Question:
The maximum volume (in cu. m) of a right circular cone having slant height \(3\) m is
The maximum volume (in cu. m) of a right circular cone having slant height \(3\) m is
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When the slant height of a cone is fixed, use \(r^2+h^2=l^2\) to eliminate one variable. Then differentiate the volume function to obtain the maximum volume.
Updated On: Jul 29, 2026
- \(6\pi\)
- \(3\sqrt3\,\pi\)
- \(\dfrac{4\pi}{3}\)
- \(2\sqrt3\,\pi\)
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The Correct Option is D
Solution and Explanation
Concept:
For a cone,
\[
V=\frac13\pi r^2h
\]
and the slant height \(l\), radius \(r\), and height \(h\) satisfy
\[
l^2=r^2+h^2.
\]
Use the given slant height to express the volume in one variable and then maximize it.
Step 1: Use the relation between \(r\), \(h\), and slant height. Given \[ l=3. \] Therefore, \[ r^2+h^2=9. \] \[ r^2=9-h^2. \]
Step 2: Express volume as a function of \(h\). \[ V = \frac13\pi r^2h. \] Substituting \[ r^2=9-h^2, \] \[ V(h) = \frac{\pi}{3}(9h-h^3). \]
Step 3: Differentiate and find the critical point. \[ \frac{dV}{dh} = \frac{\pi}{3}(9-3h^2). \] For maximum volume, \[ \frac{dV}{dh}=0. \] \[ 9-3h^2=0. \] \[ h^2=3. \] \[ h=\sqrt3. \]
Step 4: Verify maximum value. \[ \frac{d^2V}{dh^2} = \frac{\pi}{3}(-6h). \] At \[ h=\sqrt3, \] \[ \frac{d^2V}{dh^2}\lt 0, \] hence the volume is maximum.
Step 5: Find the maximum volume. \[ r^2 = 9-3 = 6. \] Therefore, \[ V_{\max} = \frac13\pi(6)(\sqrt3). \] \[ = 2\sqrt3\,\pi. \] Hence, \[ \boxed{V_{\max}=2\sqrt3\,\pi} \] \[ \boxed{\text{Answer = (D)}} \]
Step 1: Use the relation between \(r\), \(h\), and slant height. Given \[ l=3. \] Therefore, \[ r^2+h^2=9. \] \[ r^2=9-h^2. \]
Step 2: Express volume as a function of \(h\). \[ V = \frac13\pi r^2h. \] Substituting \[ r^2=9-h^2, \] \[ V(h) = \frac{\pi}{3}(9h-h^3). \]
Step 3: Differentiate and find the critical point. \[ \frac{dV}{dh} = \frac{\pi}{3}(9-3h^2). \] For maximum volume, \[ \frac{dV}{dh}=0. \] \[ 9-3h^2=0. \] \[ h^2=3. \] \[ h=\sqrt3. \]
Step 4: Verify maximum value. \[ \frac{d^2V}{dh^2} = \frac{\pi}{3}(-6h). \] At \[ h=\sqrt3, \] \[ \frac{d^2V}{dh^2}\lt 0, \] hence the volume is maximum.
Step 5: Find the maximum volume. \[ r^2 = 9-3 = 6. \] Therefore, \[ V_{\max} = \frac13\pi(6)(\sqrt3). \] \[ = 2\sqrt3\,\pi. \] Hence, \[ \boxed{V_{\max}=2\sqrt3\,\pi} \] \[ \boxed{\text{Answer = (D)}} \]
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