The maximum volume (in cu. units) of the cylinder which can be inscribed in a sphere of radius 12 units is:
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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.
Step 1: {Determine the relationship between the radius and height of a cylinder inscribed within a sphere} Let the cylinder's radius be \( r \) and its height be \( h \). The equation governing the sphere is \( r^2 + \left(\frac{h}{2}\right)^2 = 12^2 \), which simplifies to:\[r^2 + \frac{h^2}{4} = 144.\]\[\Rightarrow V = 144\pi h - \frac{\pi}{4}h^3\] \[\Rightarrow \frac{dV}{dh} = 144\pi - \frac{3\pi}{4}h^2\]\[\Rightarrow \frac{dV}{dh} = 0 \Rightarrow 144\pi = \frac{3\pi}{4}h^2\]\[\Rightarrow h^2 = 48 \times 4 \Rightarrow h = 8\sqrt{3}\]\[\therefore 12^2 = r^2 + 48 \Rightarrow r^2 = 96\]\[{Maximum Volume} = \pi r^2 h = \pi \times 96 \times 8\sqrt{3} = 768\sqrt{3}\pi { cm}^3.\]The optimization process yields a maximum volume of \[ 768 \sqrt{3} \pi)\]