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 dimensions of the rectangle be \( x \) and \( y \), where \( x \) is the shorter side. The perimeter condition is: \[ 2x + 2y = 12 \quad \Rightarrow \quad x + y = 6 \] Thus, \( y = 6 - x \). When the rectangle is swept along a circular rotation with the shorter side fixed as the axis, the volume of the solid formed is given by the formula for the volume of a cylinder: \[ V = \pi x^2 y = \pi x^2 (6 - x) \] Now, to find the maximum volume, differentiate \( V \) with respect to \( x \): \[ \frac{dV}{dx} = \pi (12x - x^3) \] Set \( \frac{dV}{dx} = 0 \) to find the critical points: \[ 12x - x^3 = 0 \quad \Rightarrow \quad x(12 - x^2) = 0 \] Thus, \( x = 0 \) or \( x^2 = 12 \), so \( x = 2\sqrt{3} \). Substitute \( x = 2\sqrt{3} \) into the perimeter equation to find \( y \): \[ y = 6 - 2\sqrt{3} \] Thus, the dimensions of the rectangle are \( 2\sqrt{3} \) cm by \( 6 - 2\sqrt{3} \) cm.