A tank with a rectangular base and rectangula | vedclass

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Question

(A) Let $l$ and $b$ be the length and breadth of the tank,and $h=2 \ m$ be the depth.Volume $V = l 	imes b 	imes h = 8 \ m^{3}$.$l 	imes b 	imes 2

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Rs. $1000$

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(A) Let $l$ and $b$ be the length and breadth of the tank,and $h=2 \ m$ be the depth.Volume $V = l 	imes b 	imes h = 8 \ m^{3}$.$l 	imes b 	imes 2 = 8 \implies lb = 4 \implies b = \frac{4}{l}$.Cost of base $C_{base} = 70 	imes (lb) = 70 	imes 4 = 280$.Cost of sides $C_{sides} = 45 	imes [2h(l+b)] = 45 	imes [2 	imes 2 	imes (l + \frac{4}{l})] = 180(l + \frac{4}{l})$.Total cost $C(l) = 280 + 180(l + \frac{4}{l})$.To minimize cost,find $\frac{dC}{dl} = 180(1 - \frac{4}{l^{2}})$.Set $\frac{dC}{dl} = 0 \implies 1 - \frac{4}{l^{2}} = 0 \implies l^{2} = 4 \implies l = 2 \ m$ (since $l > 0$).Then $b = \frac{4}{2} = 2 \ m$.Check second derivative: $\frac{d^{2}C}{dl^{2}} = 180(\frac{8}{l^{3}})$. At $l=2$,$\frac{d^{2}C}{dl^{2}} = 180(1) = 180 > 0$,so cost is minimum.Minimum cost $= 280 + 180(2 + \frac{4}{2}) = 280 + 180(4) = 280 + 720 = 1000$.

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