Question

A rectangle of perimeter \(24\) cm is revolved along one of its sides to sweep out a cylinder of maximum volume. Find the dimensions of the rectangle. 

Hint:

For maximum or minimum problems, express the required quantity in terms of a single variable using given constraints, then differentiate and set the derivative equal to zero.

Answer 1

Step 1: Define the variables.
Let the length of the rectangle be \(l\) and the breadth be \(b\).
Given that the perimeter of the rectangle is \[ 2(l+b)=24 \] \[ l+b=12 \] \[ b=12-l \] When the rectangle is revolved about one of its sides, it generates a cylinder.
In this cylinder: \[ \text{Radius} = b \] \[ \text{Height} = l \] Step 2: Write the formula for volume of the cylinder.
The volume of a cylinder is \[ V=\pi r^2 h \] Substituting \(r=b\) and \(h=l\): \[ V=\pi b^2 l \] Using \(b=12-l\): \[ V=\pi (12-l)^2 l \] Step 3: Express the volume as a function of \(l\).
\[ V(l)=\pi l(12-l)^2 \] Expand the expression: \[ (12-l)^2=144-24l+l^2 \] Thus \[ V=\pi(144l-24l^2+l^3) \] Step 4: Differentiate to find maximum volume.
\[ \frac{dV}{dl}=\pi(144-48l+3l^2) \] For maximum volume \[ \frac{dV}{dl}=0 \] \[ 144-48l+3l^2=0 \] Divide by \(3\): \[ 48-16l+l^2=0 \] \[ l^2-16l+48=0 \] \[ (l-4)(l-12)=0 \] Thus \[ l=4 \quad \text{or} \quad l=12 \] But if \(l=12\), then \(b=0\), which is not possible.
Hence \[ l=4 \] Step 5: Find the breadth.
\[ b=12-l \] \[ b=12-4 \] \[ b=8 \] Step 6: Verify maximum condition.
The second derivative \[ \frac{d^2V}{dl^2}=\pi(-48+6l) \] Substitute \(l=4\): \[ \pi(-48+24)=-24\pi<0 \] Hence the volume is maximum.
Step 7: Final dimensions.
Thus the rectangle must have sides \(4\) cm and \(8\) cm to produce the cylinder of maximum volume.
Final Answer:
\[ \boxed{l=4\text{ cm},\quad b=8\text{ cm}} \]