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.

A cone inscribed in a sphere

Hints

Hint 1
Think of what you see when you cut the sphere and cone down the central axis. Add some radii to your diagram This is what we see when we cut the sphere and the cone down the central cone axis.

Hint 2
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. We want to relate the height of the cone to the radius R of the sphere and the base radius r of the cone.

Solutions

Solution

The goal is to maximize the volume of the cone, $V_{cone}=\pi r^{2}{\frac {h}{3}}$. However, it has to fit inside a sphere of radius $R$, so in particular, we should have $0\leq r\leq R$, and $0\leq h\leq 2R$. We will use a constraint to eliminate one of the variables, for example $h$. it is important to draw a sketch to determine the relationship between $r,h,$ and $R$ that forms a constraint. We will determine the height of the cone by first finding the length of the green side labeled $x$. In the diagram, we draw the cross-section of the two shapes, which reveals an equilateral triangle inscribed in a circle of fixed radius $R$. We want to express $h$ in terms of the variable $r$ (radius of cone base) and the fixed radius $R$. To do so, note that $h=R+x$, where $x$ is a side length (shown in green) of a right angle triangle with other sides $r,R$. Hence $x={\sqrt {R^{2}-r^{2}}}$ so we have that $h=R+x=R+{\sqrt {R^{2}-r^{2}}}$ now we can express the cone volume as $V_{cone}={\frac {\pi }{3}}r^{2}h={\frac {\pi }{3}}r^{2}\left(R+{\sqrt {R^{2}-r^{2}}}\right)={\frac {\pi }{3}}\left(Rr^{2}+{\sqrt {R^{2}r^{4}-r^{6}}}\right)$. (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, $r$, and a fixed (constant) $R$ . We find critical points by setting $V_{cone}'(r)=0$. Computing the derivative, we have $V_{cone}'(r)={\frac {\pi }{3}}\left(2Rr+{\frac {4R^{2}r^{3}-6r^{5}}{2{\sqrt {R^{2}r^{4}-r^{6}}}}}\right)=0$. We now have to solve for $r$: $2Rr+{\frac {4R^{2}r^{3}-6r^{5}}{2{\sqrt {R^{2}r^{4}-r^{6}}}}}=0$ $2Rr=-{\frac {2R^{2}r^{3}-3r^{5}}{r^{2}{\sqrt {R^{2}-r^{2}}}}}$ Either $r=0$(which is trivial) or else we can cancel out several factors of $r$ $2R={\frac {3r^{2}-2R^{2}}{\sqrt {R^{2}-r^{2}}}}$ $2R{\sqrt {R^{2}-r^{2}}}=(3r^{2}-2R^{2})$ Square both sides: $4R^{2}(R^{2}-r^{2})=(3r^{2}-2R^{2})^{2}=9r^{4}-12r^{2}R^{2}+4R^{4}$ $4R^{4}-4R^{2}r^{2}=9r^{4}-12r^{2}R^{2}+4R^{4}$ Simplifying algebraically: $-4R^{2}r^{2}=9r^{4}-12r^{2}R^{2}$ $8R^{2}r^{2}=9r^{4}$ Finally, we arrive at the radius of the cone base, $8R^{2}=9r^{2}$ so that $r={\frac {2{\sqrt {2}}}{3}}R$. We can then (optionally) find the cone height from the formula above, namely $h=R+{\sqrt {R^{2}-r^{2}}}=R+{\sqrt {R^{2}-(8/9)R^{2}}}={\frac {4R}{3}}$ and we can also find the volume of the optimal cone $V_{cone}={\frac {\pi }{3}}r^{2}h={\frac {\pi }{3}}\cdot {\frac {8}{9}}R^{2}\cdot {\frac {4R}{3}}={\frac {32\pi }{81}}R^{3}$. We still need to justify that this is the largest and not the smallest cone, i.e. that the critical point is a local maximum. The best way to do this is to observe that $V_{cone}(r)=0$ at the two endpoints of the interval $r=0,r=R$, and that $V_{cone}(r)$ is positive inside that interval. Hence, since there is only one critical point in this interval, it has to be a local maximum.