A cylinder is inscribed in a right circular cone of height 6.5 and radius (at the base) equal to 8.

Suppose r is the base radius of the cylinder inscribing the cone. Let h be the height.

Using similar triangles on the largest cross-section, 

\(\dfrac{h}{8 - r} = \dfrac{6.5}8\\ h = \dfrac{13}2 - \dfrac{13}{16}r\)

 

Then the volume is \(\pi r^2 h = \dfrac{13}{16} \pi r^2 (8 - r)\).

Solving \(\dfrac{d}{dr} \dfrac{13}{16} \pi r^2 (8 - r) = 0\) gives r = 16/3 or r = 0 (rej.)

 

Note that 

\(\dfrac{d^2}{dr^2}\Big|_{r = \frac{16}3} \dfrac{13}{16} \pi r^2 (8 - r) = -13\pi < 0\). Then maximum volume is attained when r = 16/3.

 

Now, you can calculate the height and get your answer.