Question:
A closed cylinder of given volume will have least surface area when the ratio of its height and base radius is
A closed cylinder of given volume will have least surface area when the ratio of its height and base radius is
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For a closed cylinder of fixed volume, the surface area becomes minimum when the height is equal to twice the radius.
Updated On: Jun 22, 2026
- \(2:1\)
- \(1:2\)
- \(2:3\)
- \(3:2\)
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The Correct Option is A
Solution and Explanation
Step 1: Write the formulas for volume and surface area.
Let the radius of the cylinder be \(r\) and height be \(h\).
Volume of a closed cylinder is \[ V=\pi r^2 h \] Since the volume is fixed, \[ h=\frac{V}{\pi r^2} \] Total surface area of a closed cylinder is \[ S=2\pi r^2+2\pi rh \] Substituting the value of \(h\), \[ S=2\pi r^2+2\pi r\left(\frac{V}{\pi r^2}\right) \] \[ S=2\pi r^2+\frac{2V}{r} \]
Step 2: Differentiate to minimize the surface area.
Differentiate \(S\) with respect to \(r\): \[ \frac{dS}{dr}=4\pi r-\frac{2V}{r^2} \] For minimum surface area, \[ \frac{dS}{dr}=0 \] Therefore, \[ 4\pi r=\frac{2V}{r^2} \] \[ 4\pi r^3=2V \] \[ 2\pi r^3=V \]
Step 3: Use the volume relation.
We know that \[ V=\pi r^2 h \] Substitute \(V=2\pi r^3\): \[ \pi r^2 h=2\pi r^3 \] Cancelling \(\pi r^2\), \[ h=2r \] Hence, \[ h:r=2:1 \]
Step 4: Final conclusion.
Therefore, the required ratio is \[ \boxed{2:1} \]
Let the radius of the cylinder be \(r\) and height be \(h\).
Volume of a closed cylinder is \[ V=\pi r^2 h \] Since the volume is fixed, \[ h=\frac{V}{\pi r^2} \] Total surface area of a closed cylinder is \[ S=2\pi r^2+2\pi rh \] Substituting the value of \(h\), \[ S=2\pi r^2+2\pi r\left(\frac{V}{\pi r^2}\right) \] \[ S=2\pi r^2+\frac{2V}{r} \]
Step 2: Differentiate to minimize the surface area.
Differentiate \(S\) with respect to \(r\): \[ \frac{dS}{dr}=4\pi r-\frac{2V}{r^2} \] For minimum surface area, \[ \frac{dS}{dr}=0 \] Therefore, \[ 4\pi r=\frac{2V}{r^2} \] \[ 4\pi r^3=2V \] \[ 2\pi r^3=V \]
Step 3: Use the volume relation.
We know that \[ V=\pi r^2 h \] Substitute \(V=2\pi r^3\): \[ \pi r^2 h=2\pi r^3 \] Cancelling \(\pi r^2\), \[ h=2r \] Hence, \[ h:r=2:1 \]
Step 4: Final conclusion.
Therefore, the required ratio is \[ \boxed{2:1} \]
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