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.

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

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

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

Now, .

Since at ,V''<0.

The volume of cone is maximum at cm when constructed this way.

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

The radius of the right circular cone when constructed this way  

   =  

    cm.

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

=π r²h

cm³

i believe that the answer is

b. y-1=-2(x-4)

hope this !