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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.
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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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