Minimum volume of circumscribed straight cone of radius r

Minimum volume of circumscribed straight cone of radius r

Should it be an circumscribed cone? Suppose the circumscribed section is △ PAB, PA and Pb are conical bus bars, AB is the diameter of the bottom circle, C is the tangent point of the ball on the bus bar PA, O is the center of the ball, ah is the height of the cone, ah is the radius of the bottom circle R, Op = x, PC = √ (x ^ 2-r ^ 2), x > R, RT △ PCO

Let a sphere with radius r have an inscribed positive cone, and the volume of the cone is the largest when the ratio of height to bottom radius is emergency

Conical volume = bottom area * height * 1 / 3 = square of radius * 3.14 * height * 1 / 3
(R + x) (r square - x square) / 3 = volume of cone
Find the maximum value of X, and the ratio of height to bottom radius is (R + x): [(r square - x square) arithmetic square root]

For a cone circumscribed to a sphere with radius r, the ratio of the side area to the sphere area is 3:2, and the radius r of the cone bottom surface is obtained

Let t be the angle between the side and the bottom of the cone,
Then bus length L = R / cos (T)
R = r*tan(t/2)
Conical side area S1 = pi * L * r = pi * r / cos (T) * r
Surface area of ball S2 = 4 * pi * R ^ 2 = 4 * pi * R ^ 2 * Tan (T / 2) ^ 2
s1/s2 = 1/cos(t)/4tan(t/2)^2 = 3/2
=>1/cos(t) * (1+cos(t)/(1-cos(t)) = 6 (tan(t/2)^2 = (1-cos(t))/(1+cos(t))
=>Cos (T) = 1 / 2 or 1 / 3
=>Tan (T / 2) = sqrt ((1-cos (T)) / (1 + cos (T))) = √ 3 / 3 or √ 2 / 2
=>R = R / Tan (T / 2) = √ 2R or √ 3R

Given that the radius of the ball is r, the radius of the bottom surface of the inner cylinder of the ball is r, and the height is h, then what are the values of R and h, the volume of the inner cylinder is the largest?

It is known from the meaning of the question that the center of the ball is at the midpoint of the height on the inner cylindrical axis, then:
R ²= r ²+ (h/2) ² Namely H ²＝ 4R ²- 4r ²
The following basic inequality is used to find the maximum volume
Because the volume of the inscribed cylinder v = π R ² h. That is, V ²= π ² r ² r ² h ²
So V ²= π ² r ² r ² (4R ²- 4r ²)
＝π ²/ 4 *(2r ²) (2r ²) (4R ²- 4r ²)
Also (2R) ²) (2r ²) (4R ²- 4r ²) ≤{[(2r ²)+ (2r ²)+ (4R ²- 4r ²)]/ 3} ³= 64(R ²)³/ 27 (if and only if 2R) ²＝ 4R ²- 4r ² 3R ²＝ 2R ² (equal sign when)
Therefore, when r = √ 6 * r / 3, H = 2 √ 3 * r / 3, V ² There is a maximum π ²/ four × 64(R ²)³/ 27＝16π ² (R ²)³/ twenty-seven
That is, the volume of the inscribed cylinder has the maximum value: 4 √ 3 × πR ³/ nine

Given that the radius of the ball is r, the radius of the bottom surface of the inscribed cylinder of the ball is R and the height is h, what are the values of R and h, the volume of the inscribed cylinder is the largest? Please answer with the knowledge of inequality

Obviously, the cylinder satisfying the condition is divided into two parts by a plane passing through the center of the circle and parallel to the bottom
Then the area of the cylinder bottom = π R ²
h=2√(R ²- r ²)
V=πr ²* 2√(R ²- r ²)= 4π√[(r ²/ 2) ² (R ²- r ²)]
According to mean inequality
(R ²/ 3) ³= [(r ²/ 2+r ²/ 2+R ²- r ²)/ 3] ³ ≥(r ²/ 2) ² (R ²- r ²)
When R ²/ 2=R ²- r ² Take the equal sign when
At this time, r = √ 6R / 3, H = 2 √ 3R / 3

Find the maximum volume of the inscribed cylinder of the ball with radius r, and find the bottom radius when the cylinder volume is the largest

Let the radius of the bottom of the cylinder be r,
Then the height of the ball center to the bottom (i.e. half of the height of the cylinder) is d,
Then d=
R2−r2，
Then the height of the cylinder is h = 2
R2−r2
Then the volume of the cylinder v = π r2h ≤ 1
2π（r2+h）
V is the maximum value when and only when R2 = H
That is R2 = 2
R2−r2
That is, R=
2(
1 + R2 − 1),
The cylinder volume is taken as the maximum