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I'm a freshman in college with no prior experience in non linear optimization. This is a question I encountered in an MCQ format.

Problem

$\ x_i \epsilon Z $

$\ i = {1,2,3,4,5} $

$\sum_{i=1}^5 x_i = 10$

$\sum_{i=1}^5 x_i^2 = 40$

Find the maximum and minimum value that $\ x_i $ can take.

Attempt

I replaced $x_i$ with $k_i + m$ and found

$\ M = \sum_{i=1}^5 k_i = 10-5m$

$\ P = \sum_{i=1}^5 k_i^2 = 40 + 5m^2 - 20m$

Solving $M^2=P$ yielded $m = 1,3$ which generated the solution set

{1,1,1,1,6}

{3,3,3,3,-2}

indicating that 6 and -2 were the required values, however I'm clueless as to why the math worked out in this manner.

Queries

I'd like to know why $M^2=P$ yielded the critical condition, how many solution sets there are to this problem, as well as a general approach to such questions.

This is my first post on this forum, thank you in advance.

Glaxionaizer
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  • Welcome, @Glaxionaizer! – amWhy Aug 17 '16 at 16:29
  • I got $-2$ to $6$ too by running it through a basic python script but where did you get the idea to set $M^2=P$? Is that what you were taught in class? I don't understand where it comes from. – Jam Aug 17 '16 at 16:50

1 Answers1

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An Interpretation

If you applied the technique of Lagrange multipliers, then you would know that, at an optimal point $\left(x_1,x_2,x_3,x_4,x_5\right)=\left(y_1,y_2,y_3,y_4,y_5\right)$, it must hold that $y_1=y_2=y_3=y_4$. Then, you can set $m$ to be the common value of $y_1,y_2,y_3,y_4$. Write $y_5=m+t$. Then, you have $$t=\sum_{i=1}^5\,\left(y_i-m\right)=\sum_{i=1}^5\,y_i -5m=10-5m$$ and $$t^2=\sum_{i=1}^5\,\left(y_i-m\right)^2=\sum_{i=1}^5\,y_i^2-2m\,\sum_{i=1}^5\,y_i+5m^2=40-20m+5m^2\,.$$ Therefore, $$40-20m+5m^2=t^2=(10-5m)^2\,.$$

Batominovski
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