Write an inequality for the time limit (using 4 minutes for standard and 10 minutes for deluxe), one for the total number of backpacks, and one for the "at least 150%" requirement relating $s$ and $d$.

Rearrange each inequality so it tells you either an upper bound or a lower bound for $s$ in terms of $d$. Then think: for a given $d$, what range of $s$ is allowed?

To make $d$ as large as possible, you want $s$ to be as small as possible (because both use machine time). Use the lower bound on $s$ that is active when $d$ is large, plug it into the time inequality, and solve for $d$. Then remember $d$ must be an integer and check nearby values.

In Desmos, let $x$ represent deluxe backpacks and $y$ represent standard backpacks. Enter these three equations (the boundaries of the inequalities):

• y = (840 - 10x)/4 (from $4y + 10x = 840$)
• y = 120 - x (from $y + x = 120$)
• y = 1.5x (from $y = 1.5x$).

Turn each boundary into an inequality in Desmos:

• y <= (840 - 10x)/4 (time limit)
• y >= 120 - x (total backpacks)
• y >= 1.5x (150% condition) The overlapping shaded region represents all $(x,y)$ pairs that satisfy every condition.

Look toward the right side of the feasible region (where $x$ is largest). Identify where the lines y = (840 - 10x)/4 and y = 1.5x intersect inside the constraints; use Desmos to see that $x$-coordinate and then consider the greatest integer $x$ less than or equal to that value that still lies in the shaded region.

Let

• $s$ = number of standard backpacks
• $d$ = number of deluxe backpacks

Translate the conditions:

1. Stitching time: Each standard takes 4 minutes, each deluxe 10 minutes, and total time is at most 840 minutes:

2. Total backpacks: At least 120 backpacks in total:

3. Standard vs. deluxe: Standards must be at least 150% of deluxe, i.e., $s \ge 1.5d$ (or $2s \ge 3d$).

Rearrange each inequality to describe what values of $s$ are allowed for a given $d$.

1. From the time limit:

This is an upper bound on $s$.

2. From total backpacks:

3. From the 150% condition:

So for any given $d$, $s$ must satisfy:

• $s \ge 120 - d$
• $s \ge 1.5d$
• $s \le \dfrac{840 - 10d}{4}$.

The lower bound on $s$ is the larger of $120 - d$ and $1.5d$.

Find when those two are equal:

• For $d < 48$, we have $120 - d > 1.5d$, so the total-backpack condition sets the minimum $s$.
• For $d > 48$, we have $1.5d > 120 - d$, so the 150% condition $s \ge 1.5d$ sets the minimum $s$.

To maximize $d$, we will be in the region $d \ge 48$, where $s$ is effectively forced to be at least $1.5d$.

When $d$ is as large as possible, we want:

• $s$ as small as allowed (so we can fit more deluxe backpacks in the time limit), so take $s = 1.5d$.
• The time limit $4s + 10d \le 840$ will be just used up (approximately equal), so set $4s + 10d = 840$.

Substitute $s = 1.5d$ into the time equation:

So based on time and ratio, $d$ cannot exceed $52.5$.

Because $d$ must be an integer, test the integers near $52.5$ in the next step to confirm which value(s) are feasible.

Check all conditions for $d = 52$.

1. From $s \ge 1.5d$:

Choose $s = 78$ (the smallest $s$ allowed by the ratio).

2. Time check:

3. Total backpacks:

So $d = 52$ works.

Now test $d = 53$ with the smallest allowed $s$.

• $s \ge 1.5 \cdot 53 = 79.5$, so the smallest integer $s$ is $80$.
• Time: $4(80) + 10(53) = 320 + 530 = 850 > 840$, which breaks the time limit.

Therefore, no value $d \ge 53$ is possible, and the maximum number of deluxe backpacks is 52.