1
$\begingroup$

Let $a,b,c,d$ be constants in the interval $[-1,1]$. Let $A$ be the minimum of the following three numbers: $\max\{2-a-c, |b-d|\},$ $\max\{2-b-c, 2-a-d, 2-b-d\},$ $\max\{2+b-c, 2-a+d, 2+b+d\}.$ Is there a neater way to express $A$?

  • 0
    Related: http://math.stac$k$exchange.com/questions/110808/a-minimization-problem-for-a-function-involving-maximum2012-02-21

2 Answers 2

1

If each of $b,d$ are replaced by their negatives $-b,-d$ then the first max is unchanged, and the second and third max get switched, since the numbers considered get switched (in the same order as written). So you can include $b>d$ as an assumption, and thus drop the absolute value in the first max, making it $\max \{2-a-c,\ b-d \}$. (This isn't much of a simplification.)

0

use this:

$\max\{x,y\}=\frac{x+y+|x-y|}{2}$

and

$\min\{x,y\}=\frac{x+y-|x-y|}{2}$