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 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.