Since the Hungarian Method only minimises, a profit table must first be converted into an equivalent loss table by subtracting every entry from the matrix's largest value, and the method is then applied exactly as for a minimisation problem. …

Conversion: the largest entry in the profit table is $M=30$. Converted loss matrix $c'_{ij}=30-p_{ij}$:

Row reduction (row minimums $0,5,0$):

Column reduction (column minimums $0,0,0$ — already the smallest in each column): matrix unchanged.

Cover zeros and test: zeros are at $W_1J_1$, $W_2J_3$, $W_3J_2$ — three zeros, no two sharing a row or column, so they are already independent; the minimum cover is 3 lines $=n$ → optimal immediately.

Assignment: $W_1\to J_1$, $W_2\to J_3$, $W_3\to J_2$ — a valid one-to-one mapping.

Profit computed from the original profit matrix: $W_1J_1=30$, $W_2J_3=25$, $W_3J_2=30$; total $=30+25+30=85$. …

Independent verification: after finding the converted-matrix assignment, recompute the total using the ORIGINAL profit matrix (never the converted loss matrix) and compare against a few alternative …

Reading the final answer straight off the converted loss matrix's zero cells (a cost saving, not a profit) instead of going back to the original profit matrix; and forgetting to reconvert at all, i.e. running the Hungarian Method directly on the profi …