The sum of the surface areas of a sphere and a cube is given. Show that when the sum of their volumes is least, the diameter of the sphere is equal to the edge of the cube.

Explanation

Let the length of one edge of cube be x units and radius of sphere be r units.
`therefore` Surface area of cube `=6x^(2)`
and surface area of sphere=`4pir^(2)`
Also, `6x^(2)+4pir^(2)=k` [constant, given]
`rArr 6x^(2)=k-4pir^(2)`
`rArr x^(2)=(k-4pir^(2))/(6)`
`rArr x=[(k-4pir^(2))/(6)]^(1//2)` ...............(i)
Now, volume of cube `=x^(3)`
and Volume of sphere `=4/3pir^(3)`
Let sum of volume of the cube and volume of the sphere be given by
`S= x^(3)+4/3pir^(3)=[(k-4pir^(2))/(6)]^(3//2) + 4/3pir^(2)`
`=2pir[(k-4pir^(2))/6]^(1//2)+4pir^(2)`...........(ii)
`=-2pir{(k-4pir^(2))/(6)}^(1//2)`
`rArr 4r^(2)=(k-4pir^(2))/(6) rArr 24r^(2)=k-4pir^(2)`
`rArr 24r^(2)+4pir^(2)=k rArr r^(2)[24+4pi]=k`
`therefore r=0 or r=sqrt(k/(24+4pi]=k`
We know that, `r ne 0`
`therefore r=1/2sqrt(k/(6+pi)]`
Again, differentiating w.r.t. r in Eq. (ii), we get
`(d^(2)S)/(dr^(2)) = d/(dr)[-2pir(k-4pir^(2)/(6))^(1//2) + 4pir^(2)]`
`-2pi[r.1/2((k-4pir^(2))/(6))^(-1//2).(-8pir)/(6)+((k-4pir)/6)^(1//2).1]+4pi.2r`
`-2pi[r.1/(2sqrt(k-4pir^(2))/(6)).(-8pir)/(6)+sqrt(k-4pir^(2))/(6)]+8pir`
`-2pi[(-8pir^(2)+12(k-(4pir^(2))/6))/(12sqrt((K-4pir^(2))/6))]+8pir`
`=-2pi[(-48pir^(2)+72k-48pir^(2))/(72sqrt((k-4pir^(2))/(6)))]+8pir=-2pi[(-96pir^(2)+72k)/(72sqrt(k-4pir^(2))/(6))]+8pir gt0`
For `r=1/2sqrt(k/(6+pi))`, then the sum of their volume is minimum.
For `r=1/2sqrt(k/(6+pi))` `x=[(k-4pi.1/4k/(6+pi))/(6)]^(1//2)`
`=[((6+pi)k-pik)/(6(6+pi))]^(1//2)=[k/(6+pi]]^(1//2)=2r`
Since, the sum of their volume is minimum when x=2r.
Hence, the ratio of an edge of cube to the diameter of the sphere is `1:1`