How to predict the maxima and minima of a tidal wave? Assume that a planet of mass $m$ and radius $r$ is orbiting a massive star of mass $M$ at a orbital distance $R$. Assume that the planet is covered in a ideal, non viscous ocean on a friction less surface. Hence the ocean will align itself along the equipotential surface generated by the two masses. Let the maximum height of the water bulge from the planet's surface be $H$ and minimum height of the water bulge be $h$.

What are the mathematical relations between the above values?

My effort at a solution
I found a related question on Physics SE. It references Hale Bradt, Tidal distortion: Earth tides and Roche lobes which is too complex for me too understand.
It would greatly help me if someone can simplify the article and present me with the formulas connecting the above quantities (without proof if needed).
I am a high school student and I need this for a Olympiad Problem which I am trying to solve.
I understand Olympiad level Physics and Astronomy.
 A: Equation 20 in that article seems to provide what you are looking for. It describes the displacement (from spherical) of the equipotential at the surface of the planet (where $\phi$ is the polar coordinate with 0 pointing towards the other mass) .
Using the terms you have defined the equation is
$$ dr=\frac{m} {M} \frac{r^4}{2 R^3} \left( 3 \cos^2 \phi - 1\right) $$
The maximum bulge will be towards the star 
$$dr(\phi =0) = H = \frac{m} {M} \frac{r^4}{R^3} $$
While on the perpendicular axis the sphere will be flattened
$$dr(\phi =90^\circ) = H = - \frac{m} {M} \frac{r^4}{2R^3} $$
You can see in that article how they have used this to calculate the 54cm value for earth tides.
To get the actual radius of the planet at a particular position you would add dr to r. 
A: I can't actually bring up the article you guys are referencing, so i thought i would 'roll my own' version of the solution.  Apologies if it turns out to be just a paraphrasing of the reference - but the links are timing out for me.   Anyway... 
Consider the following diagram which depicts two bodies, the earth with mass M and centre C, and the moon with mass ‘m’ and centre D.  Also consider three points on the earth; E which is arbitrary and A and B which reside in the central plane (more so alaong the line ABCD).  The whole system is assumed to rotate about a centre of gravity (which is not depicted – but would lie presumably somewhere between C and B).

Now blue vectors = gravitation force, red vectors = centrifugal force and green vectors = resultant of blue and red i.e. the resultant tidal force.  The potential at point E can now be determined.  Note that the gravitational and centrifugal force must balance at the centre of the earth otherwise the system wouldn't remain stable in its rotation (i.e. the moon would crash into the earth etc).  Couple this fact with the inverse square law of the gravitational force (i.e. the changing length of the blue vector) and the consistency of the centrifugal force  (i.e. the red vector) that delivers a wonderful balancing act producing tidal forces pointing in opposite directions.
From the cosine rule you have
$${{q}^{2}}={{R}^{2}}+{{r}^{2}}-2Rr\cos \left( \theta  \right)\Rightarrow \frac{1}{q}=\frac{1}{R\sqrt{1+{{\left( \frac{r}{R} \right)}^{2}}-2\frac{r}{R}\cos \left( \theta  \right)}}$$ 
Now assume $r<<R$ so $r/R$  is very small.  Recall (or if you like prove) the expansion $${{\left( 1+x \right)}^{-1/2}}=1-\frac{1}{2}x+\frac{3}{8}{{x}^{2}}+O\left( {{x}^{3}} \right),\,\,\left| x \right|<1$$ hence
$$\frac{1}{q}=\frac{1}{R}\left( 1-\frac{1}{2}\left( {{\left( \frac{r}{R} \right)}^{2}}-2\frac{r}{R}\cos \left( \theta  \right) \right)+\frac{3}{8}{{\left( {{\left( \frac{r}{R} \right)}^{2}}-2\frac{r}{R}\cos \left( \theta  \right) \right)}^{2}}+O\left( {{\left( \frac{r}{R} \right)}^{3}} \right) \right)$$
Expanding, collecting like powers and dropping any terms of the order ${{\left( \frac{r}{R} \right)}^{3}}$ or higher then we have the approximation
$$\frac{1}{q}\simeq \frac{1}{R}\left( 1+\frac{r}{R}\cos \left( \theta  \right)+\frac{1}{2}{{\left( \frac{r}{R} \right)}^{2}}\left( 3{{\cos }^{2}}\left( \theta  \right)-1 \right) \right)$$
Hence the potential at E is 
$${{V}_{E}}=-\frac{Gm}{q}\simeq -\frac{Gm}{R}\left( 1+\frac{r}{R}\cos \left( \theta  \right)+\frac{1}{2}{{\left( \frac{r}{R} \right)}^{2}}\left( 3{{\cos }^{2}}\left( \theta  \right)-1 \right) \right)$$
Now the force is the gradient of this potential and so the first term contributes nothing.  The second term is simply the gravitational force due to the moon (consider taking the gradient along the direction $r\cos \left( \theta  \right)$).  Hence the tidal force originates from the last term, so we label it as the tidal potential
$${{V}_{TE}}=-\frac{Gm{{r}^{2}}}{2{{R}^{3}}}\left( 3{{\cos }^{2}}\left( \theta  \right)-1 \right)$$
Note that $g=\frac{GM}{{{r}^{2}}}$ and hence 
$${{V}_{TE}}=-g\frac{m}{M}\frac{{{r}^{4}}}{2{{R}^{3}}}\left( 3{{\cos }^{2}}\left( \theta  \right)-1 \right)$$
Potential energy (or a change in it / gained or lost) is nothing more than a measure of the work needed to move an object a certain distance.  Therefore, the potential above can be used to represent a change in distance simply by noting that $-{{V}_{TE}}/g$ represents the displacement performed by the tidal force at the surface of the earth.  Hence 
$$\Delta r=\frac{m}{M}\frac{{{r}^{4}}}{2{{R}^{3}}}\left( 3{{\cos }^{2}}\left( \theta  \right)-1 \right)$$
The rest then easily follows...i think.
