Since the question asks for the largest possible number of large cups, think about whether making any small cups helps or hurts that goal. What happens to the resources available for large cups if you increase or decrease $s$?

Try setting $s=0$ and rewrite both inequalities in terms of $\ell$ only. Then figure out the largest integer $\ell$ that satisfies both of those inequalities at the same time.

Each inequality will give you an upper bound on $\ell$. The true maximum $\ell$ must be less than or equal to both bounds. Which of those two bounds is smaller?

Use $x$ for the number of small cups and $y$ for the number of large cups. In Desmos, type:

• 9x + 15y <= 900
• 30x + 42y <= 3600
• x >= 0
• y >= 0 This will shade the region of all $(x,y)$ pairs that satisfy the coffee and time limits.

Because $y$ represents large cups, you want the point in the shaded region with the greatest possible $y$-value. Visually, this is the highest point in the shaded region (closest to the top of the graph) that still lies within the intersection of all the inequalities.

Add the vertical line x = 0 in Desmos. Find where this line intersects the boundary of the shaded region (in particular, the line 9x + 15y = 900). Click that intersection point: the $y$-coordinate shown is the maximum number of large cups you can make when $x=0$, and it still satisfies both constraints.

Let $s$ be the number of small cups and $\ell$ be the number of large cups.

• Coffee constraint: each small uses 9 oz and each large uses 15 oz, and there are at most 900 oz total.
  • This gives the inequality

• Time constraint: each small takes 30 seconds and each large takes 42 seconds, and there are at most 3{,}600 seconds.
  • This gives the inequality

We also know $s \ge 0$ and $\ell \ge 0$, and we want to make $\ell$ as large as possible.

We want to maximize $\ell$, the number of large cups.

Every small cup uses some coffee and time that could instead be used for large cups. If we reduce $s$, we free up resources that can only help (or at least not hurt) our ability to make more large cups.

So, to maximize $\ell$, it is best to use the smallest possible value of $s$, which is $s=0$ (no small cups). Substitute $s=0$ into both inequalities to get constraints involving only $\ell$:

• From coffee:

• From time:

Now we just need the largest $\ell$ that satisfies both of these inequalities.

Solve each inequality for $\ell$:

• From the coffee constraint:

• From the time constraint:

To satisfy both constraints, $\ell$ must be less than or equal to both upper bounds, so it cannot exceed the smaller one. The smaller upper bound is 60, so the greatest possible integer value of $\ell$ is 60.