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a.
Volume of a cylinder:
V = π * r² * h
Being r its radius and h its height.
For a cylinder of radius r, its height h can be express using the properties of similar triangles:
=
Then,
h = H * (R- r)/R
Replacing h in the expression for volume, derivating that expression, equating to zero and solving for r, we can find the radius of the cylinder of maximum volume:
V = π * r² * H (R-r)/R = π* H * (Rr² - r³)/R
We derivate using the following properties of derivation:
f(x) = x^n, then, f'(x) = n*x^(n-1)
f(x) = g(x) + h(x), then, f'(x) = g'(x) + h'(x)
f(x) = K* g(x), then, f'(x) = K* g'(x) for K = constant
We get:
dV/dr = (πH/R)*(2Rr - 3r²)
(πH/R)*(2Rr - 3r²) = 0
Solving for r, we have:
r = 2R/3
h = H(R-r)/R = H(R - 2/3 R)/R = H/3
V = π * (2 R/3)² * H/3 = 4/9 *(1/3 * π * R² * H) = 4/9 * Volume of the cone
b. The surface Area is found using the following expression:
A = 2πrh
We simplify using the expressions found previously:
A = 2π r * H(R-r)/R = 2πH(Rr - r²)/R
Derivating:
dA/dr =(2πH/R)*(R - 2r)
(2πH/R)*(R - 2r) = 0
r = R/2
h = H(R- R/2)/R = H/2
A = 2π*R/2 *H/2 = 1/2 * π * R * H
