Question

A right triangle whose hypotenuse is 3 centimeters long is revolved about one of its legs to generate a right circular cone. Find the radius, height, and volume of the right cone that will have the greatest volume when constructed this way.

Answer 1

Answer:

The height of the right circular cone when constructed this way is $\sqrt3$ cm.

The radius of the right circular cone when constructed this way is $\sqrt6$ cm.

The volume of the right circular cone when constructed this way is $6\sqrt3 \pi$ 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

$\Rightarrow h=\sqrt3$

Now, $V''|_{h=\sqrt3}=-6\pi (\sqrt3)$.

Since at $h=\sqrt3$,V''<0.

The volume of cone is maximum at $h=\sqrt3$ cm when constructed this way.

The height of the right circular cone when constructed this way is $\sqrt3$ cm.

The radius of the right circular cone when constructed this way  $r=\sqrt{9-(\sqrt3)^2$

   =  $\sqrt{9-3}$

   $=\sqrt6$ cm.

The volume of the  right circular cone when constructed this way is

=π r²h

$=\pi (\sqrt6)^2\sqrt3$

$=6\sqrt3 \pi$ cm³