The two cones problem

After A. Oldknow and others

Cardboard cones

Single cone

We cut a sector of a certain angle from a cardboard circle, roll it into a cone and find the volume. 
We assume the radius is our unit

R = 1

The basis of the cone is  called   r   and its height  h  . These relate by

\(\(\(Clear[f, sol, volume]\)\(
\)
    \)\)

sol[angle_] := 
      Solve[{r ==  R angle/2/π, r^2 + h^2 == R^2}, {r, 
          h}]

volume[angle_] :=  1/3 π r^2 h /. Last[sol[angle]]

Plot[volume[angle], {angle, 0, 2 π}, 
      AspectRatio -> Automatic]

Two cones

Roll another cone from the remaining cardboard and add

Plot[volume[(2 Pi  -  angle)], {angle, 0, 2 π}, 
      AspectRatio -> Automatic]

Summing:

f[angle_] := 
      volume[angle] + volume[(2 π  -  angle)]

\(Plot[f[angle], {angle, 0, 2\ π}
, 
      AspectRatio -> Automatic]\)

It seems this function has a single hump:

NSolve[D[f[angle], angle] == 0]

Computing the derivative reveals a double hump:

f[angle] /. First[%]

and a smaller volume at the center

f[π] // N

which is in fact a local minimum!

Controle by plotting f'

The default plotdoes not give much information

Plot[f'[angle], {angle, 0, 2 Pi}]

Vertical stretching gives the full information:

Show[%, PlotRange -> {-0.02, 0.02}, 
      Ticks -> {{0, Pi/2, Pi, 3  Pi/2, 2 Pi}, None}]