Hello, grs!

A fascinating problem . . . hope I got it right.

A hollow cylinder's internal radius is expanding at a constant rate $\displaystyle (b)$ (the external radius is fixed),

and the cylinder is lengthening at the same constant rate $\displaystyle (b).$

In terms of the internal radius, what should the length (height) of the cylinder be

in order for the combined internal surface area of the cylinder

and the areas of each end of the cylinder to remain constant. Code:

: R :
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h |///| : |///|
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: r :

The outer radius is $\displaystyle R$ (constant).

The inner radius is $\displaystyle r$.

The height is $\displaystyle h.$

The lateral area of the cylinder is: .$\displaystyle 2\pi rh$

The ends are "rings", each with area: .$\displaystyle \pi(R^2-r^2)$

The total surface area is: .$\displaystyle S \;=\;2\pi rh + 2\pi(R^2-r^2) \;=\;2\pi rh + 2\pi R^2 - 2\pi r^2$

The total area is to be a constant: .$\displaystyle 2\pi rh + 2\pi R^2 - 2\pi r^2 \;=\;K$

Differentiate with respect to time: .$\displaystyle 2\pi r\,\frac{dh}{dt} + 2\pi h\,\frac{dr}{dt} - 4\pi r\,\frac{dr}{dt} \;=\;0$

Since $\displaystyle \frac{dh}{dt} = \frac{dr}{dt} = b$, we have: .$\displaystyle 2\pi r(b) + 2\pi h(b) - 4\pi r(b) \:=\:0 $

Solve for $\displaystyle h\!:\;\;2\pi h - 2\pi r \:=\:0 \quad\Rightarrow\quad\boxed{ h \:=\:r}$