A(t) has a maximum at the same point as A(t) - 34 (evident). Further, A(t) has a maximum at the same point where 1/(A(t) - 34) has a minimum (also evident and take into account that A(t)-34 > 0).
1/(A(t) - 34) = 1 + 0.22*(t-4)^2.
This clearly has a minimum at t=4 ( (t-4)^2>=0 and =0 only for t=4 ).
t=0 corresponding to 7am, so t=4 corresponding to 7+4=11am.
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