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Question

(A) Let the side of the square base be $x \ cm$ and the height of the tank be $h \ cm$.The volume of the tank is $V = x^2 h = 4000$.Thus,$h = \frac{40

Vedclass · Accepted Answer

side $= 20 \ cm$,height $= 10 \ cm$

Vedclass · Answer

(A) Let the side of the square base be $x \ cm$ and the height of the tank be $h \ cm$.The volume of the tank is $V = x^2 h = 4000$.Thus,$h = \frac{4000}{x^2}$.The surface area $S$ of an open tank is given by $S = x^2 + 4xh$.Substituting $h$ in the surface area formula: $S = x^2 + 4x(\frac{4000}{x^2}) = x^2 + \frac{16000}{x}$.To find the minimum surface area,differentiate $S$ with respect to $x$: $\frac{dS}{dx} = 2x - \frac{16000}{x^2}$.Set $\frac{dS}{dx} = 0$ for critical points: $2x = \frac{16000}{x^2} \implies x^3 = 8000 \implies x = 20 \ cm$.Now,find the height: $h = \frac{4000}{20^2} = \frac{4000}{400} = 10 \ cm$.Since $\frac{d^2S}{dx^2} = 2 + \frac{32000}{x^3} > 0$ at $x = 20$,the surface area is minimum at $x = 20 \ cm$ and $h = 10 \ cm$.

An open tank with a square bottom is to contain $4000 \ cm^3$ of liquid. Find the dimensions of the tank such that the surface area of the tank is minimum.

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