Question:
If an open cylinder of given surface area has maximum volume, then its radius is
If an open cylinder of given surface area has maximum volume, then its radius is
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For optimization problems involving cylinders, first express the volume in terms of a single variable using the surface area constraint, then differentiate and apply the second derivative test.
Updated On: Jun 26, 2026
- Height of the cylinder
- Height of the cylinder \(/2\)
- \(2\) times Height of the cylinder
- \(3\) times Height of the cylinder
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The Correct Option is A
Solution and Explanation
Step 1: Write the surface area of an open cylinder.
For an open cylinder (open at the top), the surface area consists of \[ \text{Curved Surface Area}+\text{Area of one base}. \] Hence, \[ S=2\pi rh+\pi r^2. \] Since the surface area is given and fixed, \[ 2\pi rh+\pi r^2=S. \] Therefore, \[ h=\frac{S-\pi r^2}{2\pi r}. \]
Step 2: Express the volume in terms of \(r\).
The volume of the cylinder is \[ V=\pi r^2h. \] Substituting the value of \(h\), \[ V = \pi r^2 \left( \frac{S-\pi r^2}{2\pi r} \right). \] \[ V = \frac{r(S-\pi r^2)}{2}. \] \[ V = \frac{Sr-\pi r^3}{2}. \]
Step 3: Differentiate and find the critical point.
\[ \frac{dV}{dr} = \frac{1}{2} \left( S-3\pi r^2 \right). \] For maximum volume, \[ \frac{dV}{dr}=0. \] Hence, \[ S-3\pi r^2=0. \] \[ S=3\pi r^2. \]
Step 4: Find the relation between \(r\) and \(h\).
Using \[ S=2\pi rh+\pi r^2, \] and \[ S=3\pi r^2, \] we get \[ 2\pi rh+\pi r^2=3\pi r^2. \] Dividing by \(\pi r\), \[ 2h+r=3r. \] \[ 2h=2r. \] \[ h=r. \]
Step 5: Verify maximum condition.
\[ \frac{d^2V}{dr^2} = -\frac{6\pi r}{2} = -3\pi r. \] Since \(r\gt 0\), \[ \frac{d^2V}{dr^2}\lt 0. \] Hence the volume is maximum.
Step 6: Final conclusion.
Therefore, \[ \boxed{r=h} \] i.e., the radius is equal to the height of the cylinder.
For an open cylinder (open at the top), the surface area consists of \[ \text{Curved Surface Area}+\text{Area of one base}. \] Hence, \[ S=2\pi rh+\pi r^2. \] Since the surface area is given and fixed, \[ 2\pi rh+\pi r^2=S. \] Therefore, \[ h=\frac{S-\pi r^2}{2\pi r}. \]
Step 2: Express the volume in terms of \(r\).
The volume of the cylinder is \[ V=\pi r^2h. \] Substituting the value of \(h\), \[ V = \pi r^2 \left( \frac{S-\pi r^2}{2\pi r} \right). \] \[ V = \frac{r(S-\pi r^2)}{2}. \] \[ V = \frac{Sr-\pi r^3}{2}. \]
Step 3: Differentiate and find the critical point.
\[ \frac{dV}{dr} = \frac{1}{2} \left( S-3\pi r^2 \right). \] For maximum volume, \[ \frac{dV}{dr}=0. \] Hence, \[ S-3\pi r^2=0. \] \[ S=3\pi r^2. \]
Step 4: Find the relation between \(r\) and \(h\).
Using \[ S=2\pi rh+\pi r^2, \] and \[ S=3\pi r^2, \] we get \[ 2\pi rh+\pi r^2=3\pi r^2. \] Dividing by \(\pi r\), \[ 2h+r=3r. \] \[ 2h=2r. \] \[ h=r. \]
Step 5: Verify maximum condition.
\[ \frac{d^2V}{dr^2} = -\frac{6\pi r}{2} = -3\pi r. \] Since \(r\gt 0\), \[ \frac{d^2V}{dr^2}\lt 0. \] Hence the volume is maximum.
Step 6: Final conclusion.
Therefore, \[ \boxed{r=h} \] i.e., the radius is equal to the height of the cylinder.
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