The rocket equation relates the final velocity of a rocket to the rocket's mass before and after the maneuver. The velocity of the rocket after the maneuver is the velocity of its exhaust gases multiplied by the natural logarithm of the ratio of its initial mass to its final mass. Incidentally, this calculation is done for the perfect case, your rocket is floating in deep space and has no gravitational field or atmosphere to work against. (from Tsiolkovsky)

Vf = Ve ln (Mi/Mf) where

Vf = final velocity of rocket,
Ve = exhaust velocity (assuming a perfectly designed and efficient engine),
Mi = initial mass (Rocket + propellant),
Mf = final mass (Rocket mass)

The total amount of energy available in the propellant is directly related to the amount of fuel. Not all this energy will go into final rocket velocity, some of it will be used to push unburned propellant during acceleration or wind up as waste heat in the exhaust.

Et = K (Mi-Mf)

Et = Energy available

K = A proportionality constant depending on the type and efficiency of the propulsion system

(Mi-Mf) = Mass of propellant expended in the maneuver

The total amount of kinetic energy imparted to the rocket at the end of the maneuver(from Newton's Laws) is:

Er = 1/2 Mf Vf**2

Er = Total rocket kinetic energy
Mf = Final mass of rocket
Vf = Final velocity of rocket

We know the more fuel you carry, the faster you go, but there is a very harsh law of diminishing returns at work here. The more propellant you carry the more of it's energy is expended in pushing unburned fuel instead of pushing rocket. So what we want to do is find the optimum amount of propellant to burn in a maneuver so that the maximum fuel energy is transferred to the rocket, and the minimum is used to push unburned propellant. I.e., we want to maximize the energy efficiency of the fuel.

Ee = Er/Et,
Ee = Energy efficiency,
Er/Et = Ratio of kinetic energy achieved to total energy available

and substituting from above

Ee = 1/2 Mf ( Ve ln (Mi/Mf)**2 / K(Mi-Mf)

and defining the mass of the rocket after the maneuver to unity (1 ton, 1 kiloton, 1 whatever)

Ee = 1/2 (Ve ln (Mi)**2 / K(Mi-1)

and collecting terms and simplifying

Ee = (Ve**2/2K) (ln (Mi) )**2 / (Mi-1)

And combining all the constants into one called "C = Ve**2/ 2K", we can create a relation between energy efficiency and intial mass

Ee = C (ln (Mi) )**2 / (Mi -1)

Note we do not need to know the value of capital C, it is irrelevant
to the value of Mi which gives maximum efficiency.

This equation will have a maximum at some value, which could be determined by taking the derivative and setting it to 0. However ,for those of us rusty with our calculus, the answer can also be calculated algebraically by iteration. It turns out that energy efficiency is maximized at

Ee = 0.648 C which occurs when Mi = 4.92, implying a final rocket velocity Vf = 1.593 Ve.

In plain English, this means that for maximum efficiency (which will never exceed about 65% of C), you need to burn almost 4 tons of propellant for every ton of rocket and your final velocity will only be about 60% more than the velocity of your exhaust gases. You can go faster if you like, but you will be using fuel inefficiently, that is, lifting this fuel up to your rocket will consume a lot of extra energy and logistical infrastructure which could be better spent on the rocket itself.

Keep in mind this entire analysis is for a simple one-way fly-by mission. If we want to decelerate after we get to our destination, the second stage will have to be configured with these same equations and physical constraints in mind. And every drop of propellant we have to carry to explore our destination will require enormous amounts of fuel just to get it there, and slow it down after it gets there. If you want to turn around and go home, and stop after you get back, we need to add even more fuel.

The only way to get around having to build impossible-to-engineer rockets, multiple-stage ships which are more than 99.99999+…% propellant by mass, is to have extraordinarily high values of Ve, values which cannot be achieved by chemical, fission or even fusion engines. Only direct matter-energy conversion technologies, such as matter-antimatter propulsion, with exhaust velocities on the order of double-digit percentages of complete E=Mc**2 conversion (Ve approximately on the order of the speed of light ) will ever get us to a rocket Vf around 0.5c, the slowest speed considered still feasible for manned interstellar exploration.