"Is there a reason why this line is wrong?"
Yes, because you nested max inside find. Take a look at the output of max: what is it? (hint: it will be a row vector of the maximum value from each column of your matrix). As its description states, find returns the indices of non-zero elements, so it will return the indices of whichever of those vector values is non-zero (which could be none). Note that find does NOT know anything about your matrix A (or its size or indices etc), because its input is a row vector (the output of max).
One easy alternative is to get rid of the find entirely:
>> A = zeros(8,6);
>> A(8,1) = 1;
>> A(6,2) = 1;
>> A(6,3) = 1;
>> A(6,4) = 1;
>> A(7,5) = 1;
>> A(6,6) = 1
A =
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 1 1 1 0 1
0 0 0 0 1 0
1 0 0 0 0 0
>> [~,idr] = max(A,[],1)
idr =
8 6 6 6 7 6
>> idc = 1:size(A,2)
idc =
1 2 3 4 5 6
If you really must use find then probably the simplest solution is to generate a logical matrix as the input to find (note if the maximum values occurs more than once in one column this will return all of their indices, use cumsum to return only the first one):
>> [idr,idc] = find(bsxfun(@eq,A,max(A,[],1)))
idr =
8
6
6
6
7
6
idc =
1
2
3
4
5
6
