Question

Find the dimensions of the right circular cylinder of maximum volume that can be placed inside a sphere of radius \(R.\)

Short answer

In conclusion, the dimensions of the right circular cylinder of maximum volume that can fit inside a sphere of radius R are given by: Radius of the cylinder (r): \(-\frac{3}{2\text{sin}(\varphi)\text{'}}\text{cos}(\varphi) R\) Height of the cylinder (h): \(-3\text{sin}(\varphi) R\) These dimensions depend on the angle (\(\varphi\)) and the radius of the sphere (R). The maximum volume of the cylinder inside the sphere can be expressed as: Maximum volume (Vmax): \(\frac{27\pi}{8(\text{sin}(\varphi)\text{'})^3}R^3\)

Step-by-step solution

Variables and equations

Let the height of the right circular cylinder be h and the radius be r. The volume of a cylinder is given by the formula \(V_\text{cylinder} = \pi r^2 h\). Observe that the center of the sphere coincides with the center of the cylinder. Let \(\varphi\) be the angle from the central axis of the cylinder to the hypotenuse of the right-triangle with one vertex on the central axis of the cylinder, another one on the circular face of the cylinder, and the other one lying on the surface of the sphere. Let d be the distance from the vertex of the central axis of the cylinder to the opposite vertex lying on the sphere surface. Then, we have the following relationship: \[R^2 = (d\text{cos}(\varphi))^2 + (d\text{sin}(\varphi))^2\] Dividing by \$R^2\$, we get: \[1 = (\frac{d\text{cos}(\varphi)}{R})^2 + (\frac{d\text{sin}(\varphi)}{R})^2\] Now, we need to find the dimensions in terms of R and d

Write cylinder's dimensions in terms of R and d

We have: \[r = d\text{cos}(\varphi)\] \[h = 2d\text{sin}(\varphi)\] Now, substituting these dimensions back into the volume formula, we get: \[V_\text{cylinder} = \pi(d\text{cos}(\varphi))^2(2d\text{sin}(\varphi)) = 2\pi d^3\text{cos}^2(\varphi)\text{sin}(\varphi)\]

Differentiate the volume function with respect to d

To find the maximum volume, we need to differentiate the function with respect to d: \[\frac{\text{d}V_\text{cylinder}}{\text{d}d} = 6\pi d^2\text{cos}^2(\varphi)\text{sin}(\varphi) + 4\pi d^3\text{cos}(\varphi)\text{sin}^2(\varphi)\text{'}\]

Set the first derivative to zero and solve for d

Next, set the first derivative to zero and solve for d: \[6\pi d^2\text{cos}^2(\varphi)\text{sin}(\varphi) + 4\pi d^3\text{cos}(\varphi)\text{sin}^2(\varphi)\text{'} = 0\] When we divide by \(\pi d^2\text{cos}(\varphi)\text{sin}(\varphi)\), we get: \[6 + 4d\text{sin}(\varphi)\text{'} = 0\] \[\implies d = -\frac{3}{2\text{sin}(\varphi)\text{'}}\] To find the critical points of the volume function, we need to look at its second derivative and check for the sign.

Calculate the second derivative and substitute the critical point

The second derivative of the volume function with respect to d is: \[\frac{\text{d}^2V_\text{cylinder}}{\text{d}^2} = 12\pi d\text{cos}^2(\varphi)\text{sin}(\varphi) + 12\pi d^2\text{cos}(\varphi)\text{sin}^2(\varphi)\text{'}\] Now, we substitute the critical point back into the second derivative: \[\frac{\text{d}^2V_\text{cylinder}}{\text{d}^2} = 12\pi-\frac{3}{2\text{sin}(\varphi)\text{'}} (2\text{cos}(\varphi)\text{sin}(\varphi) - \frac{3}{2\text{sin}(\varphi)\text{'}} \text{cos}(\varphi)\text{sin}(\varphi)\text{'})\] Since this expression is always negative, the volume function has a maximum for the given critical point.

Find the maximum volume of the cylinder

Now, we substitute the critical point back into the volume function to find the maximum volume: \[V_\text{max} = 2\pi\left(-\frac3{2\text{sin}(\varphi)\text{'}}\right)^3(\text{cos}^2(\varphi)\text{sin}(\varphi)) = \frac{27\pi}{8(\text{sin}(\varphi)\text{'})^3}\] Since the dimensions of the cylinder are dependent on the value of R, the maximum volume for any sphere of radius R is given by: \[V_\text{max} = \frac{27\pi}{8(\text{sin}(\varphi)\text{'})^3}R^3\] Thus, the dimensions of the right circular cylinder with maximum volume that can be placed inside a sphere of radius R are as follows: \[r = d\text{cos}(\varphi) = -\frac3{2\text{sin}(\varphi)\text{'}}\text{cos}(\varphi) R\] \[h = 2d\text{sin}(\varphi) = -3\text{sin}(\varphi) R\]

Key concepts

Cylinder and Sphere Geometry

When dealing with geometry involving cylinders and spheres, we often need to think about how the shapes fit together or interact spatially. In this exercise, we are looking at a right circular cylinder inside a sphere. A right circular cylinder is one where the sides are perpendicular to its circular bases, resembling a soup can.
The center of the sphere coincides with the center axis of the cylinder, meaning the sphere engulfs the cylinder. Understanding this spatial relationship is key to solving optimization problems involving these shapes.

• The cylinder's radius and height need to be such that they maximize volume without exceeding the bounds of the sphere.
• The conditions ensure the entire cylinder is within the sphere's volume.
• This interplay requires visualizing the cylinder's maximum diameter fit within the sphere’s diameter. It's like fitting as big a can as possible inside a beach ball.

With the sphere having a radius \(R\), the spatial fit gives important equations, leading to a system that relates the dimensions of the cylinder and sphere.

Differentiation in Calculus

Differentiation is a powerful tool in calculus used to find rates of change. In optimization problems, it's instrumental for determining maximum or minimum values of a function. Here, it helps find the maximum volume of the cylinder within the sphere.
We start by forming an equation for the volume of the cylinder, which depends on its radius \( r \) and height \( h \):

• The volume of a cylinder is \( V = \pi r^2 h \). In this scenario, both \( r \) and \( h \) are constrained by the sphere's size.
• Next, substitute the equations involving \( r \) and \( h \) in terms of a single variable related to the sphere's radius, \( R \).

Differentiation of this volume function with respect to the chosen variable lets us discover where the volume reaches its peak. It's essential to apply the product and chain rules accurately during differentiation to keep track of all terms.

Critical Points in Calculus

Critical points are where the derivative of a function is zero or undefined. These points help identify where a function’s minimum or maximum values occur.

• By setting the first derivative of the volume function equal to zero, we find the critical points related to the cylinder’s maximum volume.
• At these points, the change of the volume with respect to the chosen variable halts, indicating a local extreme point.
• To confirm whether a critical point is a maximum or minimum, the second derivative test is applied.

In our problem, calculating the second derivative helps verify that the critical point corresponds to a maximum volume for the cylinder. This step involves substituting the critical point back to ascertain a negative second derivative, confirming the peak.
Understanding critical points thus ensures you choose the most optimal cylinder dimensions inside any given sphere radius.