In a min heap, maximum element is present in one of the leaf nodes.
If index of heap starts with $1,$ indices of leaves are $\lfloor n/2 \rfloor +1,\lfloor n/2 \rfloor +2,\lfloor n/2 \rfloor +3 \ldots n$.
So, we have to perform linear search on at most $n/2 +1$ elements to find the maximum element. (we cannot perform binary search since it is not guaranteed that leaves are in sorted order) and that multiplied by some constant $c$ will be the time complexity of the optimal algorithm. (Here, $c$ includes the cost of all operations which includes comparison, index shift etc. for a single maximum element compare)
In asymptotic terms, $c * (n/2 +1)$ is $\Theta(n).$