Mathematical mind boggler, very confusing possible new mathematical breakthrough

Question

I'm not trying to make a joke here but I am very confused I been trying to figure this out for like 6 hours straight now got about 20 notepads opened up here, 15 calculators and I cant crunch it I'm always getting too much excess in the end.

Lets explain some variables here we got to work with. Say we got

2566 min points / 2566 max points

0 min xp / 4835 max xp

There is 2 types of jobs that need to use both variables (points and xp)

Job (1) subtracts 32 points per click and adds 72 xp per click.

Job (2) subtracts 10 points per click and adds 14 xp per click.

I'm trying to figure out how to calculate the excess properly. So it would waste the minimal amount of Job(1)'s to still have enough points to do as much Job(2)'s as it possibly can and still reach max xp.

Thats the thing I dont want to run Job1's until there are no more points left because in doing so, the Job1's will exceeds the maximum XP (2566) and I will never get to do any Job2's.

I want to get the maximum possible Job2's in then using proper calculation achieve or overflow the MaxXP of 2566 with Job1's to always achieve max XP. Pretty much my situation is that I need to get 2566 MaxXP to be able to continue completing jobs. While keeping that in mind I want to place most priority on job2's and only use Job1's to achieve the necessary MaxXP of 2566 to reset the min points to max to redo the process all over. I am trying to automate this.

Here is my equations

amountOfJob1s = (minPoints / 32)

amountOfJob2s = (minPoints / 10)

excessXP = (amountOfJob1s * 72) - maxXP

if excessXP < 0 then break

Results

mustDoJob1s = ???

mustDoJob2s = ???

Thank you if anyone can help me figure this out so I can put a good equation here I'd appreciate it.

Either this is not mathematically possible or I just can't crunch it I do believe I have enough variables.

Answer 1

Let job1 be the amount of job1 and job2 be the amount of job2. We are left with two equations and two unknowns:

job1 * 32 + job2 * 10 = 2566
job1 * 72 + job2 * 14 = 4835

So:

job1 = 45.683...
job2 = 110.411...

Given job1 as the higher xp/point ratio and you wanna go over 4835 xp, round job1 up, compute job2 and round it down.

job1 = 46

job1 * 32 + job2 * 10 = 2566
job2 = 109.4

job2 = 109

Check:

job1 * 32 + job2 * 10 = 2562 points
job1 * 72 + job2 * 14 = 4838 xp

Done.

Two unknowns is hardly a 'new mathematical breakthrough' :)

Answer 2

I assume you want to get as much "XP" as possible, while spending no more than 2566 "points" by "clicking" an integer number of times {n1, n2} on each of two "jobs". Here is the answer in Mathematica:

In[8]:= Maximize[{72 n1 + 14 n2, n1 >= 0, n2 >= 0, 
  32 n1 + 10 n2 <= 2566}, {n1, n2}, Integers]

Out[8]= {5956, {n1 -> 80, n2 -> 0}}

Or, maybe you need to spend exactly 2566 points? Then the best you can do is:

In[9]:= Maximize[{72 n1 + 14 n2, n1 >= 0, n2 >= 0, 
  32 n1 + 10 n2 == 2566}, {n1, n2}, Integers]

Out[9]= {5714, {n1 -> 78, n2 -> 7}}

Is this what you wanted?

Answer 3

Let a be the number of Job 1 and b the number of Job 2.

XP = 72 a + 14 b
P = 32 a + 10 b

You appear to want to solve for a and b, such that XP <= 4835, P <= 2566 and b is as large as possible.

72 a + 14 b <= 4835
32 a + 10 b <= 2566

b will be largest when a = 0, i.e.

b <= 4835 ÷ 14, => b <= 345
b <= 2566 ÷ 10, => b <= 256 

As b must be both below 345 and 256, it must be below 256.

Substitute back in:

72 a + 14 × 256 <= 4835, => a <= ( 4835 - 14 × 256 ) ÷ 72, => a <= 17
32 a + 10 × 256 <= 2566, => a <= ( 2566 - 10 × 256 ) ÷ 32, => a <= 0

so a = 0, XP is 2560 and points used is 3584.

Alternatively, you can solve for the closest satisfaction of the two inequalities

72 a + 14 b <= 4835                (1)
32 a + 10 b <= 2566                (2) 
b <= ( 2566 - 32 a ) ÷ 10          (3) rearrange 2
72 a <= 4835 - 1.4 ( 2566 - 32 a ) (4) subst 3 into 1
27.2 a <= 1242.6
a <= 45.68

so choose a = 45 as the largest integer solution, giving b = 112, XP is 4808, points used is 2560

For either of these, there's no computer programming required; if the constants associated with the two jobs change, then the formulas change.

For harder to solve examples, the relevant area of mathematics is called linear programming