$DIR='/someDir';
$FILE='someFile';
$DIR2='/someOtherDir';
$out = `echo cp "$DIR/$FILE $DIR2/$FILE" . "_dev1"`;
print $out,$/;
#
# output
#
cp /someDir/someFile /someOtherDir/someFile . _dev1
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The cp command is not getting the parameters you expected. The double-quotes are not needed within back-quotes, unless you want them surrounding the parameters to cp in case of spaces in the directory or file names. In that case you'd need to escape them with backslashes. Use:

`cp $DIR/$FILE $DIR2/${FILE}_dev1`
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The braces around the variable name (the 2nd instance of $FILE) distinguish the variable from the following text.