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-

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 - 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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