000

Solving for y and z in these equations, we get:

3y + 2z ≤ 8,000
y + z ≤ 1,500

Now, let's prioritize filling the remaining space with Type B containers. Suppose we fill the ship with 1,000 Type B containers:

3(1000) + 2z ≤ 8,000
z ≤ 500

Now, we have reached the container limit again and only need to consider the weight constraint:

3,000 + 2z ≤ 8,000
2z ≤ 5,000
z ≤ 2,500

Since z must be a whole number, the maximum value for z is 2,500. However, this would mean that we are not using any Type C containers, which contradicts our assumption of including all three types of containers.

So, let's try one more approach: instead of maximizing the number of Type B containers first, let's maximize the number of Type C containers while still having some Type A and Type B containers on board.

Suppose we start with 500 Type A containers and 500 Type B containers. This contributes 17,500 tons to the weight and leaves room for 1,000 more containers:

4(500) + 3(500) + 2z ≤ 10,000
1,000 + z ≤ 2,000

Solving for z in these equations, we get:

2z ≤ 2,500
z ≤ 1,250

Now, we have reached the container limit again and only need to consider the weight constraint:

2(1250) ≤ 2,500

This time, we can see that we are able to include all three types of containers without violating either constraint. The optimal solution is to load 500 Type A containers, 500 Type B containers, and 1,000 Type C containers. This results in a total weight of 20(500) + 15(500) + 10(1000) = 10,000 + 7,500 + 10,000 = 27,500 tons.