Find the optimal sliding door width

Perl, 190 180 154 133 128 117 bytes

includes +1 for -p

use POSIX;$m=1E4;for$q(80,100,120){($m,@z)=($p,$n,ceil$_/$n)if$m>($p=(150,200,220)[$x++]*($n=ceil$_/$q))}$_="@z $m"

Commented:

use POSIX;                                  # for ceil()
$m = 1E4;                                   # init min price to 10k
for $q (80,100,120) {                       # iterate widths
    ($m,@z) = ($p,$n, ceil $_/$n)           # update min, output
    if $m > (                               #
       $p = (150,200,220)[$x++]             # grab price
          * ( $n = ceil $_/$q )             # times nr of doors needed
    )
}
$_="@z $m"

---

• Save 11 bytes by inlining and splitting hash to two arrays

• Save 5 bytes by using -p (thanks to @dev-null)

• Save 18 bytes by using POSIX::ceil and 3 more by using list syntax for hash (thanks to @msh210)

JavaScript (ES6), 101 bytes

t=>[[80,150],[100,200],[120,220]].map(([w,p])=>[n=-~(~-t/w),-~(~-t/n),n*p]).sort((a,b)=>a[2]-b[2])[0]

-~(~-a/b) is the same as Math.ceil(a/b) in 31-bit integers.

PowerShell, 137 135 bytes

param($a)$j=9e9;60..120|%{if((($c=[math]::ceiling($a/$_))*($p=(220,(200,150)[$_-le80])[$_-le100]))-lt$j){$j=($k=$c)*$p;$i=$_}}
$k;$i;$j

Output is newline-separated.

We take input $a, set our cost $j to 9000000000 (a large number that's way more than we would ever need). Next, we loop from 60..120 with |%{...}. Each iteration we calculate the $p price of the current item with a pseudo-ternary statement, then calculate the $c ceiling of $a/$_. If the current total is smaller than the smallest total we've seen ($j), save all these variables: $j (the total), $k (the number of doors required), and $i (the door width), and continue the loop. Once the loop is finished, just output the best values.

_{Edit -- Saved two bytes by moving the $c and $p assignments into the if conditional}