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Let the length and breadth of the rectangular sheet be $l$ and $b$ respectively. The perimeter is given as 300 cm, so we have: $2(l + b) = 300$ $l + b = 150$ $b = 150 - l$
When the sheet is rolled along its length $l$, the circumference of the base of the cylinder is $l$, and the height of the cylinder is $b$. Thus, $2\pi r = l$ $r = \frac{l}{2\pi}$ The volume $V$ of the cylinder is given by: $V = \pi r^2 h = \pi (\frac{l}{2\pi})^2 b = \pi \frac{l^2}{4\pi^2} (150 - l) = \frac{l^2(150 - l)}{4\pi}$
To maximize the volume, we need to find the critical points by taking the derivative of $V$ with respect to $l$ and setting it to zero. $\frac{dV}{dl} = \frac{1}{4\pi} \frac{d}{dl} (150l^2 - l^3) = \frac{1}{4\pi} (300l - 3l^2)$ Set $\frac{dV}{dl} = 0$: $300l - 3l^2 = 0$ $3l(100 - l) = 0$ So, $l = 0$ or $l = 100$. Since $l = 0$ would result in zero volume, we consider $l = 100$.
To verify that $l = 100$ gives a maximum volume, we take the second derivative of $V$ with respect to $l$: $\frac{d^2V}{dl^2} = \frac{1}{4\pi} (300 - 6l)$ Evaluate at $l = 100$: $\frac{d^2V}{dl^2} = \frac{1}{4\pi} (300 - 600) = \frac{-300}{4\pi} = \frac{-75}{\pi}$ Since $\frac{d^2V}{dl^2} < 0$, the volume is maximized when $l = 100$.
When $l = 100$, we have $b = 150 - l = 150 - 100 = 50$. Therefore, the dimensions of the rectangular sheet are length $l = 100$ cm and breadth $b = 50$ cm.
Final Answer: Length = 100 cm, Breadth = 50 cm
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