Fluctuating Crime Coefficient

Algebra Level 5

In the anime Psycho Pass , the crime coefficient is an index used to gauge a person's tendency of committing crimes. The index is recorded as a non-zero integer.

Inspectors from the Public Safety's Bureau actively hunt down citizens with a crime coefficient of over 100 since this indicates a high likelihood of committing crimes. To do so, they rely on the Dominator, a gun-like tool capable of reading a person's mental state and gauging their intent of committing crimes. To targets with a crime coefficient of over 100 but not exceeding 300, an energy burst which paralyzes the target is fired while for targets with crime coefficients exceeding 300, the target is exterminated.

All was well and good order was maintained in the society until the emergence of Shogo Makishima, a criminally asymptomatic individual. Such individuals are free to vary their crime coefficient by will, therefore rendering the Dominator obsolete. (Imagine robbing a bank and having the Dominator cries: Crime coefficient: 0. Not a target for enforcement action! )

With Makishima on the loose and wrecking havoc as he pleased, the Public Safety Bureau came up with an alternative strategy. They successfully altered the Dominator so that it was able to compute a new "fluctuation coefficient". The fluctuation coefficient equaled the average of sum of all differences in crime coefficients between successive encounters. The Dominator was programmed specifically to exterminate Makishima (it won't work on ordinary citizens) if the "fluctuation coefficient" exceeded a certain positive real number $K$ .

To illustrate this, suppose that for a law-abiding citizen, the Dominator records crime coefficients of $30,10,20$ in $3$ separate encounters. On the fourth encounter, if the citizen has a crime coefficient of 15, the three differences between the four successive encounters would be $30-10=20,20-10=10,20-15=5$ and the average of the sum (so-called "fluctuation coefficient") would be $\frac{20+10+5}{3} \approx 11.7$ .

The Public Safety's Bureau had 7 previous encounters with Makishima. Makishima's crime coefficients on the seven encounters were recorded as: $x,100,2x,120,3x,18,3x$ , where $x$ is an unknown non-negative integer.

The Public Safety Bureau is determined to exterminate Makishima on the $8^\text{th}$ encounter, no matter what it takes. What is the maximum $K$ allowed to be programmed into the Dominator to ensure a successful extermination?

Submit your answer to 2 decimal places.

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Reference: Anime: Psycho Pass .

This question is part of the problem set Mathematics in Anime

The answer is 46.29.

1 solution

Zk Lin
Mar 1, 2016

Let Makishima’s crime coefficient on the $8$ th encounter be $y$ and denote $f(x,y)$ as the fluctuation coefficient.

Note that $f(x,y)=\frac{|100-x|+|100-2x|+|120-2x|+|120-3x|+|18-3x|+|18-3x|+|3x-y|}{7}$ . We can rephrase the question as “find maximum $K$ such that $f(x,y) \geq K$ .”

Since $|3x-y|$ is non-negative, $f(x,y)$ is minimized only when $y=3x$ . Therefore, we are now seeking to minimize $g(x)=f(x,3x)$ .

Note that $g(x)=\frac{|100-x|+2|50-x|+2|60-x|+3|40-x|+6|6-x|}{7}$ .

For $0 \leq x \leq 6$ , we have

$g(x)=\frac{100-x+2(50-x)+2(60-x)+3(40-x)+6(6-x)}{7}=\frac{476-14x}{7}$ . The minimum of this function is attained when $x=6$ . A quick computation reveals that $g(6)=56$ .

For $7 \leq x \leq 40$ , we have

$g(x)=\frac{100-x+2(50-x)+2(60-x)+3(40-x)+6(x-6)}{7}=\frac{404-2x}{7}$ . The minimum of this function is attained when $x=40$ . A quick computation reveals that $g(40)=46.2857$ .

For $41 \leq x \leq 50$ , we have

$g(x)=\frac{100-x+2(50-x)+2(60-x)+3(x-40)+6(x-6)}{7}=\frac{164+4x}{7}$ . The minimum of this function is attained when $x=41$ . A quick computation reveals that $g(41)=46.8571$ .

For $51 \leq x \leq 60$ , we have

$g(x)=\frac{100-x+2(x-50)+2(60-x)+3(x-40)+6(x-6)}{7}=\frac{8x-36}{7}$ . The minimum of this function is attained when $x=51$ . A quick computation reveals that $g(51)=53.1429$ .

For $61 \leq x \leq 100$ , we have

$g(x)=\frac{100-x+2(x-50)+2(x-60)+3(x-40)+6(x-6)}{7}=\frac{12x-276}{7}$ . The minimum of this function is attained when $x=61$ . A quick computation reveals that $g(61)=65.1429$ .

For $x \geq 101$ , we have

$g(x)=\frac{x-100+2(x-50)+2(x-60)+3(x-40)+6(x-6)}{7}=\frac{14x-476}{7}$ . The minimum of this function is attained when $x=101$ . A quick computation reveals that $g(101)=134$ .

Therefore, the maximum permissible $K$ is $46.2857 \approx \boxed{46.29}$ , attained when $x=40,y=120$ .

Moderator note:

Once we realize that we want to minimize $f(x, 3x)$ , is there a better approach than testing all of the values?

Calvin Lin Staff - 5 years, 3 months ago

Lol I did like this too.

Trung Đặng Đoàn Đức - 5 years, 2 months ago

Fluctuating Crime Coefficient

Reference: Anime: Psycho Pass .

The answer is 46.29.

1 solution

Moderator note:

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