Cylindrical Container Optimization Minimizing Material Usage For 200 Cm³ Volume
In the realm of mathematical optimization, a classic problem involves determining the most efficient way to construct a container. This article delves into such a problem, specifically focusing on a closed cylindrical container designed to hold a fixed volume of liquid. Our primary objective is to discover the dimensions – the radius and height – that minimize the amount of material required to manufacture the container. This is a practical problem with applications in various industries, from packaging and manufacturing to engineering and design. By employing calculus and optimization techniques, we can arrive at the optimal dimensions that balance the container's volume with the surface area, leading to significant cost savings and resource efficiency. Understanding the mathematical principles behind this optimization problem allows us to make informed decisions in real-world scenarios, contributing to sustainable and economical solutions. So, let's embark on this mathematical journey to explore the intricacies of cylindrical container optimization and unlock the secrets of efficient design.
The core of this optimization problem lies in the need to construct a closed cylindrical container with a specific volume while using the least amount of material possible. We are given that the container must hold 200 cubic centimeters (cm³) of liquid. Our task is to find the radius (r) and height (h) of the cylinder that will achieve this volume using the minimum surface area. This challenge involves a delicate balance between the dimensions of the cylinder. A taller, narrower cylinder might have a smaller circumference but a larger height, while a shorter, wider cylinder might have a larger circumference but a smaller height. Our goal is to find the sweet spot where the combination of radius and height results in the smallest possible surface area, thus minimizing material usage. This problem exemplifies a common engineering challenge where optimizing resources is crucial for both cost-effectiveness and sustainability. By applying mathematical principles, we can determine the ideal dimensions for the cylindrical container, ensuring that we meet the volume requirement while minimizing the material footprint. This optimization process not only saves resources but also contributes to environmentally conscious design practices.
Setting Up the Equations: Volume and Surface Area
To mathematically address the problem of minimizing material usage for our cylindrical container, we must first establish the fundamental equations that govern its volume and surface area. The volume (V) of a cylinder is given by the formula: V = πr²h, where r represents the radius of the base and h represents the height of the cylinder. In our specific scenario, we know that the volume must be 200 cm³, so we can write: πr²h = 200. This equation provides a crucial link between the radius and height, allowing us to express one variable in terms of the other. Next, we need to consider the surface area (A) of the closed cylindrical container. This includes the areas of the two circular bases (top and bottom) and the lateral surface (the curved side). The area of each circular base is πr², and since there are two bases, their combined area is 2πr². The lateral surface area is given by 2πrh, which represents the circumference of the base multiplied by the height. Therefore, the total surface area of the closed cylinder is: A = 2πr² + 2πrh. This equation represents the quantity we want to minimize – the amount of material used to construct the container. By expressing both the volume and surface area in terms of the radius and height, we have laid the groundwork for applying optimization techniques to find the dimensions that minimize material usage.
Expressing Height in Terms of Radius
With the equations for volume and surface area established, our next step is to simplify the problem by expressing one variable in terms of the other. In this case, it is convenient to express the height (h) in terms of the radius (r) using the volume constraint. We know that the volume of the cylinder must be 200 cm³, so we have the equation: πr²h = 200. To isolate h, we can divide both sides of the equation by πr², resulting in: h = 200 / (πr²). This equation provides a direct relationship between the height and the radius, allowing us to eliminate h from the surface area equation. By expressing h in terms of r, we reduce the surface area equation to a function of a single variable, which is a crucial step in the optimization process. This simplification allows us to use calculus techniques, such as finding the derivative and setting it to zero, to determine the value of r that minimizes the surface area. The ability to manipulate equations and express variables in terms of each other is a fundamental skill in mathematical problem-solving, and it plays a key role in optimizing various real-world scenarios.
Substituting into the Surface Area Equation
Now that we have expressed the height (h) in terms of the radius (r) as h = 200 / (πr²), we can substitute this expression into the surface area equation. Recall that the surface area equation is given by: A = 2πr² + 2πrh. By replacing h with 200 / (πr²), we obtain a new equation for the surface area that is solely a function of r: A(r) = 2πr² + 2πr(200 / (πr²)). This substitution is a critical step in simplifying the optimization problem. It transforms the surface area equation from a function of two variables (r and h) into a function of a single variable (r). This simplification makes it possible to apply single-variable calculus techniques to find the minimum surface area. By performing the substitution, we have effectively reduced the complexity of the problem, making it more amenable to mathematical analysis. The resulting equation, A(r) = 2πr² + 2πr(200 / (πr²)), represents the surface area of the cylindrical container as a function of its radius. Our next task is to find the value of r that minimizes this function, which will give us the optimal radius for the container.
Simplifying the Surface Area Equation
After substituting the expression for height (h) in terms of radius (r) into the surface area equation, we obtained: A(r) = 2πr² + 2πr(200 / (πr²)). The next step is to simplify this equation to make it easier to work with. We can start by simplifying the second term: 2πr(200 / (πr²)) = 400πr / (πr²) = 400 / r. Therefore, the simplified surface area equation becomes: A(r) = 2πr² + 400 / r. This simplified equation is much more manageable for differentiation, which is the key to finding the minimum surface area. By simplifying the equation, we have reduced the number of terms and eliminated the π symbol from the denominator of the second term. This makes the equation less cluttered and easier to differentiate. The simplified surface area equation, A(r) = 2πr² + 400 / r, represents the relationship between the surface area of the cylindrical container and its radius in a clear and concise form. This simplification is a crucial step in preparing the equation for optimization, allowing us to proceed with the calculus-based approach to find the optimal radius.
