Question

The maximum volume (in cu. units) of the cylinder which can be inscribed in a sphere of radius 12 units is:

• A. \( 384 \sqrt{3} \pi \)
• B. \( 768 \sqrt{3} \pi \)
• C. \( 768 \pi / \sqrt{3} \)
• D. \( 1152 \pi / \sqrt{3} \)

Hint:

To maximize the volume of a cylinder inscribed in a sphere, use the relation between radius and height of the sphere, and then differentiate to find the maximum.

Answer 1

The Correct Option is B

Step 1: {Find the relation between radius and height of the cylinder inscribed in a sphere}
Let the radius of the cylinder be \( r \) and height \( h \). The equation for the sphere is \( r^2 + \left(\frac{h}{2}\right)^2 = 12^2 \), or: \[ r^2 + \frac{h^2}{4} = 144. \] \[\Rightarrow V = 144\pi h - \frac{\pi}{4}h^3\]
\[\Rightarrow \frac{dV}{dh} = 144\pi - \frac{3\pi}{4}h^2\] \[\Rightarrow \frac{dV}{dh} = 0 \Rightarrow 144\pi = \frac{3\pi}{4}h^2\] \[\Rightarrow h^2 = 48 \times 4 \Rightarrow h = 8\sqrt{3}\] \[\therefore 12^2 = r^2 + 48 \Rightarrow r^2 = 96\] \[{Volume} = \pi r^2 h = \pi \times 96 \times 8\sqrt{3} = 768\sqrt{3}\pi { cm}^3.\] By solving the optimization problem, the maximum volume comes out to be
\[ 768 \sqrt{3} \pi)\]

Answer 2

Step 1: Understanding the problem
We are asked to find the maximum volume of a cylinder that can be inscribed in a sphere with a radius of 12 units. The cylinder's axis is aligned with the sphere's axis, and the diameter of the cylinder must fit within the sphere's diameter.

Step 2: Volume of the Cylinder
The volume \( V \) of a cylinder is given by the formula: \[ V = \pi r^2 h \] where \( r \) is the radius of the base of the cylinder and \( h \) is the height of the cylinder.

Step 3: Relationship between the dimensions of the cylinder and the sphere
For the cylinder to be inscribed in the sphere, the diameter of the cylinder must be equal to the diameter of the sphere at its widest point. The sphere's radius is 12 units, so the sphere's diameter is 24 units.
Let the radius of the base of the cylinder be \( r \), and let the height of the cylinder be \( h \). Since the cylinder is inscribed, the relationship between the radius and the height of the cylinder can be derived using the Pythagorean theorem in the right triangle formed by the radius of the sphere, the radius of the cylinder, and half the height of the cylinder.
The relation is: \[ r^2 + \left( \frac{h}{2} \right)^2 = 12^2 \] which simplifies to: \[ r^2 + \frac{h^2}{4} = 144 \] Multiplying the equation by 4 to clear the fraction: \[ 4r^2 + h^2 = 576 \] This is our key equation that relates \( r \) and \( h \).

Step 4: Maximizing the Volume
Now, we need to maximize the volume \( V = \pi r^2 h \), subject to the constraint \( 4r^2 + h^2 = 576 \). We can solve this using optimization techniques such as Lagrange multipliers or directly by substitution.
From the constraint \( 4r^2 + h^2 = 576 \), solve for \( h \): \[ h^2 = 576 - 4r^2 \] Thus, \[ h = \sqrt{576 - 4r^2} \] Substitute this expression for \( h \) into the volume formula: \[ V = \pi r^2 \sqrt{576 - 4r^2} \] To maximize this, take the derivative of \( V \) with respect to \( r \) and set it equal to zero.

Step 5: Differentiating the Volume Expression
The derivative of \( V = \pi r^2 \sqrt{576 - 4r^2} \) with respect to \( r \) is computed using the product rule and chain rule: \[ \frac{dV}{dr} = \pi \left( 2r \sqrt{576 - 4r^2} + r^2 \cdot \frac{d}{dr} \left( \sqrt{576 - 4r^2} \right) \right) \] First, differentiate \( \sqrt{576 - 4r^2} \): \[ \frac{d}{dr} \left( \sqrt{576 - 4r^2} \right) = \frac{-8r}{2\sqrt{576 - 4r^2}} = \frac{-4r}{\sqrt{576 - 4r^2}} \] Now substitute this into the derivative: \[ \frac{dV}{dr} = \pi \left( 2r \sqrt{576 - 4r^2} - \frac{4r^3}{\sqrt{576 - 4r^2}} \right) \] Set \( \frac{dV}{dr} = 0 \) to find the critical points: \[ 2r \sqrt{576 - 4r^2} = \frac{4r^3}{\sqrt{576 - 4r^2}} \] Multiply both sides by \( \sqrt{576 - 4r^2} \) to clear the denominator: \[ 2r (576 - 4r^2) = 4r^3 \] Simplify: \[ 1152r - 8r^3 = 4r^3 \] \[ 1152r = 12r^3 \] \[ r^2 = 96 \] Thus, \[ r = \sqrt{96} = 4\sqrt{6} \] Now substitute this value of \( r \) back into the equation for \( h \): \[ h^2 = 576 - 4r^2 = 576 - 4 \times 96 = 576 - 384 = 192 \] \[ h = \sqrt{192} = 8\sqrt{3} \] Thus, the dimensions of the cylinder are: \[ r = 4\sqrt{6}, \quad h = 8\sqrt{3} \]

Step 6: Maximum Volume
Finally, substitute these values into the formula for the volume of the cylinder: \[ V = \pi r^2 h = \pi (4\sqrt{6})^2 (8\sqrt{3}) = \pi \times 96 \times 8\sqrt{3} = 768 \sqrt{3} \pi \] Step 7: Conclusion
The maximum volume of the cylinder that can be inscribed in the sphere is \( 768 \sqrt{3} \pi \).