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$$4\pi r^2 + \frac{4}{3}\pi r^3 = \frac{16}{3}\pi m^3.$$

This is all I got:

$$4 r^2 + \frac{4}{3}r^3 = \frac{16}{3}m^3.$$

How to simplify the equation and solve it "by inspection"?

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  • $\begingroup$ $r=1=m~~ \text{or} ~~r=0=m{}{}{}{}{}{}{}$ $\endgroup$ – Mikasa Jan 16 '14 at 8:30
  • $\begingroup$ @B. S. how did you get that??? $\endgroup$ – Joshua Jan 16 '14 at 8:32
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    $\begingroup$ by inspection,as you wanted. :D $\endgroup$ – Mikasa Jan 16 '14 at 8:32
  • $\begingroup$ @B. S. but I dont know what it means "by inspection" in the first place:( $\endgroup$ – Joshua Jan 16 '14 at 8:34
  • $\begingroup$ Is "m" a constant given ? If yes, is it positive or negative ? $\endgroup$ – Claude Leibovici Jan 16 '14 at 8:35
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There are multiple solutions to this equation since $m$ can take on different values. The best thing one can do in this case is assume that $m=r$ and solve the equation. So, $$4 r^2 + \frac{4}{3}r^3 = \frac{16}{3}r^3$$ $$r^2\left(4-4r \right)=0$$ $$\therefore \ r=0=m \ \text{or} \ r=1=m$$

Note that these are not the only answers. If we took $r=m^{3}$, we get an additional solution, $r=-4 \implies m=-4^{1/3}$

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Just consider a function defined by the LHS. Since it is a polynomial which contains a third power of "$r$", the function will start at $-\infty$ and will grow up to $\infty$.

The function is zero for $r=-3$ and $r=0$. Its derivative cancels for $r=0$ and $r=-2$; for $r=-2$, the value of the function is $\dfrac{16}{3}$ and a check of the second derivative shows that this point is a maximum.

Now, solving your equation can be seen as a search of the intersection of the function and an horizontal line corresponding to $y=\dfrac{16 m^3}{3}$. So, what we can say is that,

if $m < 0$, the solution for "$r$" will be smaller than $-3$
if $m = 0$, the solutions for "$r$" are $-3$ and $0$
if $0 < m < 1$, there will be two solutions, one such than $-3 < r < -2$ and the other such that $-2< r < 0$
if $m > 1$, there will be a unique solution such that $r > 0$.

All the above can be done by a visual inspection of the graph at the function defined by the LHS.

Is this what you expect ? If not, please clarify.

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  • $\begingroup$ I think it is overkilled. This question is about the volume of a tank. 4πr2 is the volume of a circular cylinder, 4/3πr3 is the volume of a sphere, and 16/3πm3 is the volume of the tank. $\endgroup$ – Joshua Jan 16 '14 at 9:06
  • $\begingroup$ so I think r must not be negative. $\endgroup$ – Joshua Jan 16 '14 at 9:06
  • $\begingroup$ @Joshua. Then your problem was not clear enough ! By the way, I forgot yhe case where m=1; in this case, there are two roots corresponding to r=-2 with a second root r > 0. $\endgroup$ – Claude Leibovici Jan 16 '14 at 9:19
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    $\begingroup$ @John. Thanks for editing. $\endgroup$ – Claude Leibovici Jan 16 '14 at 9:19

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