Finding the Derivative of the Surface Area Function
To find the minimum surface area of the cylindrical container, we need to employ calculus techniques, specifically finding the derivative of the surface area function. Our simplified surface area equation is: A(r) = 2πr² + 400 / r. To find the derivative, A'(r), we differentiate each term with respect to r. The derivative of 2πr² with respect to r is 4πr. The derivative of 400 / r, which can be rewritten as 400r⁻¹, with respect to r is -400r⁻², or -400 / r². Therefore, the derivative of the surface area function is: A'(r) = 4πr - 400 / r². This derivative represents the rate of change of the surface area with respect to the radius. By finding the points where the derivative is equal to zero, we can identify potential minimum or maximum points of the surface area function. The process of differentiation is a fundamental tool in calculus and is essential for optimization problems. By finding the derivative of the surface area function, we have taken a crucial step towards determining the optimal radius that minimizes the material required to construct the cylindrical container.
Setting the Derivative to Zero and Solving for Radius
Having found the derivative of the surface area function, A'(r) = 4πr - 400 / r², our next step is to find the critical points by setting the derivative equal to zero and solving for the radius (r). This is a standard technique in calculus for finding potential minimum or maximum points of a function. Setting the derivative to zero gives us the equation: 4πr - 400 / r² = 0. To solve for r, we first add 400 / r² to both sides: 4πr = 400 / r². Next, we multiply both sides by r²: 4πr³ = 400. Now, we divide both sides by 4π: r³ = 100 / π. Finally, we take the cube root of both sides to find the value of r: r = ∛(100 / π). This value of r represents the radius at which the surface area function has a critical point. To determine whether this critical point corresponds to a minimum or maximum, we can use the second derivative test or analyze the behavior of the function around this point. However, in this context, we are seeking to minimize the surface area, so we expect this critical point to represent a minimum. The solution r = ∛(100 / π) is a crucial result, as it provides the optimal radius for the cylindrical container that minimizes material usage.
Calculating the Optimal Radius
In the previous step, we derived the expression for the optimal radius (r) as r = ∛(100 / π). Now, we need to calculate the numerical value of this expression. Using a calculator, we can approximate the value of π as 3.14159. Therefore, 100 / π ≈ 100 / 3.14159 ≈ 31.831. Taking the cube root of this value, we get: r ≈ ∛31.831 ≈ 3.175 centimeters. This value represents the radius of the cylindrical container that will minimize the amount of material used to construct it, while still holding a volume of 200 cm³. The optimal radius is a crucial parameter in determining the overall dimensions of the container. It provides a starting point for calculating the corresponding optimal height and ultimately designing the most efficient container. By calculating the numerical value of the optimal radius, we have moved closer to the final solution of the optimization problem.
Determining the Optimal Height
Now that we have calculated the optimal radius (r) to be approximately 3.175 centimeters, we can determine the corresponding optimal height (h) using the equation we derived earlier: h = 200 / (πr²). Substituting the value of r into this equation, we get: h = 200 / (π(3.175)²) ≈ 200 / (3.14159 * 10.080625) ≈ 200 / 31.676 ≈ 6.314 centimeters. This value represents the optimal height of the cylindrical container that, in conjunction with the optimal radius, will minimize the surface area and thus the material used. The optimal height is the final piece of the puzzle in determining the dimensions of the most efficient cylindrical container. By calculating both the optimal radius and the optimal height, we have successfully solved the optimization problem. These dimensions provide a practical guide for constructing a cylindrical container that meets the volume requirement of 200 cm³ while minimizing material usage. The results demonstrate the power of calculus and optimization techniques in solving real-world engineering and design challenges.
Verifying the Minimum Surface Area
To ensure that the dimensions we have calculated indeed correspond to a minimum surface area, we can employ the second derivative test. The second derivative of the surface area function, A''(r), can be found by differentiating the first derivative, A'(r) = 4πr - 400 / r², with respect to r. The derivative of 4πr with respect to r is 4π. The derivative of -400 / r², which can be rewritten as -400r⁻², with respect to r is 800r⁻³, or 800 / r³. Therefore, the second derivative of the surface area function is: A''(r) = 4π + 800 / r³. To apply the second derivative test, we evaluate A''(r) at the optimal radius, r ≈ 3.175 cm: A''(3.175) ≈ 4π + 800 / (3.175)³ ≈ 4π + 800 / 31.975 ≈ 4π + 25.019. Since 4π is positive and 25.019 is positive, their sum is also positive. A positive second derivative indicates that the surface area function has a local minimum at the given radius. This confirms that the dimensions we calculated, with a radius of approximately 3.175 cm and a height of approximately 6.314 cm, do indeed minimize the surface area of the cylindrical container. The second derivative test provides a rigorous mathematical verification of our solution, giving us confidence that we have found the optimal dimensions for the container.
In conclusion, we have successfully determined the dimensions of a closed cylindrical container that minimizes the amount of material required to hold 200 cm³ of liquid. By applying calculus and optimization techniques, we found the optimal radius to be approximately 3.175 centimeters and the optimal height to be approximately 6.314 centimeters. These dimensions were derived by setting up equations for the volume and surface area of the cylinder, expressing the height in terms of the radius, finding the derivative of the surface area function, and setting the derivative to zero to find critical points. The second derivative test was used to verify that the calculated dimensions indeed correspond to a minimum surface area. This problem highlights the practical applications of calculus in optimizing real-world scenarios, such as minimizing material usage in manufacturing and design. The solution not only provides the most efficient dimensions for the container but also demonstrates the importance of mathematical principles in achieving resource efficiency and cost savings. By understanding and applying optimization techniques, we can make informed decisions in various industries, contributing to sustainable and economical solutions.
