Question

A cylinder and a cone start with the same radius and height. The radius of the cone is then tripled, and the height of the cone is cut in half. The radius of the cylinder stays the same, but the height of the cylinder is doubled. Which change produces a greater increase in volume (i.e., which figure’s volume increases by a larger factor)? Justify your answer. Write “pi” for ^ and “r^2” for r squared

Answer 1

Cone:
Original cone = (1/3)π(h)r^2
Changed cone = (1/3)π(h/2)(3r)^2
= (1/2)(1/3)π(h)9r^2
= (9/2) * Original cone
=4.5 * Original cone

Cylinder:
Original cylinder = π(h)r^2
Changed cylinder = π(2h)r^2
=2 * Original cylinder

Therefore the cone is the greatest relative increase in volume.

Answer 2

Answer:

Change produced in cone is greater than change produced in cylinder.

Step-by-step explanation:

Given : A cylinder and a cone start with the same radius and height.

We have to find which change produces a greater increase in volume.

We know Volume of cylinder = $\pi r^2h$

and Volume of cone = $\frac{1}{3}\pi r^2h$

Where r is radius and h is height.

Let r' and h' denotes new radius and new height

Consider Cylinder first ,

The radius of the cylinder stays the same, but the height of the cylinder is doubled.

that is r' = r and h' = 2h

Then Volume of new cylinder becomes,

Volume of cylinder = $\pi (r')^2h'=2\pi r^2h$

$Change\ produced = \frac{new-original}{original}$

that is

$Change\ produced\ in\ cylinder = \frac{2\pi r^2h-\pi r^2h}{\pi r^2h}=1$

Now, Consider Cone, we have,

The radius of the cone is then tripled, and the height of the cone is cut in half.

r' = 3r and  h' = $\frac{h}{2}$

Thus, The volume of new cone becomes,

Volume of cone = $\frac{1}{3}\pi (r')^2h'=\frac{1}{3}\pi (3r)^2\frac{h}{2}$

$Change\ produced = \frac{new-original}{original}$

that is

$Change\ produced\ in\ cone =\frac{\frac{1}{3}\pi (3r)^2\frac{h}{2}-\frac{1}{3}\pi r^2h}{\frac{1}{3}\pi r^2h} =\frac{7}{2}=3.5$

Thus, Change produced in cone is greater than change produced in cylinder.