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If the sum of the surface areas of cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum? - Mathematics

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प्रश्न

If the sum of the surface areas of cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum?

योग
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उत्तर

Let x be the edge of the cube and r be the radius of the sphere.

Surface area of cube = 6x2

And surface area of the sphere = 4πr2

∴ 6x2 + 4πr2 = K  ......(constant)

⇒ r = `sqrt(("K" - 6x^2)/(4pi)`  .....(i)

Volume of the cube = x3 and the volume of sphere = `3/4 pi"r"^3`

∴  Sum of their volumes (V) = Volume of cube + Volume of sphere

⇒ V = `x^3 + 4/3 pi"r"^3`

⇒ V = `x^3 + 4/3 pi xx (("K" - 6x^2)/(4pi))^(3/2)`

Differentiating both sides w.r.t. x, we get

`"dV"/"dx" = 3x^2 + (4pi)/3 xx 3/2("K" - 6x^2)^(1/2) (- 12x) xx 1/((4pi)^(3/2)`

= `3x^2 + (2pi)/((4pi)^(3/2)) xx (-12x) ("K" - 6x^2)^(1/2)`

= `3x^2 + 1/(4pi^(1/2)) xx (-12x) ("K" - 6x^2)^(1/2)`

∴ `"dV"/"dx" = 3x^2 - (3x)/sqrt(pi) ("K" - 6x^2)^(1/2)`  ....(ii)

For local maxima and local minima, `"dV"/"dx"` = 0

∴ `3x^2 - (3x)/sqrt(pi) ("K" - 6x^2)^(1/2)` = 0

⇒ `3x[x - ("k" - 6x^2)^(1/2)/sqrt(pi)]` = 0

x ≠ 0

∴ `x - ("K" - 6x^2)^(1/2)/sqrt(pi)` = 0

⇒ x = `("K" - 6x^2)^(1/2)/sqrt(pi)`

Squaring both sides, we get

x2 = `("K" - 6x^2)/pi`

⇒ `pix^2 = "K" - 6x^2`

⇒ `pix^2 + 6x^2` = K

⇒ `x^2(pi + 6)` = K

⇒ x2 = `"K"/(pi + 6)`

∴ x = `sqrt("K"/(pi + 6)`

Now putting the value of K in equation (i), we get

`6x^2 + 4pir^2 = x^2(pi + 6)`

⇒ `6x^2 + 4pi"r"^2 = pix^2 + 6x^2`

⇒ `4pi"r"^2 = pi"r"^2`

⇒ 4r2 = x2

∴ 2r = x

∴ x:2r = 1:1

Now differentiating equation (ii) w.r.t x, we have

`("d"^2"V")/("dx"^2) = 6x - 3/sqrt(pi) "d"/"dx" [x("K" - 6x^2)^(1/2)]`

= `6x - 3/sqrt(pi)[x * 1/(2sqrt("K" - 6x^2)) xx (-12x) + ("K" - 6x^2)^(1/2) * 1]`

= `6x - 3/sqrt(pi) [(-6x^2)/sqrt("K" - 6x^2) + sqrt("K" - 6x^2)]`

= `6x - 3/sqrt(pi) [(-6x^2 + "K" - 6x^2)/sqrt("K" - 6x^2)]`

= `6x + 3/sqrt(pi) [(12x^2 - "K")/sqrt("K" - 6x^2)]`

Put x = `sqrt("K"/(pi + 6)`

= `6sqrt("K"/(pi + 6)) + 3/sqrt(pi)[((12"K")/(pi + 6) - "K")/sqrt("K" - (6"K")/(pi + 6))]`

= `6sqrt("K"/(pi + 6)) + 3/sqrt(pi) [(12"K" - pi"K" - 6"K")/sqrt((pi"K" + 6"K" - 6"K")/(pi + 6))]`

= `6sqrt("K"/(pi + 6)) + 3/sqrt(pi) [(6"K" - pi"K")/sqrt((pi"K")/(pi + 6))]`

= `6sqrt("K"/(pi + 6)) + 3/(pisqrt("K"))[(6"K" - pi"K") sqrt(pi + 6)] > 0`

So it is minima.

Hence, the required ratio is 1 : 1 when the combined volume is minimum.

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अध्याय 6: Application Of Derivatives - Exercise [पृष्ठ १३८]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 6 Application Of Derivatives
Exercise | Q 31 | पृष्ठ १३८

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