GLENDALE, Ariz. — The longest-tenured active player on the Dodgers’ roster will be sticking around a little longer.
Max Muncy agreed to a contract extension that could keep him with the team through the 2028 season. The contract guarantees him $10 million – a $7 million salary in 2027 with a $3 million buyout if the Dodgers do not exercise a club option for 2028 at another $10 million. The 35-year-old Muncy will earn $10 million in 2026 after the Dodgers exercised a club option for 2026 in his previous contract.
Originally drafted by the Oakland A’s, Muncy was briefly out of baseball when the A’s released him in March 2017. The Dodgers signed him as a free agent a month later and Muncy spent the season in Triple-A. He returned to the majors with the Dodgers in 2018 and emerged as a key piece of their lineup, playing first base, second base and eventually third base.
In eight seasons with the Dodgers, Muncy has hit .232 with an .843 OPS and 209 home runs, including four seasons with at least 35. A two-time All-Star, Muncy has been part of the Dodgers’ World Series title-winning teams in 2020, 2024 and 2025 and is the franchise’s all-time leader in postseason home runs (16).
Muncy has been slowed by injuries the past two seasons and hasn’t played more than 100 games since 2023.