Optimizing Dimensions Of Cylindrical Can Minimum Surface Area Problem
Introduction
In the realm of optimization problems, a classic challenge lies in determining the most efficient dimensions for a container. Specifically, we delve into finding the dimensions of the lightest open-top right circular cylindrical can that can hold a volume of 6,859 cm³. This exploration involves a blend of geometry and calculus, where we aim to minimize the surface area (and thus the material used) while maintaining a fixed volume. Understanding the principles behind this optimization can be applied to various real-world scenarios, from manufacturing to packaging design. This guide provides a step-by-step approach to solving this problem, ensuring clarity and practical application. By mastering this, one gains a deeper understanding of optimization techniques and their significance in engineering and design.
Problem Statement: Minimizing Material for a Cylindrical Can
Our central question revolves around a specific scenario: What are the ideal dimensions—radius and height—of an open-top right circular cylindrical can that holds a volume of 6,859 cm³, while using the least amount of material? This is a quintessential optimization problem where we seek to minimize the surface area of the can, which directly corresponds to the amount of material required. The constraint we have is the fixed volume. This type of problem is not just an academic exercise; it has profound implications for industries where material costs are a significant factor. By optimizing the dimensions, manufacturers can reduce material usage, leading to cost savings and environmental benefits. The problem's complexity arises from the interplay between the radius and the height, both influencing the volume and surface area in different ways. The challenge is to find the sweet spot where the surface area is minimized for the given volume. To solve this, we need to formulate mathematical relationships for volume and surface area, and then employ calculus to find the minimum surface area. This involves setting up equations, finding derivatives, and solving for critical points, a process that demonstrates the power of calculus in solving real-world optimization problems.
Setting Up the Equations: Volume and Surface Area
To tackle this optimization challenge, we first need to establish the mathematical foundation. We'll begin by defining the formulas for the volume and surface area of an open-top right circular cylinder. 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 case, the volume is fixed at 6,859 cm³. This gives us our constraint equation:
πr²h = 6,859.
Next, we need to formulate the equation for the surface area (SA) of the open-top cylinder. Since the can has no top, the surface area consists of the circular base and the lateral surface. The area of the circular base is πr², and the lateral surface area is 2πrh. Therefore, the total surface area is:
SA = πr² + 2πrh.
Our goal is to minimize this surface area. However, we have two variables, 'r' and 'h'. To solve this optimization problem, we need to express the surface area in terms of a single variable. We can achieve this by using the volume constraint equation to express 'h' in terms of 'r' (or vice versa) and substituting it into the surface area equation. This step is crucial as it transforms the problem into a single-variable optimization, making it solvable using basic calculus techniques.
Expressing Height in Terms of Radius
Having established the volume and surface area equations, the next crucial step in our optimization process is to express the height ('h') of the cylinder in terms of its radius ('r'). This allows us to reduce the surface area equation to a single variable, making it amenable to calculus-based minimization. We start with the volume constraint equation:
πr²h = 6,859.
To isolate 'h', we divide both sides of the equation by πr², resulting in:
h = 6,859 / (πr²).
This equation now gives us the height of the cylinder as a function of its radius. This is a critical step because it allows us to eliminate 'h' from the surface area equation, which currently depends on both 'r' and 'h'. By substituting this expression for 'h' into the surface area equation, we will obtain a new equation for the surface area that is solely a function of 'r'. This transformation is a standard technique in optimization problems involving constraints, as it simplifies the problem and allows us to apply single-variable calculus methods. The ability to manipulate equations and express variables in terms of others is a fundamental skill in optimization and mathematical problem-solving.
Substituting into the Surface Area Equation
Now that we have expressed the height ('h') in terms of the radius ('r'), we can substitute this expression into the surface area (SA) equation. This substitution is a pivotal step in our optimization journey, as it reduces the problem to minimizing a function of a single variable. Recall the surface area equation for the open-top cylinder:
SA = πr² + 2πrh.
We substitute h = 6,859 / (πr²) into this equation:
SA = πr² + 2πr[6,859 / (πr²)].
Simplifying the equation, we get:
SA = πr² + (2 * 6,859) / r,
which can be further written as:
SA(r) = πr² + 13,718 / r.
Now we have the surface area as a function of 'r' alone. This is a significant achievement because it transforms our problem into a standard calculus optimization problem: finding the minimum of a single-variable function. The function SA(r) represents the total material needed for the can as a function of its radius. Our next step is to use calculus to find the value of 'r' that minimizes this function. This involves finding the derivative of SA(r) with respect to 'r', setting it equal to zero, and solving for 'r'.
Finding the Derivative of the Surface Area
With the surface area expressed as a function of the radius, SA(r) = πr² + 13,718 / r, we now embark on the calculus-based optimization process. The first step is to find the derivative of SA(r) with respect to 'r'. This derivative, denoted as SA'(r), will give us the rate of change of the surface area with respect to the radius. Setting this derivative equal to zero will help us identify the critical points, which are potential points of minima or maxima.
To find the derivative, we differentiate each term of the equation:
SA(r) = πr² + 13,718r⁻¹.
