Given a right circular cone, you put an upside-down cone inside it so that its vertex is at the center of the base of the larger cone, and its base is parallel to the base of the larger cone. If you choose the upside-down cone to have the largest possible volume, what fraction of the volume of the larger cone does it occupy? (Let H and R be the height and radius of the large cone, let h and r be the height and radius of the small cone. Use similar triangles to get an equation relating h and r. The formula for the volume of a cone is V = 1/2pir^2h.)

Pretty sure I figured it out, 4/27. I found this by simplifying:

((1/3pi (h - 2/3 h))(4/9 r^2)) / (1/3 pi r^2 h)

How did you get this? Please help I am stuck.

Draw the cones. If the large cone has height H and radius R, the small cone has height h and radius r, so that

r/R = 1 - h/H

The ratio of the two volumes is v/V = r^2h/R^2H = (r/R)^2 h/H

if we maximize that ratio as a function of u = h/H, we get

f(u) = (1-u)^2 * u
= (u - 2u^2 + u^3)
f'(u) = (1-4u+3u^2)
= (h-1)(3h-1)

Clearly, when u=1, f(u) = 0 a minimum
when u = 1/3, f(h) = 4/9*1/3 = 4/27

so, h/H = 1/3 for max volume ratio

Okay, letting x be height and b radius of smaller,

then x/b=h/r, and
volumeof smaller is 1/3pi(h-x)b^2, or
1/3pi(h-x)(x^2r^2/h^2=
1/3pi((x^2r^2/h)-(x^3r^2/h^2).
Now differinate with respect to x,
and set to 0, so
((2xr^2/h)-(3x^2r^2/h^2)=0,
so you get x=2/3h.
So, fraction is ((1/3pi(h-2/3h)(4/9r^2)/(1/3pir^2h)=
4/27

Ah, the cone conundrum! Let's see if we can solve it with a touch of humor.

First, let's call our large cone "Biggie" and our small cone "Smalls." We'll put on our thinking caps and get to work!

Since Smalls is an upside-down cone inside Biggie, their bases are parallel. This means that the ratios of their corresponding dimensions are the same, following the wondrous world of similar triangles.

So, if we compare the heights of Biggie and Smalls (H and h respectively), and their radii (R and r), we get the following ratio:

(H - h) / R = h / r

We can simplify that equation to:

r = (R * h) / (H - h)

Now, let's calculate the volume of both cones. Remember, the volume of a cone is V = (1/3) * π * r^2 * h.

For Biggie, we have:

V_biggie = (1/3) * π * R^2 * H

For Smalls, since we've swapped it with the equation above:

V_smalls = (1/3) * π * [(R * h) / (H - h)]^2 * h

To find out the fraction of the volume that Smalls occupies, we'll divide V_smalls by V_biggie:

(V_smalls / V_biggie) = [(R * h) / (H - h)]^2 * h / (R^2 * H)

To make things clearer, let's call this ratio "Mr. Volume Fraction." So, we have:

Mr. Volume Fraction = [(R * h) / (H - h)]^2 * h / (R^2 * H)

Now, we could simplify this even more, but let's not get too serious here. Let's just say that Mr. Volume Fraction is a bit of a math whiz, juggling all those numbers for us.

So, at last, we've arrived at the answer to our conical conundrum. The upside-down cone, Smalls, occupies a volume given by Mr. Volume Fraction.

Remember, though, friends, that the values of H, R, h, and r will determine how much Smalls fills up Biggie. So, if you're working with specific values, plug them in and let Mr. Volume Fraction do his thing!

And there you have it, a mixture of math and humor to lighten up the cone confusion. Happy calculating!

To solve this problem, we need to use similar triangles to relate the heights and radii of the large and small cone, and then use the formula for the volume of a cone to find the fraction of volume occupied by the smaller cone.

Let's start by visualizing the right circular cone and the upside-down cone inside it. We can see that the two cones are similar, which means their corresponding sides are proportional.

Let the height and radius of the large cone be H and R, respectively. Similarly, let the height and radius of the small cone be h and r, respectively.

Now, let's consider the similar triangles formed by the two cones. We have two similar triangles: one formed by the vertical heights of the two cones, and the other formed by the radii of the two cones.

By comparing the corresponding sides of the similar triangles, we can set up the following proportion:

H / h = R / r

Cross-multiplying this proportion gives us:

Hr = hR

Next, we can solve this equation for h:

h = (Hr) / R

Now, we can substitute this expression for h into the formula for the volume of a cone:

V_small = (1/2) * π * r^2 * h

Substituting the value of h we found earlier, we get:

V_small = (1/2) * π * r^2 * [(Hr) / R]

Simplifying this equation, we have:

V_small = (1/2) * π * r^3 * H / R

On the other hand, we know the volume of the large cone is given by:

V_large = (1/3) * π * R^2 * H

Now, let's find the fraction of the volume of the larger cone occupied by the smaller cone:

Fraction = V_small / V_large

Substituting the respective formulas for the volumes:

Fraction = [(1/2) * π * r^3 * H / R] / [(1/3) * π * R^2 * H]

Simplifying this equation, we get:

Fraction = (3/2) * (r^3 / R^3)

Therefore, the fraction of the volume of the larger cone occupied by the smaller cone is (3/2) * (r^3 / R^3).