What is the smallest possible height of the cylinder?

Question

A cylindrical container must be constructed to hold two spheres of diameters $30$ cm and $20$ cm. The base must fit perfectly to a sphere of $30$ cm in diameter. What is the smallest possible height of the cylinder to fit the two spheres?

Comments: Placing the sphere with diameter $30$ and then with diameter $20,$ the upper sphere will move $5$ cm sideways to the cylinder wall. How to find the difference in heights?

Ps: Sorry for any mistakes in English.

Answer 1

$GE = 15 + 10 = 25$

$EK = EI - GJ = 15-10 = 5$

$GK = \sqrt{25^2 - 5^2} = 10\sqrt6$

So, height of the cylinder $= BI + GK + JA = 15 + 10\sqrt6 + 10 = 25 + 10\sqrt6 \approx 49.49 \lt 50$

Answer 2

Extension from properties of circles : When two spheres touch each other, the point of tangency lies on the line joining the centers of two spheres.

For minimum height of second sphere above the bigger sphere, the smaller sphere is tangent to both the larger sphere as well as to the wall of the cylinder (at one point). Use tangency to find distances (Pythagoras theorem).

I got

$ 25+10\sqrt{6} $

Answer 3

Hint:

The equation of smaller circle will be $$(x+10)^2+(y-k)^2=100$$ You can find $k$ by the fact that the big circle touches it. The minimum height will be $h+10$.

$$k=15+10\sqrt 6\\ h=25+10\sqrt 6$$