A right circular cone is to be inscribed in another right circular cone of volume ......

Question

 A right circular cone is to be inscribed in another right circular cone of volume 2m^3  and altitude 2m, with the same axis and with the vertex of the inner cone touching the base of the outer cone.

What must be the altitude of the inscribed cone in order to have the largest possible volume? 

Answer 1

V = (1/3)πR²H = 2m³ (everything's capital to represent that of the larger cone)

(1/3)πR²(2m) = 2m³

(π/3)R² = m²

R² = (3/π)m²

R = m√(3/π)

⇒ D = 2m√(3/π) is the diameter of the base of the larger cone

Now when looking at v = (1/3)πr²h for the smaller cone, we will derive the volume with respect to the height to find the height that yields the optimal volume. Before we can do that, we need to replace r with something in terms of h. You can tell that they have a linear relationship, but it let's first lay down the (height, diameter) coordinate pairs.

We have (h = 0, d = 2m√(3/π)) since the inscribed cone is upside-down compared to the larger one.

And (h = 2m, d = 0)

If you plot these points, they form a line with a slope of -√(3/π) and a d-intercept of 2m√(3/π). Therefore you get a line of d = -√(3/π)h + 2m√(3/π).

We know the radius will always be half of this, to give us: r = (-h/2)√(3/π) + m√(3/π) = (-h/2 + m)√(3/π)

We plug this back into our volume formula:

v = (1/3)π[(-h/2 + m)√(3/π)]²h

v = (1/3)π(-h/2 + m)²√(3/π)²h

v = (-h/2 + m)²h as the constant factors cancel

v = (h²/4 - mh + m²)h

v = h³/4 - mh² + m²h

Now we take the dv/dh

dv/dh = (3/4)h² - 2mh + m²

Here, I personally decided to plot this (letting m = 1) just to confirm that the derivative after h = 0 is at first positive, then it passes zero, then it's negative before reaching h = 2 (which is the tallest height possible 2m, m = 1). This is to make sure that the height that gives the optimal volume is at that h-intercept, which it is.

So we just set dv/dh = 0 and solve for h:

0 = (3/4)h² - 2mh + m² Here, I suggest we use the quadratic formula

h = [2m ± √(4m² - 3m²)]/(3/2)

h = (2/3)[2m ± √(m²)]

h = (2/3)(2m ± m)

Try both options:

h = (2/3)(2m + m)

h = (2/3)(3m) = 2m

We know a height of 2m yields a diameter of 0, so this is one is incorrect.

h = (2/3)(2m - m)

h = (2/3)m = 2m/3

This one yields a positive volume and so is the correct choice! Yay calculus :)

Answer 2

Since the inner cone has the same axis and height; there's no difference.

v=bh/3

2=r^2pi*2/3

2/(2/3)/pi=r^2

3/pi=r^2

.9549.......=r^2

^.5=.9772.......