Question

A right 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).

Answer 1

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:

$\frac{H}{R}$ = $\frac{h}{R-r}$

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

Answer 2

A) The volume of the maximum-volume cylinder has been proved to be four ninths the volume of the cone.

B) The dimensions of the cylinder with maximum lateral surface area are;

r = R/2 and h = H/2

Formula for volume of a cone is;

V = ⅓πr²h

Now, we are told that a right 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

Thus;

Volume of cone; V = ⅓πR²H

Now, if the radius and height of the cylinder are r and h respectively, then it means that;

Volume of cylinder; V = πr²h

Now, we can use similar triangles formed to get the relationship;

H/h = R/(R - r)

Making h the subject gives;

h = H(R - r)/R

Thus, volume of cylinder is now;

V = πr²(H(R - r)/R)

>> V = πH(r² - r³/R)

Differentiating with respect to r gives;

dV/dr = πH(2r - 3r²/R)

For max volume, Let's equate to 0 to find r.

Thus;

At dV/dr = 0;

πH(2r - 3r²/R) = 0

2r - 3r²/R = 0

2r = 3r²/R

r = 2R/3

Putting 2R/3 for R in H/h = R/(R - r) gives;

h = H/3

Thus;

Volume of cylinder = πr²h = π(2R/3)²(H/3)

Volume of cylinder = (4/9)πR²H

Thus,it has been proved that the volume of the maximum-volume cylinder is four ninths the volume of the cone.

B) Formula for the surface area of cylinder is given by;

A = 2πrh

From earlier, we saw that;h = H(R - r)/R

Thus;

A = 2πrH(R - r)/R

A = 2πH(r - r²/R)

Differentiating with respect to r gives;

dA/dr = 2πH(1 - 2r/R)

At max lateral surface area, dA/dr = 0

Thus;

2πH(1 - 2r/R) = 0

>> 1 - 2r/R = 0

r = R/2

Thus;

h = H(R - R/2)/R

h = H/2

Thus;

A = 2π(R/2) × (H/2)

A = ½πRH

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