मराठी

An open box with square base is to be made of a given quantity of cardboard of area c2. Show that the maximum volume of the box is cc363 cubic units

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प्रश्न

An open box with square base is to be made of a given quantity of cardboard of area c2. Show that the maximum volume of the box is `"c"^3/(6sqrt(3))` cubic units

बेरीज
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उत्तर


Let x be the length of the side of the square base of the cubical open box and y be its height.

∴ Surface area of the open box

c2 = x2 + 4xy

⇒ y = `("c"^2 - x^2)/(4x)`  ....(i)

Now volume of the box, V = x × x × y

⇒ V = x2y

⇒ V = `x^2(("c"^2 - x^2)/(4x))`

⇒ V = `1/4 ("c"^2x - x^3)`

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

`"dv"/"dx" = 1/4 ("c"^2 - 3x^2)` ....(ii)

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

∴ `1/4 ("c"^2 - 3x^2)` = 0

⇒ c2 – 3x2 = 0

⇒ x2 = `"c"^2/3`

∴ x = `sqrt("c"^2/3) = "c"/sqrt(3)`

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

`("d"^2"V")/("dx"^2) = 1/4 (- 6x)`

= `(-3)/2 * "c"/sqrt(3) < 0`  ...(maxima)

Volume of the cubical box (V) = x2y

= `x^2(("c"^2 - x^2)/4x)`

= `"c"/sqrt(3)[("c"^2 - "c"^2/3)/4]`

= `"c"/sqrt(3) xx (2"c"^2)/(3 xx 4)`

= `"c"^3/(6sqrt(3))`

Hence, the maximum volume of the open box is `"c"^3/(6sqrt(3))` cubic units.

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पाठ 6: Application Of Derivatives - Exercise [पृष्ठ १३७]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
पाठ 6 Application Of Derivatives
Exercise | Q 29 | पृष्ठ १३७

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