To find the dimensions (radius and height) of a right circular cylinder of maximum volume that can be inscribed in a sphere of radius $R = 10$ cm, we can set up the problem using calculus.

Step 1: Set Up Variables and Relationships

Let:

• $r$ = radius of the cylinder's base.
• $h$ = height of the cylinder.

The cylinder is inscribed in a sphere, so its height and radius are constrained by the sphere’s radius. From the geometry of the problem, if we slice through the center, we get a right triangle with:

• Hypotenuse (diameter of the sphere) = $2R = 20$ cm,
• One leg = height of the cylinder $h$,
• The other leg = $2r$.

By the Pythagorean theorem: $h^2 + (2r)^2 = (2R)^2$ Substitute $R = 10$: $h^2 + 4r^2 = 400$

Solving for $h$: $h = \sqrt{400 - 4r^2}$

Step 2: Set Up the Volume of the Cylinder

The volume $V$ of a cylinder is given by: $V = \pi r^2 h$ Substitute $h = \sqrt{400 - 4r^2}$ into the volume formula: $V = \pi r^2 \sqrt{400 - 4r^2}$

Step 3: Differentiate to Maximize the Volume

To find the maximum volume, differentiate $V$ with respect to $r$ and set the derivative to zero.

1. Differentiate $V$ with respect to $r$: $\frac{dV}{dr} = \pi \cdot \left(2r \sqrt{400 - 4r^2} + r^2 \cdot \frac{-4r}{\sqrt{400 - 4r^2}}\right)$ Simplify: $\frac{dV}{dr} = \pi \left(2r \sqrt{400 - 4r^2} - \frac{4r^3}{\sqrt{400 - 4r^2}}\right)$ Set $\frac{dV}{dr} = 0$ and solve for $r$.

2. Simplifying the equation above (details omitted here) leads to: $r = \sqrt{\frac{400}{6}} = \frac{10}{\sqrt{3}} \approx 5.77 \text{ cm}$

Step 4: Find the Corresponding Height

Using $r = \frac{10}{\sqrt{3}}$ in the equation for $h$: $h = \sqrt{400 - 4 \left(\frac{10}{\sqrt{3}}\right)^2} = \sqrt{400 - \frac{400}{3}} = \sqrt{\frac{800}{3}} \approx 10.77 \text{ cm}$

Conclusion

The dimensions of the right circular cylinder with maximum volume that can be inscribed in a sphere of radius 10 cm are approximately:

• Radius $r \approx 5.77$ cm
• Height $h \approx 10.77$ cm

Would you like a deeper breakdown of any part?

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1. How is the relationship between cylinder and sphere derived?
2. How does the Pythagorean theorem apply in 3D geometry for this problem?
3. What is the significance of setting the derivative to zero in optimization problems?
4. Could we solve this using Lagrange multipliers instead of substitution?
5. How does the result change if the sphere's radius changes?

Tip: For inscribed shapes, use geometry to establish relationships between dimensions before applying calculus.