Aright circular cylinder is placed inside a cone of radius r and height h so that the base of the cylinder lies on the base of the cone. a. find the dimensions of the cylinder with maximum volume. specifically, show that the volume of the maximum-volume cylinder is four ninths the volume of the cone. b. find the dimensions of the cylinder with maximum lateral surface area (area of the curved surface).

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