Optimal soda can


a. Classical problem Find the radius and height of a cylindrical soda can with a volume of 354 cm³ that minimize the surface area.

Verified step by step guidance

1. Start by identifying the formulas involved: The volume of a cylinder is given by \( V = \pi r^2 h \) and the surface area is given by \( A = 2\pi r^2 + 2\pi rh \).
2. Given that the volume \( V = 354 \text{ cm}^3 \), express the height \( h \) in terms of the radius \( r \) using the volume formula: \( h = \frac{354}{\pi r^2} \).
3. Substitute \( h \) from the volume equation into the surface area formula to express the surface area \( A \) solely in terms of \( r \): \( A = 2\pi r^2 + \frac{708}{r} \).
4. To find the radius that minimizes the surface area, take the derivative of \( A \) with respect to \( r \), set it equal to zero, and solve for \( r \). This involves finding \( \frac{dA}{dr} = 4\pi r - \frac{708}{r^2} \) and setting \( \frac{dA}{dr} = 0 \).
5. Solve the equation \( 4\pi r - \frac{708}{r^2} = 0 \) to find the critical points. Then, use the second derivative test or analyze the behavior of \( \frac{dA}{dr} \) to confirm that the solution gives a minimum surface area. Finally, use the value of \( r \) to find \( h \) using the expression \( h = \frac{354}{\pi r^2} \).

Key Concepts

Volume of a Cylinder

The volume of a cylinder is calculated using the formula V = πr²h, where r is the radius and h is the height. In this problem, the volume is given as 354 cm³, which serves as a constraint for determining the optimal dimensions of the can.

Surface Area of a Cylinder

The surface area of a cylinder is given by the formula A = 2πr² + 2πrh, where the first term accounts for the areas of the two circular bases and the second term accounts for the lateral surface area. Minimizing this surface area while maintaining a fixed volume is the core objective of the problem.

Optimization Techniques

Optimization in calculus involves finding the maximum or minimum values of a function. In this case, techniques such as setting up a function for surface area in terms of one variable (using the volume constraint) and applying derivatives to find critical points are essential for solving the problem.