Question

How to determine height of the cylinder with maximum volume engraved in a sphere with radius `R`?

Answer 1

`h=sqrt((4R^2)/3` where `R` is the radius of the sphere

Explanation:

This problem is really interesting, but will definitely require some visualization to figure out. Let's start by using the formula for the volume of a cylinder:

We know that the area of a circle is:

`A=pir^2`

Therefore, the volume is:

`V=pir^2h`

This cylinder, however, is engraved in a sphere. Its cross-sectional area and height are, therefore, restricted by the sphere as shown below:

Now, imagine cutting the sphere in half two times. After the first and second cuts, we will see:

If we focus on the cross section in the last cut, we can use the Pythagorean theorem to find a relationship between `1/2h`, `r`, and `R`. Specifically,

`(1/2h)^2+r^2=R^2`
`h^2/4+r^2=R^2`

Solving for `r`, we get:
`r^2=R^2-h^2/4`

Now that we have `r^2` as a function of `h`, plug it back into the original volume equation:

`V=pi(R^2-h^2/4)*h`

Simplify the expression:

`V=piR^2h-pih^3/4`

Now that we have the volume of the cylinder expressed as a function of its height, `h`, take the derivative of the volume function with respect to `h` and set it equal to 0.

(Recall: Local maximum of a function is located where its derivative equals 0).

`(dV)/(dh)=piR^2-3h^2/4`
`0=piR^2-(3h^2)/4`
`-piR^2=-(3h^2)/4`
`-4/(3pi)*piR^2=-(3pih^2)/4*-4/(3pi)`
`(4R^2)/3=h^2`

Therefore, the value of `h` that maximizes the volume of a cylinder engraved in a cylinder is:

`h=sqrt((4R^2)/3)`

Answer 2

`h = (2sqrt3)/3R`

Explanation:

Assuming a cylinder with the vertical axis coincident with the `z` axis engraved in a sphere centered at the origin and defining

`r = R sin theta` Cylinder base radius
`h = 2Rcos theta` Cylinder height

we have

`V_c = pi r^2 h` Cylinder volume

but

`V_c = pi(R sin theta)^2(2R cos theta) = 2pi R^3 sin^2theta cos theta`

so

`max_(theta) V_c` is at `theta_0` such that

`(d V_c)/(d theta) = 2piR^3(2sin theta cos^2 theta-sin^3 theta)=0`

or

`{(sin theta = 0),(3cos^2 theta-1 = 0):}`

The solutions are

`theta = 0` or `theta = pm arccos(sqrt3/3)`

so

`h = (2sqrt3)/3R`

NOTE: This is a maximum point because

`(d^2V_c)/(d theta^2) = -(8 pi R^3)/sqrt[3] < 0`

Answer 3

See the answer below:
This problem has been translated and adapted from the same problem shown in the website http://ecalculo.if.usp.br/derivadas/estudo_var_fun/probl_otimizacao/problemas/problema11.htm

Thanks to the students of Mathematics at USP, University of São Paulo who developed this course in Mathematics as a conclusion paper and posted at http://ecalculo.if.usp.br/.

Explanation:

See the answer below:

This problem has been translated and adapted from the same problem shown in the website http://ecalculo.if.usp.br/derivadas/estudo_var_fun/probl_otimizacao/problemas/problema11.htm

Thanks to the students of Mathematics at USP,University of São Paulo who developed this course in Mathematics as a conclusion paper and posted at http://ecalculo.if.usp.br/.