The sum of the surface areas of the rectangular parallelopiped with sides `x , 2x` and `x/3` and a sphere is given to be constant. Prove that the sum

We have given that, the sum of the surface areas of a reactangular parallelopiped with sides x, 2x and `x/3` and a sphere is constant.
`therefore S=2(x.2x+2x.x/3+x/3.x)+4pir^(2)=k`
`k=2[2x^(2)+(2x^(2))/(3)+(x^(2))/(6)] + 4pir^(2)`
`=2[3x^(2)]+4pir^(2)=6x^(2)+pir^(2)`
`rArr 4pir^(2)=k-6x^(2)`
`rArr r^(2)=(k-6x^(2))/(4pi)`
`rArr r=sqrt((k-6x^(2))/(4pi)`
Let V denotes the volume of both the parallelopiped and the sphere.
Then, `V=2x.x.x/3+4/3pir^(3)=2/3x^(3)+4/3pir^(3)`
`=2/3x^(3)+4/3pi.1/(8pi^(3//2))(k-6x^(2))^(3//2)`.................(ii)
On differentiating both sides, w.r.t. x, we get
`(dV)/(dx)=2/3.3x^(2)+1/(6psqrt(pi)).3/2(k-6x^(2))^(1//2).(-12x)`
`=2x^(2)-(12x)/(4sqrt(pi))sqrt(k-6x^(2))^(1//2)`..............(iii)
`therefore (dV)/(dx)=0`
`rArr 2x^(2)=(3x)/(sqrt(pi))(k-6x^(2))^(1//2)`
`rArr 4x^(4)=(9x^(2))/(pi)(k-6x^(2))`
`rArr 4pix^(2)=9kx^(2)-54x^(4)`
`rArr 4pix^(4)+54x^(4)=9kx^(2)`
`rArr x^(4)[4pi+54]=9.k.x^(2)`
`rArr x^(4)[4pi+54]=9.k.x^(2)`
`rArr x^(2)=(9k)/(4pi+54)`
`rArr x=3.sqrt(k/(4pi+54))`..............(iv) Again, differentitating Eq. (iii), w.r.t. x, we get
`(d^(2)V)/(dx^(2))=4x-3/sqrt(pi)[x.1/2(k-6x^(2))^(-1//2).(-12x)+(k-6x^(2))^(1//2).1]`
`=4x-3/sqrt(pi)[(-6x^(2)+k-6x^(2))/(sqrt(k-6x^(2)))]`
`=4x-3/sqrt(pi)[(k-12x^(2))/(sqrt(k-6x^(2)))]`
Now, `(d^(2)V)/(dx^(2))_(x=3.sqrt((k/(4pi+54))))=(4.3)sqrt(k/(4pi+54))-3/sqrt(pi)[(k-12.9.k/(4pi+54))/(sqrt(k-(6.9.k)/(4pi+54)))]`
`12sqrt(k/(4pi+54))-3/sqrt(pi)[(4kpi+54k-108k//4pi+54)/(sqrt(4kpi+54k-54k//4pi+54))]`
`=12sqrt(k/(4pi+54))-3/sqrt(pi)[(4kpi-54k)/(sqrt(4kpisqrt(4pi+54)))]`
`=12sqrt(k/(4pi+54))-6/sqrt(pi)[(k(2pi-27))/(sqrt(k)sqrt(16pi^(2)+216pi))]` [Since, `(pi-27) lt 0 rArr (d^(2)V)/(dx^(2)) gt 0, k gt 0]`
For `x=3sqrt(k/(4pi+54))`, the sum of volumes is minimum.
For `x=3 sqrt(k/(pi+54))`, then `r=sqrt((k-6x^(2))/(4pi))` [using Eq. (i)]
`=1/(2sqrt(pi))sqrt(k-6.(9k)/(4pi+54))`
`=1/(2sqrt(pi)).sqrt((4kpi+54k-54k)/(4pi+54))`
`=1/(2sqrt(pi))sqrt((4kpi)/(4pi+54))=sqrt(k)/(sqrt(4pi+54))=1/3x`
`rArr x=3r` Hence proved.
`therefore` Minimum sum of volume,
`(V)_(x=3.sqrt(k/(4pi+54)))=2/3x^(3)+4/3pir^(3)=2/3x^(3)+4/3pi.(1/3x)^(3)`
`=2/3x^(3)+4/3pi.x^(3)/27=2/3x^(3)(1+(2pi)/(27))`