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Question
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
Solution
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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