The way to calculate this is to calculate what the best strategy is for 1 serve then use this result to calculate the best strategy for 2 serves and so on.

In general for x{i} where x is speed of serve with i serves left:
Probability of successuful serve = (120 - x{i}) / 70
Probability of unsuccessful serve = (x{i} - 50)/ 70
Expectation{i} = x{i} * (120 - x{i}) / 70 + Expectation{i-1} * (x{i} - 50) / 70
We have Expextation{0} = 0
Therefore Expectation{1} = x{1} * (120 - x{1}) / 70 + 0 * (x{1} - 50) / 70
To maximise Expectation{1}, we use calculus to find where gradient is equal to 0.
dExpectation{1}/dx{1} = (120 - (2 * x{1})) / 70
0 = (120 - (2 * x{1})) / 70
x{1} = 60

So with 1 serve you should try 60mph which gives the maximum expectation of 3600/70 = 51.42857143

So for 2 serves we calculate:
Expectation{2} = x{2} * (120 - x{2}) / 70 + 3600/70 * (x{2} - 50) / 70

Finding the maximum as above we obtain the best serve speed for 2 serves remaining is 12000/140 = 85.71428571 and a expectation of 68.22157434.

I haven't got time at thae moment to calculate further values, but the above gives the method of solving these types of problems. The approximate result for 3 serves is to serve at ~94 mph with an expectation of ~77.8

In general the best strategy is to serve fastest at the beginning and decrease the serve speed as you continue to fault. I suspect that this is true of any function, not just linear, where there is a negative correlation between serve speed and success.