An open tank with a square bottom,to contain | vedclass

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(B) Let $x$ be the length of the side of the square bottom and $h$ be the height of the tank.Given volume $V = x^2 h = 4000 \ cm^3 \dots (i)$.The surf

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side of square bottom $= 20 \ cm$,height $= 10 \ cm$.

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(B) Let $x$ be the length of the side of the square bottom and $h$ be the height of the tank.Given volume $V = x^2 h = 4000 \ cm^3 \dots (i)$.The surface area $A$ of an open tank is $A = x^2 + 4xh \dots (ii)$.From $(i)$,$h = \frac{4000}{x^2}$.Substituting $h$ in $(ii)$,we get $A = x^2 + 4x \left( \frac{4000}{x^2} \right) = x^2 + \frac{16000}{x}$.Differentiating with respect to $x$,$\frac{dA}{dx} = 2x - \frac{16000}{x^2}$.For minimum surface area,set $\frac{dA}{dx} = 0$,so $2x = \frac{16000}{x^2} \Rightarrow x^3 = 8000 \Rightarrow x = 20 \ cm$.Checking the second derivative,$\frac{d^2A}{dx^2} = 2 + \frac{32000}{x^3}$.At $x = 20$,$\frac{d^2A}{dx^2} = 2 + \frac{32000}{8000} = 2 + 4 = 6 > 0$,hence the area is minimum.Calculating height,$h = \frac{4000}{20^2} = \frac{4000}{400} = 10 \ cm$.

An open tank with a square bottom,to contain $4000 \ cm^3$ of liquid,is to be constructed. The dimensions of the tank,so that the surface area of the tank is minimum,are

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