The way I understand minimal and maximal is the following.
Consider the following collection of sets:
{{1,2},{2,4},{1,6,7},{1,2,4,5,6},{8}}
The minimal members of this collection of sets is: {1,2},{2,4},{1,6,7},{8}
Meaning there are no other sets in the collection that are proper subsets of the set.
Minimal Set Example 1: in the set {1,2} this case holds true because the only proper subsets possible are {1} and {2}. Those sets {{1},{2}) are nowhere to be found in the collection. Therefore we include {1,2} as a part of the minimal collection of sets.
Minimal Set Example 2: in the set {1,6,7} this case holds true because the only proper subsets possible are {1},{6},{7},{1,6},{1,7}, and {6,7}. Those proper subsets ({1},{6},{7},{1,6},{1,7}, and {6,7}) are nowhere to be found in the collection. Therefore we include {1,6,7} as a part of the minimal collection of sets.
The maximal members of this collection of sets is: {1,6,7},{1,2,4,5,6},{8}
Meaning none of these sets are proper subsets of other sets in the collection. NOTICE: The maximal collection is not mutually exclusive of the minimum collection. See {1,6,7}.
Maximal Set Example 1: in the set {8} this case holds true because this set is not found to be a subset of any of the other sets in the collection. Meaning 8 is not found in any of the other sets in the collection. Therefore we include {8} as a part of the maximal collection of sets.
Maximal Set Example 2: in the set {1,6,7} this case holds true because this set is not found to be a subset of any of the other sets in the collection. Meaning all of the numbers of the set {1,6,7} are not found together as a part of another set in the collection. Therefore we include {1,6,7} as a part of the maximal collection of sets.
It's important to be precise in your context, here. If you are discussing a poset $P$, and wondering if $P$ has a least upper bound in the poset $P$, then you are wondering precisely if $P$ has a maximum element.
However, if you were discussing a subset of $P$, say $A$, and wondering if $A$ has a least upper bound in the poset $P$, then that's a different matter. If $A$ has a maximum element, then it will readily be the least upper bound of $A$. However, if $A$ has no maximum element, but only maximal elements--such as $A=\{a,b\}$ with $a,b$ incomparable--then the least upper bound of $A$ (if it exists) will necessarily be a member of $P$ that is not in $A$.
In your particular example with $A=\{4,5,7\}$, note that $4$ is incomparable with both $5$ and $7$, and that $5$ is greater than $7$. Hence, any upper bound of $A$ need only be an upper bound of $4$ and $5$, and any lower bound of $A$ need only be a lower bound of $4$ and $7$. The set of all upper bounds of $A$ is $\{1,2,3\}$ and this set has a least element, namely $3$, so $\sup A=3$. The only lower bound of $A$ is $8$, so $\inf A=8$.
Note that in general, $\sup A$ (if it exists) is the minimum element of the set of upper bounds of $A$, and $\inf A$ (if it exists) is the maximum element of the set of lower bounds of $A$. If $A$ has no upper (lower) bounds, then $\sup A$ ($\inf A$) doesn't exist. For example, if we didn't have the node $8$ in your example $W$, then $\inf A$ wouldn't exist. If the set of upper (lower) bounds of $A$ is non-empty, but has no minimum (maximum) element, then $\sup A$ ($\inf A$) doesn't exist. For example, suppose we were to add another node, $9$, to your example, on the same level as $8$, with $9\prec 6$, $9\prec 7$, and with $8$ and $9$ incomparable. In that case, the set of lower bounds of $A$ would be $\{8,9\}$, but since $8$ and $9$ are incomparable, that set has no maximum element, meaning $\inf A$ wouldn't exist.
Best Answer
The poset is graded or ranked (via $f:(a,b)\to a+b$). Equivalently, every maximal chain has same length. example of a chain: $(1,1)\to (1,2)\to \dots \to (1,n)\to (2,n)\to (3,n)\to \dots \to (m,n)$.
The length of the antichain the largest whitney number (number of elements with a particular rank).
PS: Writing down the hasse diagram is the best way to get the idea. Although, I have used the facts about 'graded' poset, they easy to prove once the definition is known.