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Question
The sum of the surface areas of a rectangular parallelopiped 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 the sphere. Also find the minimum value of the sum of their volumes.
Solution
It is given that, the sum of the surface areas of a rectangular parallelepiped with sides x, 2x and `x/3` and a sphere is constant.
Let S be the sum of both the surface area.
∴ S = 2`(x * 2x + 2x * x/3 +x/3 * x) + 4pi"r"^2` = k
⇒ `4pi"r"^2 = "k" - 6x^2`
⇒ r2 = `("k" - 6x^2)/(4pi)`
⇒ r = `sqrt(("k" - 6x^2)/(4pi)` .....(i)
Let V denotes the sum of the volume of both the parallelepiped and the sphere.
Then, V = `2x * x * x/3 + 4/3 pi"r"^3`
= `2/3 x^3 + 4/3 pi"r"^3`
= `2/3 x^3 + 4/3pi(("kk" - 6x^2)/(4pi))^(3/2)`
= `2/3 x^3 + 4/3 pi (("k" - 6x^2)/(4pi))^(3/2)`
⇒ V = `2/3 x^3 + 1/(6sqrt(pi)) ("k" - 6x^2)^(3/2)` ....(ii)
Differentiating w.r.t. x,
`"dV"/"dx" = 2/3 * 3x^2 + 1/(6sqrt(pi)) * 3/2 * ("k" - 6x^2)^(1/2)(-12x)`
= `2x^2 - (3x)/sqrt(pi) ("k" - 6x^2)^(1/2)` ....(iii)
Let `"dV"/"dx"` = 0
⇒ `2x^2 = (3x)/sqrt(pi) ("k" - 6x^2)^(1/2)`
⇒ `4x^4 = (9x^2)/pi ("k" - 6x^2)`
⇒ `4pix^4 = 9"k"x^2 - 54x^4`
⇒ `x^2 = (9"k")/(4pi + 54)`
⇒ x = `3sqrt("k"/(4pi + 54))` .....(iv)
Clearly this is point minima.
When x = `3sqrt("k"/(4pi + 54))`
`"r"^2 = ("k" - 6) ((9"k")/(4pi + 54))/(4pi)`
= `("k"(4pi + 54) - 54"k")/(4pi(4pi + 54))`
= `(4"k"pi)/(4pi(4pi + 54))`
= `"k"/(4pi + 54)`
⇒ r = `sqrt("k"/(4pi + 54))`
⇒ x = 3r
Also V = `2/3x^3 + 4/3 pi"r"^3`
= `2/3(3"r")^3 + 4/3 pi"r"^3`
= `18"r"^3 + 4/3 pi"r"^3`
= `(18 + 4/3 pi)"r"^3`
= `((54 + 4pi)/3)("k"/(4pi + 54))^(3/2)`
= `"k"^(3/2)/(3(4pi + 54)^(3/2)`
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