How Can You Minimize the Expression Involving Absolute Values?

  • Context: High School 
  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    2015
Click For Summary
SUMMARY

The problem presented involves minimizing the expression $|x+2| + 2|x-5| + |2x-7| + |0.5x-5.5|$ for real numbers x. The correct solutions were provided by forum members kaliprasad and greg1313, demonstrating effective strategies for handling absolute value expressions. Key techniques included breaking down the expression into piecewise functions and analyzing critical points where each absolute value term changes its behavior.

PREREQUISITES
  • Understanding of absolute value functions
  • Familiarity with piecewise functions
  • Knowledge of critical points and their significance in optimization
  • Basic calculus concepts related to minimization
NEXT STEPS
  • Study piecewise function analysis techniques
  • Learn about optimization methods in calculus
  • Explore graphical representations of absolute value functions
  • Investigate real-world applications of minimizing expressions involving absolute values
USEFUL FOR

Mathematics students, educators, and anyone interested in optimization problems involving absolute values will benefit from this discussion.

anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Here is this week's POTW:

-----

Minimize $|x+2|+2|x-5|+|2x-7|+|0.5x-5.5|$, given $x$ is a real number.

-----

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!
 
Physics news on Phys.org
Congratulations to the following members for their correct solution::)

1. kaliprasad
2. greg1313

Solution from kaliprasad:
Let the function be f(x)

we have

$f(x)=|x+2|+2|x−5|+|2x−7|+|.5x−5.5|$

we need to consider at 4 points

$x=−2,x=3.5,x=5.x=11$

now we get the f(x) as in 3 regions as

$f(x)=−3.5x+24.5$ for $−2≤x≤3.5$

$f(x)=.5x+10.5$ for $3.5≤x≤5$

$f(x)=4.5x−9.5$ for $5≤x≤11$

Clearly $f(x)$ decreases from $x = - 2$ to $3.5$ and increases from $3.5$ to $11$.

Before $-2$ $f(x)$ decreases in range from - infinity to $- 2$ and it increases in range from $11$ to infinity and hence the lowest point occurs at $x=3.5$ the value being $f(3.5)=12.25$.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K