To solve this problem, we need to calculate the maximum possible volume of a cylinder that can fit inside a sphere of radius 10 cm.

Step 1: Geometric relationship between the cylinder and the sphere

Let the radius of the sphere be $R = 10 \, \text{cm}$. For the cylinder to fit inside the sphere, its axis must be aligned with the axis of the sphere.

The cylinder will have:

• Radius $r$
• Height $h$

For the cylinder to fit inside the sphere, the height of the cylinder, $h$, and its radius, $r$, must satisfy the condition that the diagonal of the cylinder is equal to the diameter of the sphere.

The diagonal of the cylinder forms the hypotenuse of a right triangle with:

• One leg equal to the radius of the cylinder, $r$
• The other leg equal to half the height of the cylinder, $\frac{h}{2}$
• The hypotenuse equal to the radius of the sphere, $R = 10 \, \text{cm}$

By the Pythagorean theorem, we have the relation:

$r^2 + \left( \frac{h}{2} \right)^2 = R^2$

Substituting $R = 10$, this becomes:

$r^2 + \frac{h^2}{4} = 10^2$

$r^2 + \frac{h^2}{4} = 100$

Step 2: Volume of the cylinder

The volume $V$ of the cylinder is given by the formula:

$V = \pi r^2 h$

Step 3: Maximize the volume

We can solve this problem by expressing the volume as a function of one variable. From the equation $r^2 + \frac{h^2}{4} = 100$, we can solve for $r^2$:

$r^2 = 100 - \frac{h^2}{4}$

Substitute this expression for $r^2$ into the volume formula:

$V = \pi (100 - \frac{h^2}{4}) h$

Simplify:

$V(h) = \pi \left( 100h - \frac{h^3}{4} \right)$

Step 4: Find the critical points

To maximize the volume, we take the derivative of $V(h)$ with respect to $h$:

$\frac{dV}{dh} = \pi \left( 100 - \frac{3h^2}{4} \right)$

Set the derivative equal to zero to find the critical points:

$100 - \frac{3h^2}{4} = 0$

Solve for $h$:

$\frac{3h^2}{4} = 100$

$h^2 = \frac{400}{3}$

$h = \frac{20}{\sqrt{3}} \approx 11.547 \, \text{cm}$

Step 5: Calculate the corresponding radius $r$

Substitute $h = \frac{20}{\sqrt{3}}$ into the equation $r^2 + \frac{h^2}{4} = 100$:

$r^2 + \frac{ \left( \frac{20}{\sqrt{3}} \right)^2 }{4} = 100$

$r^2 + \frac{400}{12} = 100$

$r^2 + 33.33 = 100$

$r^2 = 66.67$

$r \approx 8.165 \, \text{cm}$

Step 6: Calculate the maximum volume

Now, substitute $r \approx 8.165 \, \text{cm}$ and $h \approx 11.547 \, \text{cm}$ into the volume formula:

$V = \pi (8.165)^2 (11.547) \approx \pi \times 66.67 \times 11.547$

$V \approx \pi \times 770.91 \approx 2424.9 \, \text{cm}^3$

Final Answer:

The maximum possible volume of the cylinder that can fit inside the sphere is approximately 2424.9 cm³.

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Would you like further clarification on any part of the solution? Here are some related questions to explore:

1. How would the volume change if the sphere’s radius were doubled?
2. Can the same method be used to calculate the maximum volume of a cone inside a sphere?
3. How do the relationships between $r$, $h$, and $R$ affect the maximum volume for other 3D shapes?
4. What is the significance of the Pythagorean theorem in fitting 3D objects within each other?
5. How does the optimization method for finding the maximum volume apply to other geometric optimization problems?

Tip: When maximizing volumes in geometry problems, always check the conditions (like the Pythagorean theorem for fitting shapes in spheres) before proceeding with calculations.