SOLUTION: You have 240 liters of gasoline at point A. The price of gasoline is 0 at point A and increase by 1$ every kilometer away from Point A. This provides the opportunity for you to ma

So, at this step I will find the maximum profit without considering the constrain on 60 liters.


My scenario in this case is THIS:


    I take 240 liters and move forward x kilometers. I spend x liters for this move.
    There and then I sell 240-2x liters of gasoline at the price x dollars per liter,
          which makes me the profit P(x) = x*(240-2x) dollars.

    I still have x liters of gasoline in my car to return back.

    My profit x*(240-2x) is maximal at x = 60 kilometers. It is equal to P(60) = 60*(240-120) = 7200 dollars.

    Surely, this profit is not achievable if the constrain on 60 liters is in place.

My currecnt scenario is THIS:

   I take 60 liters and move forward 10 kilometers. I spend 10 liters for this move.
   There and then I sell 60-2*10 liters of gasoline at the price 10 dollars per liter,
          which makes me the profit P(x) = 10*(60-2*10) = 10*40 = 400 dollars.

    I still have 10 liters of gasoline in my car to return back.

   Then I repeat this action 3 more times until I spent all 240 liters of gasoline.

   Thus my final profit is 4*400 = 1600 dollars.

 
Next calculation is for the SIMILAR scenario at x= 20 kilometers.

   So, I take again 60 liters and move forward 20 kilometers. I spend 20 liters for this move.
   There and then I sell 60-2*20 = 20 liters of gasoline at the price 20 dollars per liter,
          which makes me the profit P(x) = 20*(60-2*20) = 20*20 = 400 dollars, SAME AS in the previous scenario.

    I still have 20 liters of gasoline in my car to return back.

   Then I repeat this action 3 more times until I spent all 240 liters of gasoline.

   Thus my final profit is 4*400 = 1600 dollars, SAME AS in the previous scenario.


   Notice very interesting fact: in the last two scenarious the output is the same and does not depend 
   on my travel distance "x" providing I have/save enough gasoline to return back.


Step 3 - making the intermediate base at x= 20 kilometers ahead and moving forward from this base 10 kilometers ahead
            (30 kilometers in all)

My updated scenario is THIS:

    I take 60 liters of gasoline, move forward 20 miles, store 20 liters of gasoline there, then return back.

    I spent 20 liters of gasoline moving to there; I stored 20 liters of gasoline there; and used the remaining
        20 liters of gasoline to return back. 20 + 20 + 20 = 60 liters, so I am OK.

    Next I repeat this procedure 3 more times. Surely, at the last, 4-th move "to there" I do not return back,
        which save me 20 liters of gasoline that is with me at the point B.

    So I still have 4*20 + 20 = 100 liters of gasoline with me at the point B x= 20 kilometers away from A.
    Of these 100 liters I should keep 20 liters stored to return back from B to A, so I have only 80 liters in my disposal.


    After that, I take 40 liters of these 80 liters, move forward 10 kilometers, sell 20 liters, making profit
        20*30 = 600 dollars, and use the remaining 10 liters to return back to the point B.

    OK.  But I still have 80-40 = 40 liters in my disposal.
     With them, I repeat the same procedure, and I add 600 dollars more to my profit, making it 600 + 600 = 1200 dollars in all.

    After completing these moves, I spend the remaining 20 liters of gasoline to return back from B to A.