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 determine the radius of the bottom that requires the most water to cover it, we can frame the problem as an optimization question. The volume of water required to cover the bottom of the can will be maximized when the surface area of the bottom is maximized.
Let's denote the radius of the bottom by r (0 < r < 4). To find the radius that corresponds to the maximum surface area, we'll use calculus and the first and second derivative tests.
First, we need to derive the formula for the surface area of the bottom of the can. The surface area (A) of a circle is given by the formula: A = πr^2.
In this case, the radius can vary since 0 < r < 4. However, we are also given that the radius of the marble is 0 < r < 4. Therefore, to ensure that the bottom is covered by water entirely, the radius of the bottom (R) must be greater than the radius of the marble.
Hence, we have the inequality: R > r.
The surface area of the bottom can now be expressed as: A = πR^2.
Next, we'll apply the first derivative test.
1. Calculate the first derivative of A with respect to R: dA/dR = 2πR.
2. Set the derivative equal to zero to find critical points: 2πR = 0.
3. Solve for R: R = 0.
The critical point we found, R = 0, is not within the allowed range of 0 < R < 4. Thus, we disregard it.
Now, let's use the second derivative test to confirm whether we have a maximum or minimum point.
1. Calculate the second derivative of A with respect to R: d²A/dR² = 2π.
2. Evaluate the second derivative test: Since the second derivative is a constant (2π > 0), this indicates a minimum point.
Therefore, there is no maximum area for the bottom of the can. The radius that requires the most water to cover it is simply R = 4 inches, which corresponds to the entire radius of the can itself.