Step 1: Variable Definition
Let the rectangle's length be \( x \) cm and its breadth be \( (150 - x) \) cm, derived from the perimeter equation \( 2(x + \text{breadth}) = 300 \). When rolled along its length, the resulting cylinder has radius \( r \) and height \( h \) given by: \[ 2\pi r = x \implies r = \frac{x}{2\pi}, \quad h = (150 - x). \] Step 2: Cylinder Volume Formulation
The cylinder's volume \( V \) is expressed as: \[ V = \pi r^2 h = \pi \left(\frac{x}{2\pi}\right)^2 (150 - x). \] Simplified, this is: \[ V = \pi \frac{x^2}{4\pi^2} (150 - x) = \frac{x^2}{4\pi} (150 - x). \] Step 3: Volume Differentiation
Differentiating \( V \) with respect to \( x \) yields: \[ \frac{dV}{dx} = \frac{1}{4\pi} \left(2x(150 - x) - x^2\right). \] The simplified derivative is: \[ \frac{dV}{dx} = \frac{1}{4\pi} \left(300x - 3x^2\right). \] Step 4: Critical Point Identification
Setting \( \frac{dV}{dx} = 0 \) to find critical points results in: \[ 300x - 3x^2 = 0 \implies x(300 - 3x) = 0 \implies x = 0 \text{ or } x = 100. \] Step 5: Second Derivative Test for Maxima
The second derivative is calculated as: \[ \frac{d^2V}{dx^2} = \frac{1}{4\pi} (300 - 6x). \] Evaluating at \( x = 100 \): \[ \frac{d^2V}{dx^2} = \frac{1}{4\pi} (300 - 600) = \frac{-300}{4\pi} = \frac{-75}{\pi} < 0. \] This confirms that \( V \) is maximized when \( x = 100 \).
Step 6: Rectangle Dimension Calculation
With \( x = 100 \), the rectangle's length is \( 100 \) cm, and its breadth is: \[ 150 - x = 50 \, \text{cm}. \]