I understand that I should take the an initial derivative and set it same to zero, i m sorry will find the maximum.
My difficulty is just how to deal with the change $h$, the height?
How deserve to I rewrite that in terms of $x$ or $r$.
I have tried come implicitly identify this equation, but that wasn"t valuable for me.
The elevation $h$ is the distance from the apex come the circular base. That is in between $0$ and $2r$ (with $r$ the radius the the sphere), and the radius of the base grows together $h$ rises from $0$ to $r$ and then shrinks again as $h$ boosts to $2r$. A diagram might convince you that by Pythagoras $(h-r)^2+x^2=r^2$, where $x$ is the radius that the base. That allows you to eliminate $x^2$ in favour of $2rh-h^2$, therefore you have $V(h)=\frac13(2rh-h^2)h$ and thus $V"(h)=\frac13(4rh-3h^2)$, i beg your pardon vanishes in ~ $h=\frac43r$ and also thus $x=\frac\sqrt83r$, with maximal volume $V=\frac13\pi\frac89r^2\frac43r=\frac3281\pi r^3=\frac827V_\text s$, wherein $V_\text s$ is the volume that the sphere.
You are watching: Find the volume of the largest right circular cone that can be inscribed in a sphere of radius r
edited Oct 31 "12 at 1:30
reply Oct 31 "12 at 1:24
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Ok. In the picture above, there are two sketches. The one on the left is in 3d, however is kinda difficult to to express to. The one on the right is in 2d and easier to watch what"s walking on, so I"m going to use that (just imagine it"s a cross-section of the 3d pic).
The an enig of the whole difficulty is come relate $h$ and $b$. This is done using the equation because that the right half of the circle: $$x = \sqrtr^2-y^2$$
$b$ is the $x$ value once $y$ is counter from the top of the round by $h$. Thus, the equation in regards to $b$, $h$, and $r$ is: $$b = \sqrtr^2 - (r-h)^2$$
Now to relate to the volume that a cone:$$V_cone = \frac\pi 3 \cdot b^2 h$$$$V_cone = \frac\pi 3 \cdot (r^2 - (r - h)^2) h$$Simplifying:$$V_cone = \frac \pi 3 \cdot (2h^2 r-h^3)$$Differentiate:$$\fracdV_conedh = \frac \pi 3 \cdot (4r - 3h)h$$Solve:$$h = \left\0, \frac4r3\right\$$We desire the maximum, which obviously will occur at the 2nd value of $h$. Finding the maximal volume is a basic algebra difficulty now.
edited Jul 13 "15 in ~ 10:38
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answer Oct 31 "12 at 1:32
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With offered radius r of a round let the inscribed cone have height h climate remaining length without radius is (h-r) permit R it is in radius that cone climate there we obtain a best angle triangle v r as hypotanious R as adjecent and (h-r) together opposite side. Now by pythagorus organize ((h-r)^2)+(R^2)=(r^2). Currently express R in terms of h and also r. We acquire (R^2)=(2hr-hh). Substitute in the volume of cone equation V=(pi/3)(R^2)h. Now V=(pi/3)(2hr-hh)h. Identify wrt h v r as consistent because radius can not be variable and also substitute come zero to gain maxima. We acquire h=(4r/3).
answered Aug 13 "15 in ~ 10:36
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