Course:MATH102/Question Challenge/2017ConeInSphere
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Question
Find the volume to the largest cone that fits into a sphere of radius R.
Hints
Hint 1 |
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Think of what you see when you cut the sphere and cone down the central axis. Add some radii to your diagram |
Hint 2 |
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We want to eliminate one of the variables, say h, from the cone volume using a constraint. In this figure we can find a relationship between the cone height and the other quantities. Note that R is fixed. |
Solutions
Solution |
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The goal is to maximize the volume of the cone, . However, it has to fit inside a sphere of radius , so in particular, we should have , and . We will use a constraint to eliminate one of the variables, for example . it is important to draw a sketch to determine the relationship between and that forms a constraint. In the diagram, we draw the cross-section of the two shapes, which reveals an equilateral triangle inscribed in a circle of fixed radius . We want to express in terms of the variable (radius of cone base) and the fixed radius . To do so, note that , where is a side length (shown in green) of a right angle triangle with other sides . Hence so we have that
now we can express the cone volume as . (The last step, where we combined terms inside the square-root is optional, but avoids needing to use the product rule in differentiating.) We have now expressed the volume of the cone in terms of a single variable, , and a fixed (constant) . We find critical points by setting . Computing the derivative, we have . We now have to solve for :
Either (which is trivial) or else we can cancel out several factors of
Square both sides:
Simplifying algebraically:
Finally, we arrive at the radius of the cone base, so that . We can then (optionally) find the cone height from the formula above, namely
and we can also find the volume of the optimal cone .
at the two endpoints of the interval , and that is positive inside that interval. Hence, since there is only one critical point in this interval, it has to be a local maximum. |