An open tank is to be constructed with a square base and vertical sides to hold a given capacity of water. The expense of lining the tank with lead is least when the ratio of the depth and the width is

1 : 1

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1 : 2

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2 : 1

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1 : 3

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Solution

The correct option is A 1 : 2
$V = a^{2} h$
Surface area $= a^{2} + 4 a h$
$= a^{2} + \frac{4 v}{a}$
$\frac{d (a r e a)}{d a} = 2 a - \frac{4 v}{a^{2}}$
$a^{3} - 2 v$
$a = (2 v)^{\frac{1}{3}}$
$h = \frac{(v)^{\frac{1}{3}}}{(2)^{\frac{1}{3}}}$
$a : h = 2 : 1$

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An open tank is to be constructed with a square base and vertical sides to hold a given capacity of water. The expense of lining the tank with lead is least when the ratio of the depth and the width is

The correct option is A 1 : 2V=a2hSurface area =a2+4ah=a2+4vad(area)da=2a−4va2a3−2va=(2v)13h=(v)13(2)13a:h=2:1

The correct option is A 1 : 2
$V = a^{2} h$
Surface area $= a^{2} + 4 a h$
$= a^{2} + \frac{4 v}{a}$
$\frac{d (a r e a)}{d a} = 2 a - \frac{4 v}{a^{2}}$
$a^{3} - 2 v$
$a = (2 v)^{\frac{1}{3}}$
$h = \frac{(v)^{\frac{1}{3}}}{(2)^{\frac{1}{3}}}$
$a : h = 2 : 1$