An open metallic tank is to be constructed wi | vedclass

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(B) Let the side of the square base be $x \,m$ and the height of the tank be $y \,m$.The volume $V = x^2 y = 500$.Thus,$y = \frac{500}{x^2}$.The surfa

Vedclass · Accepted Answer

$10 \,m, 10 \,m, 5 \,m$

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(B) Let the side of the square base be $x \,m$ and the height of the tank be $y \,m$.The volume $V = x^2 y = 500$.Thus,$y = \frac{500}{x^2}$.The surface area $S$ of the open tank is $S = x^2 + 4xy$.Substituting $y$,we get $S = x^2 + 4x \left(\frac{500}{x^2}\right) = x^2 + \frac{2000}{x}$.Differentiating with respect to $x$: $\frac{dS}{dx} = 2x - \frac{2000}{x^2}$.Setting $\frac{dS}{dx} = 0$,we get $2x = \frac{2000}{x^2} \Rightarrow x^3 = 1000 \Rightarrow x = 10 \,m$.The second derivative $\frac{d^2S}{dx^2} = 2 + \frac{4000}{x^3}$. At $x = 10$,$\frac{d^2S}{dx^2} = 2 + 4 = 6 > 0$,so the area is minimum.The height $y = \frac{500}{10^2} = 5 \,m$.Thus,the dimensions are $10 \,m, 10 \,m, 5 \,m$.

An open metallic tank is to be constructed with a square base and vertical sides,having a volume of $500 \,m^3$. The dimensions of the tank for the minimum area of the sheet metal used in its construction are:

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