A square piece of tin of side 18 , cm is to b | vedclass

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(A) Let the side of the square to be cut off be $x \, cm$. Then,the length and the breadth of the box will be $(18-2x) \, cm$ each,and the height of t

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$3$

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(A) Let the side of the square to be cut off be $x \, cm$. Then,the length and the breadth of the box will be $(18-2x) \, cm$ each,and the height of the box will be $x \, cm$. Therefore,the volume $V(x)$ of the box is given by: $V(x) = x(18-2x)^2 = x(324 - 72x + 4x^2) = 4x^3 - 72x^2 + 324x$. To find the maximum volume,we find the derivative $V'(x)$: $V'(x) = 12x^2 - 144x + 324$. Setting $V'(x) = 0$: $12(x^2 - 12x + 27) = 0 \Rightarrow 12(x-9)(x-3) = 0$. So,$x = 9$ or $x = 3$. If $x = 9$,the side of the box $(18-2x)$ becomes $0$,which is not possible. Thus,$x = 3$. Now,we check the second derivative $V''(x) = 24x - 144$. $V''(3) = 24(3) - 144 = 72 - 144 = -72 Since $V''(3) < 0$,the volume is maximum at $x = 3 \, cm$.

$A$ square piece of tin of side $18 \, cm$ is to be made into a box without a top by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible (in $, cm$)?

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