Can in a Cone

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).

To find the volume of the largest can that can fit entirely under a cone with volume 900 cubic cm, we can follow these steps:

1. Draw pictures of at least THREE different cases:
a. In the first case, imagine a cone with a can that fits entirely inside it. The can is at its maximum size without going outside the boundaries of the cone.
b. In the second case, consider a cone with a slightly smaller can that also fits entirely inside it.
c. In the third case, visualize a cone with a significantly smaller can that fits entirely inside it.

2. Define variables:
Let's define the variables:
- V: Volume of the largest can
- r: Radius of the can's base
- h: Height of the can
- R: Radius of the cone's base
- H: Height of the cone

Note that the radius of the can's base is smaller than or equal to the radius of the cone's base, and the height of the can is smaller than or equal to the height of the cone.

3. Determine the volume of the can (V):
The volume of a can is given by the formula V = π * r^2 * h.

4. Express the height of the can (h) in terms of the radius of the can's base (r):
Since the can fits entirely under the cone, we can express the height of the can (h) as:
h = (H/R) * r

5. Substitute the expression for h into the volume equation:
V = π * r^2 * [(H/R) * r]
V = (π * H/R) * r^3

6. Express the volume of the cone (900 cubic cm) in terms of r:
The volume of a cone is given by the formula V = (1/3) * π * R^2 * H.
We know that V = 900, so we can write:
900 = (1/3) * π * R^2 * H

7. Solve for the height of the cone (H) in terms of the radius of the cone's base (R):
H = (3 * 900) / (π * R^2)

8. Substitute the expression for H into the volume equation:
900 = (1/3) * π * R^2 * [(3 * 900) / (π * R^2)]
900 = 900

9. Differentiate the expression for V with respect to r:
dV/dr = d/dx[(π * H/R) * r^3]

10. Set the derivative equal to zero to find the critical points:
dV/dr = 0

11. Solve for r to find the critical points and check if they are within the desired range:
Find the values of r that satisfy the equation dV/dr = 0 and check which ones are within the allowed range (r <= R).

12. Calculate the volume of the can (V) using the optimal value of r:
Substitute the optimal value of r into the equation V = (π * H/R) * r^3 to find the maximum volume of the can.

By following these steps, you will be able to find the optimal case for the largest can that can fit under the given cone.