Hi muskang2000,Yes, your approach is completely valid, and it lands on the same answer the rest of the thread reaches (
A, 7 hr). Let me confirm each piece so you can trust it.
Your setup is exactly right:
- H alone for
x hours, K alone for
y hours, both together for
14 - (x + y) hours.
- Total output:
3x + 4y + 7(14 - x - y) = 77, which cleanly simplifies to
4x + 3y = 21.
- Together-time =
14 - (x + y), so to minimize it you maximize
x + y. All correct.
One small thing worth making airtight: you found two sample solutions (x=
3, y=
3 and x=
0, y=
7) and picked the larger x+y by inspection. That works here, but it's even safer to show
why x=
0, y=
7 is the true maximum rather than just one of several options.
From
4x + 3y = 21, solve for x:
x = (21 - 3y)/4. Then:
x + y = (21 - 3y)/4 + y = (21 + y)/4This
increases as y increases, so you want y as large as possible. Since x can't go below
0, set x =
0, which forces y =
7. That gives x + y =
7, and together-time =
14 - 7 = 7 hours.
Notice this matches the intuition everyone else used: push the faster machine (K) to run alone as much as possible, and bring H in only for the leftover. Your algebra is just the formal version of that same idea - so it's a perfectly good method to use on test day.
Answer: Amuskang2000
Machine H worked alone for x hrs out of 14,
Machine K worked alone for y hrs out of 14,
Then, Machine H & K together worked for 14 - (x + y) hrs. We need to minimise this. For that, we need to maximise x+y.
3x+4y+7(14-x-y)=77.
Solve to get 4x + 3y = 21.
Possible solutions: x=3,y=3; x=0, y=7
x+y is maximum at x=0, y=7. Hence, 14 - (x+y) = 14 - 7 = 7.
Bunuel please could you confirm if this is also the right approach? thanks!