Finding maximum and minimum in inequality word problem

These problems limit a total with "at most" (an upper cap, $\le$) or "at least" (a lower floor, $\ge$). Because the quantities compete for that total, one is largest exactly when the other is smallest — so to maximize one variable, minimize the competing one, and to minimize one, maximize the other.

Here is a maximum. A student spends at most $20 on pens ($2 each) and notebooks ($5 each), buying at least one of each. To buy the most pens, buy as few notebooks as possible — just one: $2p + 5 \le 20$, so $2p \le 15$ and $p \le 7.5$. Rounding down, the most pens is $7$.

Here is a minimum. A club must raise at least $25 selling badges ($2 each) and mugs ($5 each), with at most $4$ mugs. To sell the fewest badges, sell as many mugs as allowed — $4$ mugs raise $20: $2b + 20 \ge 25$, so $2b \ge 5$ and $b \ge 2.5$. Rounding up, the fewest badges is $3$.

Notice the rounding rule when the variable counts whole objects: for a maximum, round the bound down (you cannot go over the cap); for a minimum, round up (you must still reach the floor).

A baker has at most $20$ hours for cakes ($3$ hours each) and pies ($2$ hours each), making at least one of each. What is the most pies?
To maximize pies, bake the fewest cakes — one, using $3$ hours: $2p + 3 \le 20$, so $2p \le 17$ and $p \le 8.5$.
Rounding down, the most pies is $8$.

A team needs at least $40$ points, scoring $5$ per goal and $2$ per assist, with at most $6$ goals. What is the fewest assists?
To minimize assists, score the most goals — $6$ goals give $30$ points: $2a + 30 \ge 40$, so $2a \ge 10$ and $a \ge 5$.
The fewest assists is $5$.

A stock must total at least $50$ items, using large packs of $8$ and small packs of $3$, with at most $5$ large packs. What is the fewest small packs?
To minimize small packs, use the most large packs — $5$ large packs give $40$ items: $3s + 40 \ge 50$, so $3s \ge 10$ and $s \ge 3.3$.
Rounding up, the fewest small packs is $4$.