i need help with a word problem

Jul 2019
3
0
USA
Emmett has just finished building a time machine. The time machine is powered by a mixture of Solution 1 and Solution 2. A gallon of Solution 1 will send the time machine back 6 years and a gallon of Solution 2 will send the time machine back 4 years. No more than 15 gallons of Solution 1 and 10 gallons of Solution 2 are available. Furthermore, the amount of Solution 2 used needs to be at least twice the amount of Solution 1 or else the time machine will explode.
Emmett wants to go at least 28 years back in time while avoiding explosions and minimizing the total amount of mixture he puts in the time machine (both solutions are very expensive). How many gallons of Solution 2 should Emmett use?
 

romsek

MHF Helper
Nov 2013
6,667
3,005
California
one has to assume that the way mixing works that

$(s_1+ s_2) ~gal \Rightarrow (6~yr/gal\cdot ~s_1~gal+4 ~yr/gal \cdot ~s_2~gal)~yr = (6s_1 + 4s_2)~yr$

we have constraints

$s_1 \leq 15~gal$

$s_2 \leq 10 ~gal$

$s_2 = 2s_1$

and our goal is

$6s_1 + 4s_2 = 28$

I'd first solve this and check if we bump up against the first two constraints. If not, we're done.

$6s_1 + 4(2s_1) = 28$

$14s_1 = 28$

$s_1 = 2$

$s_2 = 4$

Neither of these violates the first two constraints so we're done.

We use 4 gallons of solution 2.