Differentiation

(1) Let r1 be the radius of the cylinder and h1 be the height of the cylinder.
r1 / (12-h1) = 5/12
12r1 = 60 - 5h1
h1 = 12 - 2.4r1
Let volume of the cylinder be V
V = π(r1^2) h1
= π (r1^2)(12 - 2.4r1)
= π (12r1^2 - 2.4 r1^3)
differentiate w.r.t. r1
dV/dr1 = π (24 r1 - 7.2r1^2)
d/dr1(dV/dr1) = π (24 - 14.4 r1)
set dV/dr1 = 0, 24 r1 - 7.2 r1^2 = 0
r1 = 0 (reject) or r1 = 10/3cm = 3.33cm
sub r1 = 10/3 into d/dv(dV/dr1) = π[24-14.4(10/3)] = -24π<0
When r1 = 3.33cm, the volume is maximum.
height of the cylinder, h1 = 12- 2.4r1 = 12 - 2.4 (3.33) = 4cm
(2)(a)
Let r, h and V be the radius, height and volume of the cone respectively
r = √(s^2- h^2)
V = 18π = π(r^2)h/3
54 = (s^2 - h^2)h
s = √(54/h + h^2) -------(1)
(b)
differentiate eq(1) w.r.t.h
ds/dh = (1/2)[1/√(54/h + h^2)]} (-54/h^2 + 2h)
=(h - 27/h^2) /√(54/h + h^2)
d/dh(ds/dh)
= [(1 + 54/h^3)√(54/h + h^2) - (h - 27/h^2)(0.5)(2h- 54/h^2)/√(54/h + h^2)]/(54/h + h^2)
= [(1+ 54/h^3) (54/h+h^2) - (h - 27/h^2)^2 ] /(54/h + h^2)^(3/2)
Set ds/dh = 0,
h - 27/h^2 = 0
27 = h^3
h = 3cm
when h = 3,
d/dh(ds/dh)
= [(1+54/(3)^3)(54/3 + 3^2) - (3 - 27/3^2)^2]/(54/3 + 3^2)^(3/2)
= 0.5 > 0
Therefore, h = 3cm , s is minimum.
minimum slant height, s = √(3^2 + 54/3) = √18 = 3√2 = 4.243cm