Step 1: Set up the Variables
Let $P, M, C$ be the sets of students choosing Physics, Math, and Chemistry respectively.
Given: $n(P)=75, n(M)=111, n(C)=40, n(P \cup M \cup C)=150$.
Let $n(P \cap C) = k$, $n(C \cap M) = k$, and $n(P \cap M) = 2k$.
Let $n(P \cap M \cap C) = g$. We are told $g \ge 1$.

Step 2: Use the Inclusion-Exclusion Principle
$n(P \cup M \cup C) = n(P) + n(M) + n(C) - [n(P \cap M) + n(M \cap C) + n(C \cap P)] + n(P \cap M \cap C)$
$150 = 75 + 111 + 40 - [2k + k + k] + g$
$150 = 226 - 4k + g$
$4k - g = 76 \implies g = 4k - 76$

Step 3: Analyze the Constraints
Since $g \ge 1$ (at least one student chose all three):
$4k - 76 \ge 1 \implies 4k \ge 77 \implies k \ge 19.25$.
Since $k$ must be an integer, the minimum value for $k$ is $20$.

Step 4: Maximize Physics but not Math
The number of students who chose Physics but not Mathematics is:
$n(P \text{ only}) + n(P \cap C \text{ only}) = n(P) - n(P \cap M)$
$= 75 - 2k$.
To maximize this value, we need to minimize $k$.
Min $k = 20$.
Max value $= 75 - 2(20) = 75 - 40 = \mathbf{35}$.