= Solution
{c}
Against $q$, choosing row $i$ gives Player I expected payoff $(Aq)_i=\sum_ja_{ij}q_j$. The best response maximizes this number; randomizing among rows cannot exceed the largest row payoff. Player II therefore minimizes that maximum, exactly the printed optimization problem.
The <minimax theorem> for a finite <zero-sum game> states
$$
\boxed{v=\min_{q\in\Delta_n}\max_i(Aq)_i
=\max_{p\in\Delta_m}\min_j(p^{\mathsf T}A)_j.}
$$
This common number is the <value of a zero-sum game>: Player I can guarantee an expected payoff of at least $v$, and Player II can hold it to at most $v$. At optimal mixed strategies their expected transfer is $v$.
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