# The first trip to the planet Nars

MASA have built a rocket to fly to the planet Nars which contains millions of tons of fuel which the Earth has run short of. The planet is 100 light-years away from earth. MASA has enough fuel to travel for 10 light-years. For a solution MASA have created a liquid which can cause the fuel to expand when burnt thus increasing the light-years the ship can travel.

MASAs ship can contain 1000 tons of fuel. 100 tons can take the ship a single light-year. Light years possible to travel = fuel / 100.
The formula MASA has devised, for every ton of it, it will multiply the light-years possible to travel by the tons of liquid squared plus the square root of fuel. So: $$X = f/100 * (\ell^2+\sqrt{f}),$$ where $X$ = Lightyears possible to travel, $f$ = fuel in tank and $\ell$ = MASAs formula in tank.

You can put as much liquid and fuel in the tank (not exceeding 1000 tons) as you want.

Is it possible for MASA to get to the planet Nars?

If so what is the minimum fuel and formula required to get there?
If not what is the minimum sized tank capacity requires and how much fuel and formula is required?

• @Gamow thanks I have never had a clue how to do that stuff – Beastly Gerbil Mar 22 '16 at 19:02
• Am I calculating this wrong or is any amount of fuel more than 100 tons with the correct amount of formula to get to 1000 tons, possible to get to Nars? – Okx Mar 22 '16 at 19:31
• @Okx Probably I am unsure yself – Beastly Gerbil Mar 22 '16 at 20:38

Well you can do this maximizing the formula given

$max \ \frac{f}{100}\times(\ell^2+\sqrt{f})$

with the constraint;

$\ell+f=1000$

The solution gives

$f=333.347$

$\ell=666.653$

and $X$ becomes

$X=1,481,542$

so it should be much more than enough :P I am not sure this is intended answer though since it is much more bigger than 1,000.

So how much $\ell$ and $f$ is needed to reach that planet? Which is the reverse optimization;

$min\ \ell+f$

with the constraint;

$\ \frac{f}{100}\times(\ell^2+\sqrt{f})=100$

$f=13.6177$

$\ell=27.0305$

Minimum size capacity will be the sum of these values;

$40.6482$