Imagine a flat-bottomed cylinderal container with a circular cross section of radius 4 in. a marble with radius 0<r<4 inches is placed in the bottom of the can. what is the radius of the bottom that requires the most water to cover it. (include first or second derivative test to prove it)
To find the radius of the bottom that requires the most water to cover it, we need to maximize the surface area of the water while keeping the volume constant.
Let's break down the problem step by step:
1. Start with the given information:
- Radius of the circular cross-section of the container: 4 inches.
- Radius of the marble: 0 < r < 4 inches.
2. Determine the height of the water covering the marble:
- Since the marble is placed at the bottom of the can, the height of the water will be equal to the radius of the container. Let's call it 'h.'
3. Find the volume of water (V) required to cover the marble:
- The volume of a cylindrical shape is given by V = π * r^2 * h.
- In this case, h = r, so V = π * r^3.
4. Differentiate the volume function with respect to r:
- Taking the derivative of V = π * r^3 with respect to r, we get dV/dr = 3π * r^2.
5. Set the derivative equal to zero and solve for r to find critical points:
- 3π * r^2 = 0
- r^2 = 0
- r = 0
6. Determine the nature of the critical point:
- Since the problem states that 0 < r < 4, we can disregard the critical point at r = 0.
7. Find the second derivative of the volume function:
- Taking the derivative of dV/dr = 3π * r^2 with respect to r, we get d²V/dr² = 6π * r.
8. Substitute the critical point (r = 0) into the second derivative:
- d²V/dr² = 6π * 0
- d²V/dr² = 0.
9. Analyze the second derivative test for the critical point:
- Since the second derivative is equal to zero, the second derivative test is inconclusive.
Therefore, we cannot determine the maximum using the second derivative test.
However, we know that the maximum volume occurs either at the critical point (r = 0) or at the endpoints of the interval (0 < r < 4). According to the given conditions, the marble and the container cannot have a radius of zero, so the maximum volume occurs at the endpoint of the interval.
Therefore, the maximum volume of water required to cover the bottom of the container occurs when r = 4 inches, which is the radius of the container.