m2 differentiation (2)

Let angle EBD = tan x
So tan x = ED/BD = 1/BD = 1/r (Since the inscribed circle is a unit circle).
tan 2x = AD/BD = h/r
Using the tan 2x formula, we get
h/r = 2(1/r)/[1 - (1/r)^2]
h/r = (2/r)(r^2)/(r^2 - 1)
h/r = 2r/(r^2 - 1)
h = 2r^2/(r^2 - 1).
h(r^2 - 1) = 2r^2
hr^2 - h = 2r^2
r^2(h - 2) = h
so r^2 = h/(h - 2).
(b) Volume of cone = pr^2h/3 = ph^2/3(h - 2) p = pi.
dV/dh = p/2[2h(h - 2) - h^2]/(h - 2)^2
Put dV/dh = 0
2h(h - 2) - h^2 = 0
2h^2 - 4h - h^2 = 0
h^2 - 4h = 0
h(h - 4) = 0
h = 0 (rej.) or h = 4.
When h = 3, V = p(9)/(3 x 1) = 3p
When h = 4, V = p(16)/(3 x 2) = 2.67p
When h = 5, V = p(25)/(3 x 3) = 2.78p
So volume is a min. when h = 4 and equals to 2.67p.