Filled To The Rim

As shown in the figure, the volume of water displacement $V$ by a sphere of radius $r$ is given by the volume of the sphere minus the volume of the spherical cap with a height of $h$ . In equation, we have:

$\begin{aligned} V & = {\color{#3D99F6} V_{\text{sphere}}} - \color{#D61F06} V_{\text{cap}} \\ & = {\color{#3D99F6} \frac 43 \pi r^3} - \color{#D61F06} \frac {\pi h^2}3(3r-h) \\ & = \frac \pi 3 \left(4r^3 - h^2(3r - h)\right) & \small \color{#3D99F6} \text{See note: }h = \frac 83 r - 4 \\ & = \frac \pi 3 \left(4r^3 - \left(\frac 83r-4\right)^2\left(\frac 13r + 4\right)\right) \\ & = \frac \pi{81} \left(108r^3 - (8r-12)^2(r+12)\right) \\ & = \frac \pi{81} \left(44r^3-576r^2+2160r-1728 \right) \\ & = \frac {4\pi}{81} \left(11r^3-144r^2+540r-432 \right) & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }r \\ \frac {dV}{dr} & = \frac {4\pi}{81} \left(33r^2-288r+540\right) \\ & = \frac {4\pi}{27}(11r-30)(r-60) \end{aligned}$

Putting $\dfrac {dV}{dr} = 0$ , then we have $r = \dfrac {30}{11}$ or $r = 60$ . Since $\dfrac {d^2V}{dr^2}\bigg|_{r=\frac {30}{11}} < 0$ , $V$ is maximum when $r = \dfrac{30}{11}$ .

Therefore, $m + n = 30 + 11 = \boxed{41}$ .

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Note: Let the center of the sphere be $O$ , the vertex of the cone be $P$ and the point of contact of the sphere and internal wall of the cone be $R$ , Then $\triangle OPR$ is an right triangle with $OR:RP:PO = 3:4:5$ . Implying $PO= \dfrac 53 r$ and $h = r + \dfrac 53 r - 4 = \dfrac 83 r - 4$ .

Here's a nice plot of the volume of the displacement as a function of $r$ , with a clear unique maximum.

This plot assumes that the cone is deeper than $4$ , but water level is at $4$ .

Lovely problem! Thanks, Brian! This will make a great bonus question for a calculus exam!

Let me jot down a solution; I'm sure it can be streamlined.

If the center of the sphere is $a$ units above the water level (with $a\leq \frac{9}{4})$ , its distance from the cone is $r=\frac{12+3a}{5}$ by the line-point distance formula (consider a vertical cross section), and $a=\frac{5r-12}{3}$ . The height of the spherical cap below the water level is $h=r-a=\frac{12-2r}{3}$ , and the volume is $V$ $=\frac{\pi h^2}{3}(3r-h)$ . A little calculus shows that the maximum is attained when $r=\frac{30}{11}$ ; the answer is $\boxed{41}$ .

Recognizing there is a 3-4-5 right triangle in the cone's cross-section, a formula relating the radius of a sphere tangent to the sides of the extended cone and whose center is at height $p$ above the tip of the cone is $r=\frac35 p$ .The depth of the spherical cap is $h=\text{Which}\left[p\geq 10,0,p<\frac{5}{2},2 r,\text{True},4-(p-r)\right]$ . This yields a formula of the spherical cap whose sphere center is a height $p$ : $f=\begin{array}{cc} \{ & \begin{array}{cc} \frac{36 \pi p^3}{125} & 2 p<5 \\ \frac{4}{375} \pi (p-10)^2 (11 p-20) & \frac{5}{2}\leq p<10 \\ 0 & \text{True} \\ \end{array} \\ \end{array}$

Maximizing $f$ (the bit of calculus mentioned elsewhere -- computing the derivative and solving for 0) gives $p=\frac{50}{11}$ , which gives $r=\frac{30}{11}$ , which give a final answer of 41.

My plot looks similar Michael Mendrin's.

Let the radius of the circle be x, measured upwards from the apex of the cone. From the geometry of the cone we have $tan(a)=\frac{3}{4}$ , from which $sin(a)=\frac{3}{5}$ Then easy trig shows that the radius of the sphere is $\frac{3x}{5}\dots(1)$ The volume we are looking for is the spherical 'cap' lying under the water plane. I put cap in quotes because the way I'm using it it looks more like the rest of the head than a little cap on top of the head. However the formula invoked by other solvers, $V=\frac{\pi}{3}h^2(3R-h)$ , still applies. We have the radius from (1) and it is easy to see that the height of the cap is $4-\frac{2x}{5}$ . Putting this all together we see that we have to maximise

$\frac{\pi}{3}(4-\frac{2x}{5})^2(\frac{9x}{5}-(4x-\frac{2x}{5}))$

This is an easy exercise (for folk working at Calculus level 5) and the maximum occurs when $x=\frac{50}{11}$

Substituting in (1) then gives the radius as $\frac{30}{11}$ and so the answer is $\boxed{41}$

@Brian Charlesworth Sir, this is similar to a problem posted earlier.....