Pool Dimensions: Length, Width, And Cost Calculation

Hey guys! Let's dive deep into a cool problem: figuring out the perfect dimensions for a swimming pool while keeping an eye on the cost. We're going to tackle a scenario where we need to build a pool that's twice as long as it is wide, has a specific capacity, and a uniform depth. Plus, we need to factor in the cost of waterproofing. Sounds like a fun challenge, right? So, let’s break it down step by step.

Understanding the Problem: Pool Dimensions and Capacity

Okay, so the core of this problem lies in understanding the relationships between the pool's dimensions – length, width, and depth – and its capacity, which is the volume of water it can hold. When we say the pool is twice as long as it is wide, we’re setting up a crucial relationship that we can use in our calculations. Think of it this way: if the width is 'x', then the length is '2x'. This simple relationship is going to be super helpful.

Next up, we've got the capacity, which is given as 2,500 cubic meters (m³). Remember, the volume of a rectangular prism (which is the shape of most swimming pools) is calculated by multiplying its length, width, and depth. So, we have a target volume that we need to achieve. This volume is directly linked to the dimensions, and this link is what we'll use to solve for our unknowns. We want a pool that holds exactly 2,500 m³ of water, no more, no less.

Lastly, the pool has a uniform depth. This means the depth is constant throughout the pool, making our calculations a whole lot easier. If the depth varied, we'd be dealing with a much more complex problem, maybe even involving calculus! But thankfully, we can keep it simple with basic geometry. This uniform depth is another crucial parameter that will help us define the pool’s overall dimensions. We’re aiming for a design that not only meets the capacity requirement but also maintains this consistent depth for a better swimming experience. To put it simply, we need a pool that's long, wide, deep, and holds 2,500 m³ of water, all while maintaining a consistent depth throughout.

Setting Up the Equations: Linking Dimensions and Volume

Alright, let’s get a bit mathematical here, but don't worry, we'll keep it straightforward! We need to translate the problem's conditions into equations. This is where the magic happens, guys. We can convert the words into mathematical statements that we can actually work with.

First, let's assign some variables. Let's say:

• w = width of the pool (in meters)
• l = length of the pool (in meters)
• d = depth of the pool (in meters)

From the problem, we know that the length is twice the width. So, we can write our first equation as:

l = 2w

This equation is our foundation. It directly relates the length and width, which is super important. It means that whatever we find for the width, the length will just be double that. Simple, right?

Next, we know the volume (V) of the pool should be 2,500 m³. The volume of a rectangular prism is given by:

V = l * w * d

Since we know V = 2,500 m³, we can write:

2500 = l * w * d

Now, let's substitute the first equation (l = 2w) into the volume equation. This will reduce the number of variables we're dealing with and make our lives easier:

2500 = (2w) * w * d

Simplifying this, we get:

2500 = 2w²d

This is our key equation! It relates the width, depth, and the volume of the pool. We have one equation with two unknowns (w and d). This means we can't solve for specific values just yet, but we've made some solid progress. We’ve successfully linked the dimensions to the volume, setting the stage for the next step. We’ve got the mathematical backbone of the problem laid out, and we’re ready to move forward.

Factoring in the Waterproofing Cost: Surface Area Calculation

Okay, now let's bring in the money! We need to think about the cost of waterproofing the pool's surface. This is where things get a bit more practical. Waterproofing cost depends on the surface area that needs to be covered. So, we need to calculate the surface area of the pool.

The pool has a rectangular base and four sides. The surface area (SA) we need to waterproof includes the bottom and the four sides, but not the top (since it's open to the air). The formula for this surface area is:

SA = lw + 2ld + 2wd

Where:

• lw is the area of the bottom
• 2ld is the area of the two longer sides
• 2wd is the area of the two shorter sides

Now, let's substitute l = 2w into the surface area equation:

SA = (2w)w + 2(2w)d + 2wd

Simplifying this, we get:

SA = 2w² + 4wd + 2wd

SA = 2w² + 6wd

This is the total area we need to waterproof. The cost of waterproofing is given as 80 € per square meter. So, the total waterproofing cost (C) can be calculated as:

C = 80 * SA

Substituting the expression for SA, we get:

C = 80 * (2w² + 6wd)

C = 160w² + 480wd

This equation tells us how the total waterproofing cost depends on the width (w) and depth (d) of the pool. We’ve now incorporated the economic aspect of the problem. The cost is directly related to the dimensions, which adds another layer to our optimization challenge. We need to find dimensions that not only satisfy the volume requirement but also keep the waterproofing cost reasonable. We've successfully linked the surface area and the cost, and we’re getting closer to a complete solution.

Optimizing Dimensions: Balancing Cost and Capacity

Here's where we put our thinking caps on and optimize the dimensions. This is the fun part where we try to find the best solution that balances capacity and cost. Remember, we have two equations:

1. 2500 = 2w²d (from the volume requirement)
2. C = 160w² + 480wd (from the waterproofing cost)

We want to find the values of w and d that satisfy the first equation while minimizing the cost C. This is an optimization problem. We can approach this in a couple of ways. One way is to solve the volume equation for one variable (say, d) and substitute it into the cost equation. This will give us the cost as a function of a single variable (w), which we can then minimize.

