Respuesta :
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:
[tex]\frac{H}{R}[/tex] = [tex]\frac{h}{R-r}[/tex]
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

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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