Suppose that the output (how many seconds passed between the arrival of the first type and second types of monkeys) is $S$. Then we know that monkeys of the first type spent $S$ seconds in the garden and monkeys of the second type spent $T-S$ seconds.

We can calculate how many coconuts the first type can gather in $S$ seconds. Let this amount be $X(S)$.

Also, we can calculate how many coconuts the second type can open in $T-S$ seconds (assuming that there is an infinite number on the ground). Let this amount be $Y(S)$.

If $X(S)>Y(S)$, this means that the actual output is less than $S$ because monkeys of the second type do not have enough time to open all the coconuts.

If $X(S)\le Y(S)$, this means that the output can be $S$ or more because there is time to open all coconuts.

The correct output is the largest $S$ for which $X(S)\le Y(S)$. We can use binary search to find it.