Minimum element in a Max heap will always be one of the leaf nodes of the heap.
Leaf nodes in heaps range from $\left \lfloor \frac{n}{2} \right \rfloor+1$ to $n$.
so in the worst case we need to check $\frac{n}{2}$ elements.
$\therefore$ time complexity is $\Omega (n)$ for finding minimum element in a max-heap in worst case.
OR
Assume that heap is stored in array and then we apply linear search on the second half of the array since all the leaf nodes will be in second half of the array
So T.C. = $\Omega (\frac{n}{2}) = \Omega(n)$