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

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`