I have proved the decision version of my problem be $\mathcal{NP}$-complete. And I know that if I can solve the optimization version in poly-time, then I can just to compare the obtained minimum (or maximum) with target value in decision version. Thus, the decision version can be solved in poly-time as well. Since the decision version is $\mathcal{NP}$-hard, so is the optimization version, i.e., the optimization version is $\mathcal{NP}$-hard.
My question is how to prove the converse direction: if the decision version can be solved in poly-time, can the optimization version be solved in poly-time as well?
You can make a polynomial number of calls to your solver for the NP-complete problem to find the optimal solution. E.g., for the travelling salesman problem you would perform a binary search with a series of queries asking whether there is a tour with some cost $c$, gradually adjusting $c$ and bracketing the solution space until you have the optimal solution.