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In a time long long ago in a land far far away (okay, maybe not that far)...

As a reward for your clever calculating services, the High King Math offers you a reward. This kingdom molds its gold into right circular cone shapes for currency, and the King will grant mold you such a cone that follows the following criteria:

You may choose the radius ® and height (h) of the cone, but it must fit, placed upside down, in a larger cone of radius R and height H, with the bases of the two cones parallel (sketched below).

In terms of R and H, what values should you choose for your cone to get the most gold?

Place one right circular cone upside down inside a larger one so that their bases are parallel as sketched below. Find the values of r and h that maximize the area of the smaller cone.


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The volume V of the larger cone is (pi/3) HR2.

Let f be the ratio of radii: r = f R, so that h = (1-f) H.

We want to maximize the volume v of the smaller cone with respect to f.

v = (pi/3) hr2 = (pi/3) (1-f) H f2R2 = (f2 - f3) V

(f2 - f3) has an extremum of 4/27 when f = 2/3.

We know it's a maximum, because it's zero when f is 0 or 1.

So when r = 2R/3 (and h = H/3,) v reaches a maximum of 4V/27.

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