The area visible from two lighthouses with angle of vision 30 degrees, built at distance 10km from each other The distance between 2 lighthouses is 10 km.
What is the maximum area of the ocean in which both can be simultaneously visible if the angle of vision for each lighthouse is 30 degrees?But the minimum?


*

*This is how I imagined it would be (some kind of generalization)





*For maximum I see the result as infinity but has to be something mesurable (something by aproximation)----think that epsilon {violet} (deviation angle) is very small(like in limits)





*For minimum I see the result as "1 POINT" but I am not sure its the correct answer.
Think that epsilon {violet} is very small (like in limits)





*I am not sure if I imagined this properly, so don't take as guaranteed what I drew.
I think about partial sums, limits, and some aproximation. (I don't know)

*When both exterior lines are perpendicular to ground, the area is infinite.

 A: 
From the figure drawn, it is not difficult to see that when ε is $0$, the common visual area (CVA, marked as red) is infinity.
Thus, the smaller is the ε, the larger is the CVA.
Depending on the meaning of finding the minimum value of the CVA, we have 2 interpretations
(1) The right beam shines directly onto O and the left beam shines the land. Then, $CVA = 0$.
(2) The left beam shines directly onto O and the right beam is the line DF and $\angle FDO = 30^0$ (and $ε = 60^0$. Then, CVA = pink region whose area can easily calculated because triangles OFG, OFD and OGD are all right-angled.
A: This is the extended part in answering the OP’s further query.

As seen in the figure, if ε is very small, the CVA is the purple region. The answer is obviously still infinity. 
Of course, the above leads us to nowhere. To make it calculate-able, we can either assume:-
(1) OP, the “width” of the sea, is L; or
(2) the light beams generated are of limited power. It can only cover a distance of length L.
Then, CVA is approximately equal to (Edit)
$10*L – (⊿OED + ⊿OFD – ⊿OHD)$
$= 10 \cdot L – (\frac {10^2 \cdot \tan (60^0 – ε)}{2} + \frac {10^2 \cdot \tan (60^0)}{2} – \frac {10^2 \cdot \sin (60^0)}{2})$;
since ε is small and therefore ⊿OHD can be considered as  equilateral.
After further simplification and approximation, 
CVA $~= 10 \cdot L – 50 \cdot (2 \cdot \tan 60^0 - \sin 60^0)$
$= ... = 10 \cdot L – 75 \sqrt 3$
