An open box with a square base is to be made out of a given quantity of a cardboard of area c^2 square units.show the maximum volume of the box (c^2 /(6(√3))) cubic units


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Question Number 100341 by peter frank last updated on 26/Jun/20

$An open box with a square$ $base is to be made out$ $of a given quantity of$ $a cardboard of area c^{2}$ $square units . show the$ $maximum volume of the$ $box \frac{c^{2}}{6 \sqrt{3}} cubic units$
Commented by PRITHWISH SEN 2 last updated on 26/Jun/20

$Let the area of the floor = x^{2}$ $the area of the remaining 4 sides = c^{2} - x^{2}$ $∴ the height of tbe box = \frac{c^{2} - x^{2}}{4 x}$ $∴ The vol . V (x) = \frac{1}{4} (c^{2} - x^{2}) x$ $V^{^{'}} (x) = \frac{1}{4} (c^{2} - {3 x}^{2}) = 0 \Rightarrow x^{2} = \frac{c^{2}}{3}$ $V^{^{″}} (x) = - \frac{3}{2} x < 0 \forall x$ $∴ V_{\max .} = \frac{1}{4} (c^{2} - \frac{c^{2}}{3}) \frac{c}{\sqrt{3}} = \frac{c^{3}}{6 \sqrt{3}} proved .$
	Answered by bobhans last updated on 26/Jun/20
	$volume of the box = V(x)= (c−2x)^2 x V′(x)= (c−2x)^2 −4(c−2x)x = 0 (c−2x){c−2x−4x} = 0 , x = (1/6)c V_(max) = (c−(1/3)c)^2 .((1/6)c) = (((4c^2 )/9))((c/6)) = ((2c^3 )/(27)) ■$
	$volume of the box = V (x) = {(c - 2 x)}^{2} x$ $V^{'} (x) = {(c - 2 x)}^{2} - 4 (c - 2 x) x = 0$ $(c - 2 x) {c - 2 x - 4 x} = 0, x = \frac{1}{6} c$ $V_{\max} = {(c - \frac{1}{3} c)}^{2} . (\frac{1}{6} c) = (\frac{{4 c}^{2}}{9}) (\frac{c}{6})$ $= \frac{{2 c}^{3}}{27} ◼$
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