The sum of the surface areas of a cuboid with sides `x ,2x` and `x/3` and a sphere is given to be constant. Prove that the sum of their volumes is minimum, if `x` is equal to three times the radius of sphere. Also find the minimum value of the sum of their volumes.

Explanation

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))`