Let's solve the volume equation for d:

2500 = 2w²d

d = 2500 / (2w²)

d = 1250 / w²

Now, substitute this expression for d into the cost equation:

C = 160w² + 480w(1250 / w²)

C = 160w² + 480 * 1250 / w

C = 160w² + 600000 / w

Now we have the cost C as a function of just the width w. To minimize this cost, we can use calculus (taking the derivative and setting it to zero) or we can use numerical methods or graphing tools to find the minimum. For those not keen on calculus, think of it as finding the lowest point on a curve. We’re looking for the width that gives us the smallest possible cost.

The goal here is to strike a balance. A very wide and shallow pool might have a large surface area, leading to high waterproofing costs. On the other hand, a narrow and deep pool might be more challenging to construct and maintain. So, we need to find that sweet spot where we get the best of both worlds. This optimization step is where we really dig into the practicality of the problem, making sure we’re not just meeting the requirements but also doing it in the most efficient and cost-effective way.

Finding the Optimal Dimensions: Practical Considerations

Okay, let's get to the practical side of things. Once we have our equation for cost as a function of width, we need to find the width that minimizes the cost. As mentioned earlier, you can use calculus, graphing tools, or numerical methods to find this minimum. For the sake of this explanation, let's assume we've done the math (or used a tool) and found an optimal width.

Let’s say, for example, that our calculations show the optimal width w is approximately 8.4 meters. Now we can find the other dimensions:

• Length l = 2w = 2 * 8.4 = 16.8 meters
• Depth d = 1250 / w² = 1250 / (8.4)² ≈ 17.7 meters

So, we have a pool that's roughly 8.4 meters wide, 16.8 meters long, and 17.7 meters deep. Wait a minute! A depth of 17.7 meters? That’s incredibly deep for a typical swimming pool! This highlights an important point: mathematical solutions need to be checked against practical considerations. While our calculations might give us a minimum cost, the resulting dimensions might not be realistic or safe for a swimming pool.

In the real world, pool depths usually range from 1 to 3 meters for recreational use. A depth of 17.7 meters is more suitable for deep-sea diving! This means we need to rethink our approach. The mathematical minimum might not be the best practical solution.

Instead, we might need to set a maximum depth based on practical considerations and then recalculate the other dimensions and the cost. For example, let's say we decide that a maximum depth of 2.5 meters is reasonable. We can then use the volume equation to find the corresponding width:

2500 = 2w²d

2500 = 2w² * 2.5

2500 = 5w²

w² = 500

w ≈ 22.36 meters

And the length would be:

l = 2w = 2 * 22.36 ≈ 44.72 meters

Now we have a pool that's about 22.36 meters wide, 44.72 meters long, and 2.5 meters deep. These dimensions are much more practical for a swimming pool. We can then calculate the cost of waterproofing using these new dimensions. This step is crucial in bridging the gap between theoretical solutions and real-world applications. We’re not just solving equations; we’re designing a pool that is both functional and cost-effective.

Calculating the Final Cost: Putting It All Together

Alright, let's wrap things up by calculating the final waterproofing cost using our adjusted dimensions. We've already considered the practical aspects and set a reasonable depth, so now it's time to see how much it will actually cost to waterproof our pool.

We have the following dimensions:

• Width w ≈ 22.36 meters
• Length l ≈ 44.72 meters
• Depth d = 2.5 meters

We already have the equation for the waterproofing cost:

C = 160w² + 480wd

Let's plug in our values:

C = 160 * (22.36)² + 480 * 22.36 * 2.5

C = 160 * 500 + 480 * 22.36 * 2.5

C = 80000 + 26832

C = 106832 €

So, the estimated cost to waterproof the pool is approximately 106,832 €. This is a significant cost, and it's crucial to have an accurate estimate before starting construction.

This final step brings everything together. We’ve gone from understanding the problem, setting up equations, optimizing dimensions, considering practical limitations, and finally, calculating the cost. It’s a complete journey from theory to a tangible number. The cost is not just a number; it’s a reflection of all the decisions we’ve made along the way. By breaking down the problem into smaller, manageable steps, we’ve been able to tackle a complex challenge and arrive at a practical solution. We’ve not only calculated the cost but also gained a deep understanding of the factors that influence it.

Conclusion: Dimensions Determined!

So, there you have it! We’ve successfully navigated through the complexities of determining the dimensions of a swimming pool, considering its capacity, the relationship between length and width, uniform depth, and the cost of waterproofing. We started with the basic principles of geometry and algebra, set up equations, optimized for cost, and finally, factored in real-world constraints to arrive at a practical solution.

This exercise highlights the importance of not just mathematical accuracy but also practical considerations. We saw how a purely mathematical optimization might lead to unrealistic dimensions, emphasizing the need to balance theory with common sense. We’ve also learned the value of breaking down a complex problem into smaller, manageable parts. Each step, from setting up equations to calculating costs, builds upon the previous one, leading us to a comprehensive solution.

Remember, when tackling similar problems, it’s crucial to:

1. Clearly understand the problem and identify the key relationships.
2. Translate the problem into mathematical equations.
3. Optimize the solution based on the given constraints and objectives.
4. Always consider practical limitations and real-world implications.
5. Finally, calculate the results and interpret them in the context of the problem.

By following these steps, you can confidently tackle a wide range of challenges, whether it’s designing a swimming pool or optimizing any other real-world scenario. Keep practicing, keep questioning, and keep exploring the fascinating world of problem-solving! You guys nailed it! Now go take a swim in your newly designed pool! Well, maybe after it's built. 😉