KCET · Maths · Application of Derivatives
The maximum volume of the right circular cone with slant height 6 units is
- A \(4 \sqrt{3} \pi\) cu units
- B \(16 \sqrt{3} \pi\) cu units
- C \(3 \sqrt{3} \pi\) cu units
- D \(6 \sqrt{3} \pi\) cu units
Answer & Solution
Correct Answer
(B) \(16 \sqrt{3} \pi\) cu units
Step-by-step Solution
Detailed explanation
\(\because\) Slant height of the cone, \(L=6\) units.
Let the radius be \(r\) and height be \(h\).
and volume \((V)=\frac{1}{3} \pi r^2 h\)
\(=\frac{1}{3} \pi\left(L^2-h^2\right) h \quad\left[\because L^2=r^2+h^2\right]\)
\(=\frac{1}{3} \pi\left(36-h^2\right) h\)
So, \(\frac{d V}{d h}=\frac{1}{3} \pi\left(36-3 h^2\right)=0 \Rightarrow h=2 \sqrt{3}\) units
Now, \(\frac{d^2 V}{d h^2}=-2 \pi h \lt 0\) at \(h=2 \sqrt{3}\) units
Hence, the maximum volume of the cone
\(V=\frac{1}{3} \pi\left(36-h^2\right) h\)
\(=\frac{1}{3} \pi\left(36-(2 \sqrt{3})^2\right) 2 \sqrt{3}\)
\(=16 \sqrt{3} \pi\) cu units
Let the radius be \(r\) and height be \(h\).
and volume \((V)=\frac{1}{3} \pi r^2 h\)
\(=\frac{1}{3} \pi\left(L^2-h^2\right) h \quad\left[\because L^2=r^2+h^2\right]\)
\(=\frac{1}{3} \pi\left(36-h^2\right) h\)
So, \(\frac{d V}{d h}=\frac{1}{3} \pi\left(36-3 h^2\right)=0 \Rightarrow h=2 \sqrt{3}\) units
Now, \(\frac{d^2 V}{d h^2}=-2 \pi h \lt 0\) at \(h=2 \sqrt{3}\) units
Hence, the maximum volume of the cone
\(V=\frac{1}{3} \pi\left(36-h^2\right) h\)
\(=\frac{1}{3} \pi\left(36-(2 \sqrt{3})^2\right) 2 \sqrt{3}\)
\(=16 \sqrt{3} \pi\) cu units
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