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

3 Answers
Apr 26, 2017

#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:
enter image source here
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:
enter image source here
Now, imagine cutting the sphere in half two times. After the first and second cuts, we will see:
enter image source here
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)#

Apr 26, 2017

#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#

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/.

enter image source here