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Given distinct, positive integers $p, b, y$ such that

$$4p+2b+3y = 149$$

How can we minimize $2p + 3b + y$?

$$2p + 3b + y = (4p+2b+3y)-2p+b-2y = (149+b)-2(p+y)$$

Which is equivalent to maximizing $p+y$ since $b>0$. However, I am not sure where this leads us.

1 Answers1

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If we define $ r=4p,\phantom{x} s=2b,\phantom{x} t=3y $, then we have the following problem:

$$ \min\left(\frac{1}{2}r+\frac{3}{2}s+\frac{1}{3}t\right) \phantom{xx}\text{s.t.}\phantom{xx}r+s+t=149 $$

From the equation, it's obvious that we need to maximize $t$ and minimize $s$. The integer solution that fulfil this is $y=45,\phantom{x}p=3,\phantom{x}b=1$

acat3
  • 11,897
  • Since $~y = 46, b > 1~$ is viable, there is a conflict between maximizing $~t~$ and minimizing $~s.~$ I don't think the conflict can be resolved analytically. If I am right, then you have to manually explore the alternative, and verify that it is not optimal. – user2661923 Jan 11 '24 at 03:58
  • @user2661923 Which feasible solution has $y=46$? – acat3 Jan 11 '24 at 05:03
  • Typo on my part, should be $~y = 47.$ – user2661923 Jan 11 '24 at 05:05
  • @user2661923 ah yes, you're right. I will work on the solution further then – acat3 Jan 11 '24 at 06:06