# The sum of a surface area 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 a cube.

*print*Print*list*Cite

## The sum of a surface area 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 a cube.

The surface area of a sphere is given by `SA=4pir^2`

The surface area of a cube is given by `SA=6s^2`

where *r* is the radius of the sphere and * s *is the side length of the cube.

Let the sum of surface areas be *k*, so `k=4pir^2+6s^2` . Then we can solve for *s* getting `s=sqrt((k-4pir^2)/6)` .

Now the sum of volumes is `4/3pir^3+s^3` . Substituting for *s* yields

`4/3pir^3+((k-4pir^2)/6)^(3/2)` . To minimize this, we can take the derivative with respect to *r*, setting the derivative equal to zero to find the critical points,and evaluating to find the minimum.

`d/(dr)``[4/3pir^3+((k-4pir^2)/6)^(3/2)]=4pir^2+3/2((k-4pir^2)/(6))^(1/2)(-4/3pir)`` `

Setting this equal to zero we find:

` ` ` ``4pir^2+3/2((k-4pir^2)/(6))^(1/2)(-4/3pir)=0`` `

`4pir^2=2pir((k-4pir^2)/(6))^(1/2)`

`2r=((k-4pir^2)/(6))^(1/2)`

`4r^2=(k-4pir^2)/6`

`24r^2=(4pir^2+6s^2)-4pir^2` by substituting back for *k*

`24r^2=6s^2`

`4r^2=s^2`

`2r=s` Since we are dealing with lengths, we only need the positive root.

Since *2r* is the diameter of the sphere, we have shown that the sum of the volumes is minimized when the diameter of the sphere equals the side length of the cube.