Maximizing Income in Evony: A Mathematical Analysis

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SUMMARY

The discussion focuses on maximizing income in the game Evony through a mathematical analysis of the population-tax policy. The relationship between population (P), tax rate (T), and income (I) is established through the equations x=P(T/100) and I=x(T/100), leading to the income equation I=(PT^2)/(100^2). The user initially struggles with finding the ideal tax rate for maximum income but later acknowledges an error in their derived equation, realizing that it incorrectly predicts income at extreme tax rates. Ultimately, the user resolves their confusion independently.

PREREQUISITES
  • Understanding of basic algebra and functions
  • Familiarity with inverse proportional relationships
  • Knowledge of calculus, specifically differentiation
  • Basic concepts of game mechanics in Evony
NEXT STEPS
  • Research optimization techniques in mathematical modeling
  • Explore advanced calculus applications in game theory
  • Learn about population dynamics in strategy games
  • Investigate the economic models used in Evony for resource management
USEFUL FOR

Game developers, mathematicians, and players interested in optimizing income strategies within Evony or similar strategy games.

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Homework Statement


I derived this problem from back when I used to play Evony. I do not endorse this game whatsoever, and I think it has made enough of a mockery for itself with its plentiful supply of false and explicit advertisements such that I don't need to add to its shame.

Anyway, my problem was that I wanted to figure out a means to achieving maximum income through the game's population-tax policy. This is how it works:

You build houses to make room for your population. The more houses, the higher your maximum population potential (denoted P) will be. The actual population you can have will be between 0 and P. This population is dependent on the tax rate (denoted T), and it follows a simple inversely proportional function. With 100% T, you will have 0 population. With 0% T you will have P population. Of course, in both these cases you will have a 0 income so they are the extreme minimums. At, say, 50% T, you will have P/2 population so you will then make P/4 gold/hour (each population gives 1 gold/hour when taxed at 100%, so 2 people give 1 gold/hour at 50% T).

The Attempt at a Solution


The people, x, is defined by: [tex]x=P\left(\frac{T}{100}\right)[/tex] where [itex]0\leq T\leq 100[/itex]

The income, I, is defined by: [tex]I=x\left(\frac{T}{100}\right)[/tex]

Therefore by combining both equations we get: [tex]I=\frac{PT^2}{100^2}[/tex]

Now, P is a constant since the max pop is defined and not going to change (for this calculation at least) and I want to find the ideal tax rate, T, in order to get maximum income.

If I take: [tex]\frac{dI}{dT}=2.10^{-4}PT=0[/tex]
I get [itex]T=0[/itex] since [itex]P\neq 0[/itex]

So what am I doing wrong? Since I already established that the extremities of the tax rate, [itex]T=0,100[/itex] result in 0 income.

And if you don't understand or need clarification in something, don't hesitate to ask :smile:
Thanks for helping me satisfy my curiosity.
 
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Actually I just noticed that my derived equation is wrong. [tex]I\neq \frac{PT^2}{100^2}[/tex] since for [itex]T=100[/itex] the income, I, should be 0 but from this equation it would be P.
I guess I need help setting up the right equation.
 
Never mind, I solved it.
 

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