Fix this HTML-ish image sizer

[sulfericacid]

This was posted as an untested solution to a SOPW a week or so past and it was stated by a monk that there was indeed an error in the calculations. Can someone provide a working solution keeping my methods thus far intact? I know it can be done via Image::Image or another module but those aren't available in my workspace and I'd like to see how close I got to the final solution.

The question was: the user needed a way to determine the h/w of an image and output an html-resized version of it so that both h/w are less than a max value.. AND the image has to stay the same aspect ratio.

#!/usr/bin/perl

use warnings;
use strict;

my $height = "1000";
my $width = "1234";

my $maxh = "600";
my $maxw = "800";

if ($height > $maxh || $width > $maxw)
{
    my $hratio = $height / $maxh;
    my $wratio = $width / $maxw;

    my $newh = $height / $hratio;
    my $neww = $width / $hratio;

   print "Height is now $newh, Width is now $neww";  
}

RESPONSE FROM ALMUT: 
It's not quite as trivial as it might seem at first (most of us who ha
+ve tried probably got it wrong the first time :) 

Just as an example, take a width of 1000 and a height of  100. Your al
+gorithm would compute 6000x600 (W x H), although the expected result 
+would be 800x80 (i.e. max width doesn't exceed 800 and aspect ratio s
+tays at 10:1)

[FunkyMonk]

You need to divide by whichever ratio is the greatest:

rescale( 1000, 1234 );
rescale( 100, 1000 );

sub rescale
{
    my ( $height, $width ) = @_;
    my $maxh = "600";
    my $maxw = "800";

    if ($height > $maxh || $width > $maxw)
    {
        my $hratio = $height / $maxh;
        my $wratio = $width / $maxw;

        my $ratio = $hratio > $wratio ? $hratio : $wratio;

        my $newh = $height / $ratio;
        my $neww = $width  / $ratio;

        printf "H=%d  W=%d  ratios: old=%0.2f new =%0.2f\n",
            $newh, $neww, $height / $width, $newh / $neww;
    }
}