Can in a Cone

Question

Find the volume of the largest can that can fit entirely under a cone with volume 900 cubic cm.
1.draw pictures of at least THREE different cases (including the endpoints if any).
2. define all variables CLEARLY (with words) and then use those variables consistently.
3. Employ calculus to find the OPTIMAL case (which will be either a critical point or an endpoint).

Answer 1

Draw a diagram. If a can of radius r and height h is placed under a cone of radius R and height H,
r/R = 1 - h/H
so, r = R(1 - h/H)

So, the volume of the can is
v = pi r^2 h = pi R^2(1-h/H)^2 h
= pi R^2 H * h/H (1-h/H)^2

now, the volume of the cone is
V = 1/3 pi R^2 H

so the ratio of volumes is

v/V = [pi R^2 H * h/H(1-h/H)^2]/(1/2 pi R^2 H)
= 3 h/H(1-h/H)^2

If we let x = h/H, that is, the ratio of the can height to the cone height,

y = v/V =

dy/dx = 3(x-1)(3x-1)

So, we have two critical points.

when x=1, that is the can is as tall as the cone, the volume is 0

max volume occurs when h/H = 1/3, or the can is 1/3 as tall as the cone.

Answer 2

hmmm. text got lost

y = 3x(1-x)^2