MHB Problem of the Week #113 - May 26th, 2014

[Chris L T521]

Thanks to those who participated in last week's POTW! Here's this week's problem!

-----

Problem: Find three positive numbers whose sum is $27$ and such that the sum of their squares is as small as possible.

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by kaliprasad, lfdahl, magneto, Opalg, and Pranav. I'm going to post two different solutions below.

magneto's solution: [sp]We want to minimize $S(x,y,z) = x^2 + y^2 + z^2$ subject to the constraint
$x + y + z = 27$. So consider the function:
$$ f(x,y,z,\lambda) = x^2 + y^2 + z^2 - \lambda(x + y + z - 27).$$
We can take the partial derivative with respect to each variable:
$f_x(x,y,z,\lambda) = 2x - \lambda$,
$f_y(x,y,z,\lambda) = 2y - \lambda$,
$f_z(x,y,z,\lambda) = 2z - \lambda$, and
$f_\lambda(x,y,z,\lambda) = 27 - x - y - z$.

We will solve using Lagrange's multiplier method --- set each equation to $0$,
and solve the system. From equation $f_x(x,y,z,\lambda) = 0$ implies $x = \frac \lambda 2$.
Analogously, we have $y = z = \frac \lambda 2$.

$f_\lambda(x,y,z,\lambda) = 27 - 3 \cdot \frac{\lambda}2 = 0$, or $\lambda = 18$. So,
we have $x = y = z = 9$. So, if an extrema exists, it is at $(9,9,9)$. We just need to verify this
is the minimum. We can do so by seeing $S(9,9,9) = 243 < 627 = S(1,1,25)$.[/sp]

Opalg's solution (which uses no calculus!): [sp]Here's a purely algebraic way to do it – no Lagrange multipliers, no calculus!

Given $a,b,c$, use the notation $$\textstyle \sum a = a+b+c,$$ $$\textstyle\sum a^2 = a^2+b^2+c^2,$$ $$\textstyle\sum bc = bc+ca+ab,$$ $$\textstyle\sum(b-c)^2 = (b-c)^2 + (c-a)^2 + (a-b)^2.$$

Suppose that $a+b+c = k$. Then $k^2 = \bigl(\sum a\bigr)^2 = \sum a^2 + 2\sum bc$. Also, $\sum(b-c)^2 = 2\sum a^2 - 2\sum bc.$ Add those equations to get $k^2 + \sum(b-c)^2 = 3 \sum a^2.$ So $\sum a^2 = \frac13\bigl(k^2 + \sum(b-c)^2\bigr)$, and this is clearly minimised by minimising $\sum(b-c)^2.$ But the minimum value of $\sum(b-c)^2$ is $0$, which occurs when $a=b=c = \frac13k$.

In the case where $k=27$ we should take $a=b=c=9$.[/sp]