Question

Differentiate its volume with respect to the radius of the base, where the surface area is constant.

Answer 1

Concept:
For a cylindrical tumbler open from the top, the total surface area consists of:

• Curved Surface Area \(=2\pi rh\)
• Area of the circular base \(=\pi r^2\)

Hence, the fixed surface area is \[ 2\pi rh+\pi r^2=S, \] where \(S\) is a constant. The volume of the tumbler is \[ V=\pi r^2h. \] Since the surface area is constant, we first express \(h\) in terms of \(r\) and then differentiate \(V\).
Given, \[ 2\pi rh+\pi r^2=S. \] Rearranging, \[ 2\pi rh=S-\pi r^2. \] Therefore, \[ h=\frac{S-\pi r^2}{2\pi r}. \] Substitute this value of \(h\) into the volume formula: \[ V=\pi r^2\left(\frac{S-\pi r^2}{2\pi r}\right). \] Simplifying, \[ V=\frac{Sr-\pi r^3}{2}. \] Differentiate with respect to \(r\): \[ \frac{dV}{dr} =\frac{d}{dr}\left(\frac{Sr-\pi r^3}{2}\right). \] Since \(S\) is constant, \[ \frac{dV}{dr} =\frac{1}{2}\left(S-3\pi r^2\right). \] Hence, \[ \boxed{\frac{dV}{dr}=\frac{S-3\pi r^2}{2}.} \] 

Answer 2

For maximum volume, \[ \frac{dV}{dr}=0. \] Using the result from part (i), \[ \frac{S-3\pi r^2}{2}=0. \] Therefore, \[ S=3\pi r^2. \] But, \[ S=2\pi rh+\pi r^2. \] Substituting, \[ 2\pi rh+\pi r^2=3\pi r^2. \] Divide both sides by \(\pi\): \[ 2rh+r^2=3r^2. \] Simplifying, \[ 2rh=2r^2. \] Since \(r>0\), \[ \boxed{h=r.} \] Thus, for the tumbler to have the maximum possible volume while keeping the surface area fixed, its height must be equal to its radius.