What is the maximum volume of a cylinder that can be cut from a sphere with a radius of 50?

Question

I am not sure why the problem has been removed. I'd like to see different approaches. Here is my solution.

To make the solution process simpler, let r = 1 first.
x^2+y^2 = 1, constraint
v = 2pix^2y, target
Differentiate the above two equations with respect to y,
2xx'+2y = 0 => x' = -y/x ......(1)
v' = 2pi(2xx'y+x^2)......(2)
Substitute (1) into (2) and simplify,
-3y^2+(x^2+y^2) = 0 =>-3y^2 + 1 = 0
y = 1/√3, x = √(2/3)
v(max) = 2pi(2/3)(1/√3) = 4pi√3/9

For r = 50, we can use similarity theorem to get
v(max) = 50^3*(4pi√3/9) = 302,299.89

To kaksi_gu…,
Thank you.
I used the plane z=0 so that I was able to reduce a 3-D problem into a 2-D problem.

Answer 1

I GOT THE SAME ANSWER AS YOU DID - but obtained it in a different way:

First, I drew a sketch of a cylinder (radius r) inside a sphere (radius R).

V = &pi · r² · h (volume of a cylinder)

-> we need to express h in terms of r and R, so that V is a function of one variable only.

The edges of the circular base of the cylinder are touching the sphere from the inside.
h/2, r and R form a right triangle
=> (h/2)² = R² - r²
=> h = 2√(R² - r²)
=> h = 2√(2500 - r²)

so the volume of the cylinder can now be expressed as
V = 2&pi · r² · √(2500 - r²)

V is a function of r now. Differentiate it with respect to r

dV/dr = 2&pi · [2r·√(2500 - r²) + r²·1/2·(2500 - r²)^(-1/2)·(-2r)]
dV/dr = 2&pi · (2r·(2500 - r²) - r³)/√(2500 - r²)

Need to find stationary points. Put dV/dr =0.
2&pi · (2r·(2500 - r²) - r³)/√(2500 - r²) = 0
2r·(2500 - r²) - r³ = 0
r( 5000 - 3r² ) = 0
=> r = 0 (that's a minimum)
=> r = √(5000/3) (that's a maximum)
=> r = - √(5000/3) (that's impossible)

So the answer is : the Volume is max when r = √(5000/3).
Then h = 2√(2500 - r²) = 2√(2500 - 5000/3) = 57.735
and
V = &pi · r² · h = &pi · 5000/3 · 57.735 = 302,299.89

Hope this helps.

______
* I understand we are cutting a solid cylinder out of a solid sphere here, and not taking the surface of the sphere and forming a cylinder with the same surface area [= 1000pi] and max volume. If we were doing the latter, the answer would be different.

Answer 2

Maximum Volume Of A Cylinder

Answer 3

♣ I did not see original question; the question could be removed by asker only before closing it! Askers do it by 2 reasons:
1/ he receives first a correct answer, he copies it, then deletes, lest to choose a best answer, enjoying sort of moral poop;
2/ the question occurs to be so obtuse that the asker feels ashamed in our eyes;
♦ here I wonder how you are going to transform spherical surface into linear surface of a cylinder?

Answer 4

I would just take the zenith angle φ (angle down from the z-axis) as my variable. I'll assume the cylinder's axis is the z-axis. With a spherical radius of ρ, we would have a cylindrical radius of r = ρ*sin(φ) and a cylindrical height of h = ρ*cos(φ). That means that the volume of the cylinder is π(r^2)h = π((ρ*sin(φ))^2)(ρ*cos(φ)) = π(ρ^3)sin²(φ)cos(φ). You need to differentiate this with respect to φ, set equal to zero and solve, which will give you all the extreme values of the original expression, which you must then examine to determine which is the maximum.