1. Set up rectangle dimensions: Let the length be \( 2r \) and the width be \( h \). Given \( 2r \) is the cylinder base circumference and \( h \) is its height, the perimeter is \( 2r + 2h = 300 \), which simplifies to \( r + h = 150 \), so \( h = 150 - r \).
2. Formulate cylinder volume: The volume \( V \) is given by \( V = \pi r^2 h = \pi r^2 (150 - r) \).
3. Optimize \( V \): To maximize \( V \), we find the derivative with respect to \( r \): \( \frac{dV}{dr} = \pi \left[ 2r(150 - r) - r^2 \right] = \pi (300r - 3r^2) \). Setting \( \frac{dV}{dr} = 0 \) yields \( 300r - 3r^2 = 0 \), which factors to \( 3r(100 - r) = 0 \). The solutions are \( r = 0 \) or \( r = 100 \). We discard \( r = 0 \) as it results in zero volume.
4. Apply second derivative test: Calculate the second derivative: \( \frac{d^2V}{dr^2} = \pi (300 - 6r) \). At \( r = 100 \), \( \frac{d^2V}{dr^2} = \pi (300 - 600) = -300\pi \). Since \( -300\pi<0 \), \( V \) is maximized when \( r = 100 \).
5. Calculate \( h \): Substitute \( r = 100 \) into the expression for \( h \): \( h = 150 - r = 150 - 100 = 50 \).
Final Answer:
The dimensions of the rectangular sheet are \( 2r = 200 \, {cm} \) and \( h = 50 \, {cm} \).