91Ó°ÊÓ

The volume of a cylinder plays a crucial role in optimization problems within calculus. Understanding the volume equation will help us explore and solve various problems, such as finding the optimal design for containers. The volume \( V \,\) of a cylinder is calculated using:\[ V = \pi r^2 h \\]Where:

• \( V \,\) is the volume
• \( r \,\) is the radius of the base
• \( h \,\) is the height of the cylinder

For the exercise at hand, we are given a cylinder with a specific volume of 100 cm³. By accurately applying the formula, you can manage fixed capacity issues while working toward optimization of other parameters like surface area. Here, a key step is expressing the height in terms of the radius, which facilitates the further task of minimizing the surface area.

Minimizing the surface area of a cylinder is pivotal, especially in manufacturing, where material costs drive the need for optimization. A closed cylinder's surface area \( A \,\) is represented by:\[ A = 2\pi r^2 + 2\pi rh \\]This formula accounts for both the circular top and bottom, and the lateral surface area. In optimization problems, such as this exercise, substituting the expression for the height in terms of radius into this formula enables us to establish a relationship purely based on the radius \( r \,\). We then derive:\[ A = 2\pi r^2 + \frac{200}{r} \\]This equation now becomes our target function to minimize. The task of reducing the surface area to the minimum possible value can save resources, making this a practical and often used optimization problem.

Derivatives are essential tools in calculus used frequently in optimization to identify minimum and maximum values of functions. By taking the derivative of the surface area function with respect to the radius, we can gain insights into where these extrema might occur. In our problem, the derivative \( \frac{dA}{dr} \,\) tells us the rate of change of the surface area:\[ \frac{dA}{dr} = 4\pi r - \frac{200}{r^2} \\]This equation helps indicate where the surface area is increasing or decreasing:

• When the derivative is positive, the function is increasing.
• When it is negative, the function is decreasing.
• The points where it equals zero are where local minima or maxima might occur.

Applying derivatives to practical problems allows us to find critical points that signify potential optimized states.

Critical points occur where the first derivative equals zero, marking potential minima or maxima in functions. In our problem, we find critical points by solving:\[ 4\pi r - \frac{200}{r^2} = 0 \\]Upon solving, we find:

• \( r = \sqrt[3]{\frac{50}{\pi}} \,\)
• The corresponding height using this radius ensures the volume constraint is satisfied.

The second derivative test serves as a verification step for ensuring these points are indeed minima or maxima. The second derivative here is:\[ \frac{d^2A}{dr^2} = 4\pi + \frac{400}{r^3} \\]If this value is positive, which it remains across all plausible values of \( r \,\), it confirms we have found a minimum. This detailed approach ensures the minimum surface area is achieved for the given volume, providing a solid foundation for further mathematical or real-world applications.