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).
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.
hmmm. text got lost
y = 3x(1-x)^2
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 (including the endpoints, if any):
Case 1: The can is small enough to fit entirely under the cone.
- The height of the cone is larger than the height of the can.
- In this case, the can is fully contained within the cone.
Case 2: The can is partially under the cone.
- The height of the cone is equal to the height of the can.
- In this case, the can is touching the bottom of the cone.
Case 3: The can cannot fit under the cone.
- The height of the cone is smaller than the height of the can.
- In this case, the can's top will be above the surface of the cone.
2. Define all variables clearly (with words) and use those variables consistently:
Let's define the variables:
- Vc: Volume of the cone
- V: Volume of the can
- Rc: Radius of the cone's base
- R: Radius of the can's base
- Hc: Height of the cone
- H: Height of the can
3. Employ calculus to find the optimal case (which will be either a critical point or an endpoint):
We want to find the largest volume of the can that fits entirely under the cone, so we need to maximize V while satisfying the condition V ≤ Vc.
To represent the volume of the can, V, in terms of the radius, R, and height, H, we can use the formula for the volume of a cylinder: V = πR^2H.
Since the height of the cone, Hc, is larger than the height of the can, H, we can relate them as: Hc > H.
Using similar triangles, we can establish the following relationship: Rc / R = Hc / H.
From the given volume of the cone, we know: Vc = (1/3)πRc^2Hc.
Now, we can solve for one variable in terms of the other. From the second equation, we have: Rc / R = Hc / H.
=> Rc = R(Hc / H).
Substituting this value for Rc in the volume equation of the cone, we get: Vc = (1/3)π(R(Hc / H))^2Hc.
Now, we can substitute this value of Vc in the constraint equation: (1/3)π(R(Hc / H))^2Hc ≥ V.
From this equation, we can solve for H in terms of R and V to find the optimal case.
To analyze whether the optimal case occurs at a critical point or an endpoint, we can take the derivative of V with respect to R and set it equal to zero.
Differentiating V with respect to R, we get: dV / dR = 0.
By solving this equation, we can find the critical points for R and determine if they fall within the range of possible values.
Alternatively, we can analyze the endpoints of the range for R and check if these points give the maximum volume for the can.
By employing calculus, we can determine the optimal case for the largest volume of the can that can fit entirely under the given cone.