Find the diameter and height of a cylinder of maximum volume which can be cut from a sphere of radius 12 cm?

To find the dimensions of the cylinder of maximum volume that can be cut from a sphere, we need to consider that the cylinder will be inscribed in the sphere.

Let the diameter of the cylinder be d and the height be h. Since the cylinder is inscribed in the sphere, the diameter of the cylinder is equal to the diameter of the sphere.

Therefore, d = 2r = 2(12) = 24 cm

The height of the cylinder can be found using the Pythagorean theorem. The radius of the sphere is the hypotenuse of a right triangle with the height of the cylinder as one leg and half of the diameter of the cylinder as the other leg.

So, h^2 + (d/2)^2 = r^2
h^2 + (24/2)^2 = 12^2
h^2 + 144 = 144
h^2 = 0
h = 0

Therefore, the height of the cylinder is 0 cm. This means that the cylinder is actually a flat disk with a diameter of 24 cm.

Let \(h\) be the height and \(d\) be the diameter of the cylinder.
Since the cylinder is cut from the sphere of radius 12 cm, the height of the cylinder cannot be greater than 24 cm. So, \(0
From the figure, we see that
$$\frac{h}{2r}=\frac{r}{h}$$

$$\Rightarrow h^2=2r^2\Rightarrow h=2r$$

and so, the volume of the cylinder is
$$V=\pi r^2 h=\pi r^2 (2r)=2\pi r^3$$

So, to find the maximum volume, we need to find the maximum of \(V=2\pi r^3\).
Replacing \(r\) by \(\frac{h}{2}\), we get
$$V=2\pi \left(\frac{h}{2}\right)^3=2\pi \left(\frac{h^3}{8}\right)=\frac{\pi h^3}{4}$$

$$V'=\frac{3\pi h^2}{4}=0 \Rightarrow h^2=0\Rightarrow h=0$$

and
$$V''=\frac{3\pi h}{2}>0$$

for all \(h>0\). Thus, there is a relative minimum at \(h=0\). Note that \(V(0)=00\). So, the minimum is not a maximum. This means that \(V\) has no maximum on the interval \(0