1. We define the following variables:

R = radius of the resultant cone;

r = radius of the original circular piece of metal;

h = height of the cone;

t = angle of the sector cut from the circle.

We know the resulting perimeter when we cut the sector out of the cone must be the same as that of the perimeter of the base of the resulting cone.

And hence we can write the radius, R, of the resulting cone in terms of the original circles radius, r, and the angle of the cut.

Using the fact that the volume of a cone is 1/3 the area of its base times the height of the cone and the right triangle geometry of a triangle with sides h and R and hypotenuse r we can write the volume V as a function of t Again, here r is the constant radius of the original circle and R is the radius of the cone. This right triangle we obtain by taking one half of the sliced triangle obtained by cutting the resulting cone along its axis. Thus the length of its height is h, the length of its base radius is R, and the length of its lateral edge is r (the radius of the original circle.)

We seek to determine the t which maximizes V[t]. And we can do this by replacing R in a computation of the volume of the cone by an expression (from the solution of the perimeter equation for R above) involving r and t alone.

We seek to maximize V[t] with respect to t, i.e. find t the angle of the cut so that V[t], the resulting volume is maximum. Thus we take the derivative with respect to t and set it equal to zero.

And we set this derivative equal to 0 to determine the extrema.

We determine the volume at the various extrema - even though we note that the solution t = 11.4134 is out of range (for we only consider cutting for t in the interval [0, 2 Pi]). Solution 1 is t = 2 Pi, which says cut all the circle out and obtain a (minimal!) 0 volume.

We consider the case of r = 4.

Our volume formula is now defined for r = 4.

Input := 


Vr[t_] = V[t]/.{r->4};

We plot volume as a function of the angle t.

Input := 


Plot[Vr[t],{t,0, 2 Pi}]

Output =


-Graphics-

And in order to maximize this volume we find where the derivative is 0.

Input := 


Vrp[t_] = Vr'[t];

Input := 


solt = Solve[Vrp[t]==0,t]//N

Output =


{{t -> 6.28319}, {t -> 1.15299}, {t -> 11.4134}}

Again t = 11.4134 is out of range and t = 6.28319 gives the 0 volume. Therefore we select solution 2 - 1.15299 radians or 66.0612 degrees (a rather flat cone) and determine the maximum volume to be 25.7963 cm^3.

Input := 


Vr3 = Vr[t]/.solt[[2]]//N

Output =


25.7963

Input := 


t  180/Pi/.solt[[2]]//N

Output =


66.0612