Answer:
The maximum volume of cone is 138.25 m³
[tex]Radius, r=7\sqrt{\dfrac{2}{3}}[/tex] m
[tex]Height, h=\dfrac{7}{\sqrt{3}}[/tex] m
Step-by-step explanation:
A right circular cone whose hypotenuse is [tex]\sqrt{7}[/tex] m
It is revolved about one leg to generate a right circular cone.
Let radius be r m and height be h m
For right angle triangle,
[tex]r^2+h^2=7^2[/tex]
[tex]r^2=49-h^2[/tex]
Volume of generated cone [tex]=\dfrac{1}{3}\pi r^2h[/tex]
[tex]V=\dfrac{1}{3}\pi (49-h^2)h[/tex]
Differentiate w.r.t h
[tex]\dfrac{dV}{dh}=\dfrac{1}{3}\pi (49-3h^2)[/tex]
For maximum/minimum [tex]\dfrac{dV}{dh}=0[/tex]
[tex]\dfrac{1}{3}\pi (49-3h^2)=0[/tex]
[tex]h=\dfrac{7}{\sqrt{3}}[/tex]
[tex]r^2=49-\dfrac{49}{3}[/tex]
[tex]r=7\sqrt{\dfrac{2}{3}}[/tex]
Using double derivative test
[tex]\dfrac{d^2V}{dh^2}=\dfrac{1}{3}\pi (-6h)[/tex]
At [tex]h=\dfrac{7}{\sqrt{3}}[/tex]
[tex]\dfrac{d^2V}{dh^2}<0[/tex] so get maximum volume.
Dimension of cone:
[tex]Radius, r=7\sqrt{\dfrac{2}{3}}[/tex] m
[tex]Height, h=\dfrac{7}{\sqrt{3}}[/tex] m
[tex]V=\frac{1}{3}\pi\cdot\left(7\sqrt{\frac{2}{3}}\right)^{2}\cdot\frac{7}{\sqrt{3}}[/tex]
The maximum volume of cone is 138.25 m³
