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(D) Let $l$ and $b$ be the length and width of the rectangle.Given that the perimeter is $48 	ext{ cm}$.$2(l + b) = 48 \Rightarrow l + b = 24 \Righta

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$8, 16$

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(D) Let $l$ and $b$ be the length and width of the rectangle.Given that the perimeter is $48 	ext{ cm}$.$2(l + b) = 48 \Rightarrow l + b = 24 \Rightarrow b = 24 - l$ ... $(i)$When the rectangle is rotated about side $b$,the radius of the cylinder is $r = l$ and the height is $h = b$.The volume of the cylinder is $V = \pi r^2 h = \pi l^2 b$.Substituting $b$ from $(i)$: $V = \pi l^2(24 - l) = 24\pi l^2 - \pi l^3$.To find the maximum volume,differentiate with respect to $l$:$\frac{dV}{dl} = 48\pi l - 3\pi l^2$.Set $\frac{dV}{dl} = 0$ for critical points:$3\pi l(16 - l) = 0 \Rightarrow l = 0$ or $l = 16$.Since $l$ must be positive,$l = 16$.Check the second derivative: $\frac{d^2V}{dl^2} = 48\pi - 6\pi l$.At $l = 16$,$\frac{d^2V}{dl^2} = 48\pi - 96\pi = -48\pi Since the second derivative is negative,the volume is maximum at $l = 16$.Then $b = 24 - 16 = 8$.Thus,the dimensions of the rectangle are $8 	ext{ cm}$ and $16 	ext{ cm}$.

Given that the solid obtained by rotating a rectangle about one of its sides is a cylinder. If the perimeter of a rectangle is $48 \text{ cm}$ and the volume of the cylinder formed by rotating it is maximum,then the dimensions of that rectangle are:

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