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.

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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.

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