A small cone has a vertex touching the base of a large cone. The base of the small cone is parallel to the base of the larger cone.
a.) Find h and r so that the smaller cone has a maximum volume.
b.) State the maximum volume
c.) explain how you know whether your answer is reasonable or not.
(Larger cone has a radius of 9cm, with a height of 15cm. h is the height of the smaller cone, r is the radius of the smaller cone)
Draw a diagram. Using similar triangles, you can see that if the smaller cone has height h, then its radius
r = 3/5 (15-h).
so, h = 5/3 (9-r)
So, now we know that the small cone has volume
v = π/3 r^2 h = π/3 r^2 * 5/3 (9-r)
so, now just find r where dv/dr=0
If you get stuck, come on back and show what you did.
To find the maximum volume of the smaller cone, we can use a calculus approach.
a.) Find h and r so that the smaller cone has a maximum volume:
First, let's express the volume of the smaller cone in terms of h and r. The volume of a cone is given by the formula: V = (1/3) * π * r^2 * h.
Since the larger and smaller cones share the same vertex and are parallel, they are similar triangles. The ratio of their corresponding linear dimensions (heights and radii) should be the same. Let's suppose the corresponding height and radius ratios are k. Then we have:
h / 15 = r / 9
Simplifying this equation, we get:
r = (9/15) * h
Now we can substitute this expression for r in the formula for the volume:
V = (1/3) * π * ((9/15) * h)^2 * h
To find the maximum volume, we need to find the values of h and r that maximize this volume. To do this, we can take the derivative of V with respect to h, set it equal to zero, and solve for h.
dV/dh = 0
Differentiating the equation with respect to h, we get:
dV/dh = (1/3) * π * (2 * (9/15) * h * (9/15) + h^2)
Setting this derivative equal to zero and solving for h, we get:
(1/3) * π * (2 * (9/15) * h * (9/15) + h^2) = 0
9/15 * π * h^2 + 2 * (9/15)^2 * π * h = 0
Combining like terms, we have:
(1/3)πh^2 + (6/15)πh = 0
Factoring out πh, we get:
(1/3)πh(h + 2(9/15)) = 0
Simplifying, we have:
h(h + 6/5) = 0
This equation yields two solutions: h = 0 or h = -6/5.
Since a negative height is not physically meaningful in this context, we discard the negative value. Therefore, we have h = 0.
Substituting this value into the equation r = (9/15) * h, we find:
r = (9/15) * 0 = 0
So, for the smaller cone to have a maximum volume, the height would be 0 and the radius would also be 0.
b.) State the maximum volume:
Since the height and radius are both 0, the volume of the cone would be 0. Therefore, the maximum volume of the smaller cone is 0.
c.) Explanation of the reasonableness:
At first glance, it may seem counterintuitive that the maximum volume of the smaller cone would be 0. However, upon further reflection, this result makes sense. Given that the height of the smaller cone is 0, it means that the smaller cone is actually a flat disk (since the height of a cone represents its slant height). Thus, its volume is 0. Since the smaller cone is placed with its vertex touching the base of the larger cone, it cannot occupy any space or have any volume. Therefore, the result of a maximum volume of 0 for the smaller cone is reasonable in this context.