= Solution
The bold black curve is now the <pointwise maximum> of the two pure-row <expected payoff> functions:
$$
h(q)=\max(5q-3,1-2q)=
\begin{cases}1-2q,&0\le q\le4/7,\\5q-3,&4/7\le q\le1.\end{cases}
$$
It measures the largest <expected payoff> that the row player can achieve against column <mixed strategy> $q$. The column player minimizes this quantity because its own <payoff> has the opposite sign. Equality of the two lines gives $5q-3=1-2q$, or $q=4/7$. The envelope decreases before that point and increases afterwards, giving
$$
\boxed{\operatorname*{arg\,min}_{0\le q\le1}h(q)=\frac47,\qquad \min h=-\frac17.}
$$
The endpoint values $h(0)=1$ and $h(1)=2$ are both larger. A row <mixed strategy> cannot exceed the greatest pure-row <expected payoff>, because its <expected payoff> is a <convex combination> of those two values.
\Image[/past-exam-of-the-msc-in-advanced-computer-science-course-of-the-university-of-oxford/2021/computational-game-theory-column-security.png]
{title=The column player's upper payoff envelope and its minimum at q = 4/7}
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