Question

Find the dimensions of a rectangle of perimeter 12 cm which will generate maximum volume when swept along a circular rotation keeping the shorter side fixed as the axis.

Hint:

To maximize the volume, take the derivative of the volume function with respect to the variable and solve for critical points. Then use the second derivative test to verify if it’s a maximum.

Answer 1

Let the rectangle's dimensions be \( x \) and \( y \), with \( x \) being the shorter side. The perimeter equation is: \[ 2x + 2y = 12 \quad \Rightarrow \quad x + y = 6 \] This implies \( y = 6 - x \). When the rectangle rotates circularly around its shorter side (fixed as the axis), the resulting solid's volume is that of a cylinder: \[ V = \pi x^2 y = \pi x^2 (6 - x) \] To maximize this volume, we differentiate \( V \) with respect to \( x \): \[ \frac{dV}{dx} = \pi (12x - 3x^2) \] Setting \( \frac{dV}{dx} = 0 \) for critical points: \[ 12x - 3x^2 = 0 \quad \Rightarrow \quad 3x(4 - x) = 0 \] The solutions are \( x = 0 \) or \( x = 4 \). Since \( x \) represents a dimension, \( x = 4 \). Substitute \( x = 4 \) into the perimeter equation to find \( y \): \[ y = 6 - 4 = 2 \] Thus, the dimensions of the rectangle that maximize the volume are 4 cm by 2 cm.