A tank with a rectangular base and rectangula | vedclass

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(B) Let the length and breadth of the tank be $x \ m$ and $y \ m$ respectively. The height of the tank is $h = 4 \ m$.The volume of the tank is $V = x

Vedclass · Accepted Answer

$₹ 3300$

Vedclass · Answer

(B) Let the length and breadth of the tank be $x \ m$ and $y \ m$ respectively. The height of the tank is $h = 4 \ m$.The volume of the tank is $V = x 	imes y 	imes h = 36 \ m^3$.Substituting $h = 4$,we get $4xy = 36$,which implies $xy = 9$,so $y = \frac{9}{x}$.The cost function $C$ is given by the cost of the base plus the cost of the four sides:$C = 100(xy) + 50(2xh + 2yh)$Substituting $xy = 9$,$h = 4$,and $y = \frac{9}{x}$:$C(x) = 100(9) + 50(2x(4) + 2(\frac{9}{x})(4))$$C(x) = 900 + 50(8x + \frac{72}{x}) = 900 + 400x + \frac{3600}{x}$To find the minimum cost,we find the derivative $C'(x)$ and set it to zero:$C'(x) = 400 - \frac{3600}{x^2} = 0$$400 = \frac{3600}{x^2} \Rightarrow x^2 = 9 \Rightarrow x = 3 \ m$.Since $x = 3$,$y = \frac{9}{3} = 3 \ m$.The minimum cost is $C(3) = 900 + 400(3) + \frac{3600}{3} = 900 + 1200 + 1200 = ₹ 3300$.

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