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$ (A) $ Let $x$ metres be the side length of the removed squares. The height of the box is $x$, the length is $8-2x$, and the breadth is $3-2x$. Since

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$ \frac{200}{27} 	ext{ m}^3 $

Vedclass · Answer

$ (A) $ Let $x$ metres be the side length of the removed squares. The height of the box is $x$, the length is $8-2x$, and the breadth is $3-2x$. Since the dimensions must be positive, $x > 0$, $8-2x > 0$, and $3-2x > 0$, which implies $0 The volume $V(x)$ is given by: $V(x) = x(3-2x)(8-2x) = x(24 - 6x - 16x + 4x^2) = 4x^3 - 22x^2 + 24x$. To find the maximum volume, we find the derivative $V'(x)$: $V'(x) = 12x^2 - 44x + 24$. Setting $V'(x) = 0$: $12x^2 - 44x + 24 = 0 \implies 4(3x^2 - 11x + 6) = 0 \implies 4(3x-2)(x-3) = 0$. The critical points are $x = 2/3$ and $x = 3$. Since $x Using the second derivative test: $V''(x) = 24x - 44$. $V''(2/3) = 24(2/3) - 44 = 16 - 44 = -28 Thus, $x = 2/3$ is the point of local maxima. The maximum volume is: $V(2/3) = (2/3)(3 - 2(2/3))(8 - 2(2/3)) = (2/3)(3 - 4/3)(8 - 4/3) = (2/3)(5/3)(20/3) = 200/27 	ext{ m}^3$.

An open-topped box is to be constructed by removing equal squares from each corner of a $3 \text{ m}$ by $8 \text{ m}$ rectangular sheet of aluminium and folding up the sides. Find the volume of the largest such box.

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