Respuesta :
Answer:
The height of the right circular cone when constructed this way is [tex]\sqrt3[/tex] cm.
The radius of the right circular cone when constructed this way is [tex]\sqrt6[/tex] cm.
The volume of the right circular cone when constructed this way is [tex]6\sqrt3 \pi[/tex] cm³.
Step-by-step explanation:
Given that,
A right triangle whose hypotenuse is 3 cm long is revolved .
Then other two legs of the triangle will be radius and height of the cone.
Assume the height and radius of the cone be h and r respectively.
From Pythagorean Theorem :
h²+r²=3²
⇒ r²= 9 - h²
Then the volume of the cone is
V= π r²h
⇒ V= π(9-h²)h [ ∵ r²= 9 - h²]
⇒V= π(9h - h³)
Differentiating with respect to h
V'=π(9 - 3h²)
Again differentiating with respect to h
V''= π(-6h)
⇒V''= (-6πh)
To maximum or minimum ,we set V'=0
π(9 - 3h²)=0
⇒3h²=9
⇒h²=3
[tex]\Rightarrow h=\sqrt3[/tex]
Now, [tex]V''|_{h=\sqrt3}=-6\pi (\sqrt3)<0[/tex].
Since at [tex]h=\sqrt3[/tex],V''<0.
The volume of cone is maximum at [tex]h=\sqrt3[/tex] cm when constructed this way.
The height of the right circular cone when constructed this way is [tex]\sqrt3[/tex] cm.
The radius of the right circular cone when constructed this way [tex]r=\sqrt{9-(\sqrt3)^2[/tex]
= [tex]\sqrt{9-3}[/tex]
[tex]=\sqrt6[/tex] cm.
The volume of the right circular cone when constructed this way is
=π r²h
[tex]=\pi (\sqrt6)^2\sqrt3[/tex]
[tex]=6\sqrt3 \pi[/tex] cm³