The derivative SA'(r) is then:
SA'(r) = 2πr - 13,718r⁻²,
which can be rewritten as:
SA'(r) = 2πr - 13,718 / r².
This derivative is a crucial piece of the puzzle. It tells us how the surface area changes as we vary the radius. The points where SA'(r) equals zero are of particular interest because they correspond to points where the surface area is either at a minimum, a maximum, or a point of inflection. In our optimization problem, we are interested in finding the minimum surface area, so we will focus on the critical points that could represent a minimum. The next step is to set SA'(r) equal to zero and solve for 'r'.
Setting the Derivative to Zero and Solving for Radius
Having found the derivative of the surface area function, SA'(r) = 2πr - 13,718 / r², the next step in our optimization process is to find the critical points. These are the points where the derivative equals zero or is undefined. In our case, we set SA'(r) equal to zero and solve for 'r':
2πr - 13,718 / r² = 0.
To solve this equation, we first add 13,718 / r² to both sides:
2πr = 13,718 / r².
Next, we multiply both sides by r²:
2πr³ = 13,718.
Now, divide both sides by 2π:
r³ = 13,718 / (2π).
Finally, we take the cube root of both sides to solve for 'r':
r = ∛[13,718 / (2π)].
Calculating the value, we find:
r ≈ 13 cm.
This value of 'r' is a critical point of our surface area function. It represents a potential minimum (or maximum) of the surface area. To confirm that this value indeed corresponds to a minimum, we can use the second derivative test or analyze the behavior of SA'(r) around this point. However, in the context of this optimization problem, it is physically plausible that this critical point corresponds to a minimum surface area. We now have the optimal radius for our cylindrical can. The next step is to find the corresponding optimal height.
Calculating the Height
Now that we have determined the optimal radius (r ≈ 13 cm) for minimizing the surface area of our cylindrical can, the next step in this optimization problem is to calculate the corresponding height ('h'). We can find the height using the equation we derived earlier, which expresses 'h' in terms of 'r':
h = 6,859 / (πr²).
Substituting the value of r ≈ 13 cm into this equation, we get:
h = 6,859 / (π * 13²).
Calculating this, we find:
h ≈ 13 cm.
Interestingly, the optimal height is approximately equal to the optimal radius. This result is not coincidental and reflects a general principle in optimization problems involving cylinders: for a given volume, the surface area is often minimized when the height is equal to the radius (or a multiple thereof). This relationship stems from the geometry of the cylinder and how the surface area and volume scale with changes in radius and height. Now that we have both the optimal radius and the optimal height, we have effectively solved the optimization problem. We have found the dimensions of the lightest open-top right circular cylindrical can that will hold a volume of 6,859 cm³.
Verifying the Minimum Surface Area
While we have calculated the radius and height that should minimize the surface area, it is prudent to verify that this is indeed a minimum. In this optimization problem, we can use the second derivative test to confirm that our critical point corresponds to a minimum surface area. Recall the first derivative of the surface area function:
SA'(r) = 2πr - 13,718 / r².
To perform the second derivative test, we need to find the second derivative, SA''(r). Differentiating SA'(r) with respect to 'r', we get:
SA''(r) = 2π + 27,436 / r³.
Now, we evaluate SA''(r) at our critical point, r ≈ 13 cm:
SA''(13) = 2π + 27,436 / (13³).
Since both terms are positive, SA''(13) is positive. A positive second derivative at a critical point indicates that the function has a local minimum at that point. This confirms that the radius we calculated indeed minimizes the surface area. Alternatively, we could analyze the sign of SA'(r) around r = 13 cm. For r < 13 cm, SA'(r) is negative, and for r > 13 cm, SA'(r) is positive. This change in sign from negative to positive also indicates a local minimum. This verification step is crucial in optimization problems, as it ensures that we have found the desired minimum and not a maximum or an inflection point. We can now confidently state that the dimensions we calculated provide the lightest can for the given volume.
Conclusion: Optimal Dimensions for the Cylindrical Can
In conclusion, we have successfully navigated the optimization problem of finding the dimensions of the lightest open-top right circular cylindrical can that holds a volume of 6,859 cm³. Through a combination of geometric principles and calculus techniques, we determined that the optimal radius is approximately 13 cm and the optimal height is also approximately 13 cm. These dimensions minimize the surface area of the can, thereby minimizing the amount of material required to manufacture it.
Our journey involved several key steps: setting up the equations for volume and surface area, expressing height in terms of radius, substituting this expression into the surface area equation, finding the derivative of the surface area function, setting the derivative to zero to find critical points, solving for the radius, calculating the corresponding height, and verifying that the critical point indeed corresponds to a minimum using the second derivative test. This methodical approach is characteristic of optimization problems, where a clear understanding of the problem, careful formulation of equations, and application of appropriate mathematical tools are essential.
The result that the optimal height is equal to the radius is an interesting outcome that highlights a general principle in cylinder optimization. This principle can be valuable in practical applications, such as designing containers for various products, where minimizing material usage is a key consideration. The techniques and concepts explored in this problem are widely applicable in various fields, including engineering, manufacturing, and logistics. By mastering these skills, one can make informed decisions that lead to efficiency and cost savings.