Linear programming word problems for algebra students, including cabinet storage optimization and snack bar profit maximization, with graphing grids provided.

Let's solve both linear programming word problems step by step.

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Problem 1: Filing Cabinets

Objective: Maximize storage volume.

Given:
- Cabinet X:
- Cost: $10 per unit
- Floor space: 6 sq ft
- Storage: 8 cubic feet
- Cabinet Y:
- Cost: $20 per unit
- Floor space: 8 sq ft
- Storage: 12 cubic feet
- Budget: ≤ $140
- Floor space: ≤ 72 sq ft
- Want to maximize storage volume

Let:
- $ x $ = number of Cabinet X
- $ y $ = number of Cabinet Y

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#### Step 1: Define the Objective Function

We want to maximize storage volume:

$$
\text{Maximize } Z = 8x + 12y
$$

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#### Step 2: Constraints

1. Budget constraint:
$ 10x + 20y \leq 140 $

2. Floor space constraint:
$ 6x + 8y \leq 72 $

3. Non-negativity constraints:
$ x \geq 0, y \geq 0 $

Also, since we're dealing with whole cabinets, $ x, y $ are integers (but we'll solve as continuous first and then check).

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#### Step 3: Simplify Constraints

1. $ 10x + 20y \leq 140 $ → divide by 10:
$ x + 2y \leq 14 $

2. $ 6x + 8y \leq 72 $ → divide by 2:
$ 3x + 4y \leq 36 $

So the system is:

$$
\begin{cases}
x + 2y \leq 14 \\
3x + 4y \leq 36 \\
x \geq 0, y \geq 0
\end{cases}
$$

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#### Step 4: Find Corner Points

Graph the feasible region and find intersection points.

Intersection of $ x + 2y = 14 $ and $ 3x + 4y = 36 $:

Solve:
From first equation: $ x = 14 - 2y $

Substitute into second:
$$
3(14 - 2y) + 4y = 36 \\
42 - 6y + 4y = 36 \\
42 - 2y = 36 \\
2y = 6 \Rightarrow y = 3 \\
x = 14 - 2(3) = 8
$$

So point: $ (8, 3) $

Now find other corner points:

1. Intersection of $ x + 2y = 14 $ with $ x = 0 $:
$ 0 + 2y = 14 \Rightarrow y = 7 $ → $ (0, 7) $

2. Intersection of $ 3x + 4y = 36 $ with $ x = 0 $:
$ 4y = 36 \Rightarrow y = 9 $ → but check budget: $ x + 2y = 0 + 18 = 18 > 14 $ → not feasible

Wait — so $ (0,9) $ violates budget. So we need only feasible corners.

Instead, find where each line hits axes within feasible region.

List all feasible corner points:

- $ (0,0) $
- $ (0,7) $: from $ x=0 $, $ x+2y=14 $ → $ y=7 $; check floor space: $ 6(0)+8(7)=56 \leq 72 $ → OK
- $ (8,3) $: intersection of two constraints
- Where $ 3x + 4y = 36 $ meets $ x $-axis? $ y=0 $: $ 3x = 36 \Rightarrow x = 12 $; check budget: $ x + 2y = 12 + 0 = 12 \leq 14 $ → OK → $ (12,0) $

But check if $ (12,0) $ satisfies floor space: $ 6(12) + 8(0) = 72 \leq 72 $ → OK

Now check if $ (12,0) $ is feasible: yes.

Now check if any other points?

Let’s list all corner points of feasible region:

1. $ (0,0) $
2. $ (0,7) $: from budget limit when $ x=0 $
3. $ (8,3) $: intersection
4. $ (12,0) $: from floor space limit when $ y=0 $

Wait — let's verify if $ (12,0) $ satisfies budget: $ 10(12) + 20(0) = 120 \leq 140 $ → yes

And floor space: $ 6(12) = 72 $ → OK

Now check if $ (12,0) $ is on the boundary of both constraints?

- Budget: $ 10x + 20y = 120 < 140 $ → not tight
- Floor space: $ 6x = 72 $ → tight

So it's a vertex.

Now check if $ (0,7) $ is valid:

- Floor space: $ 8(7) = 56 \leq 72 $ → OK
- Budget: $ 20(7) = 140 $ → OK

So yes.

Now evaluate objective function $ Z = 8x + 12y $ at each corner:

1. $ (0,0) $: $ Z = 0 $
2. $ (0,7) $: $ Z = 8(0) + 12(7) = 84 $
3. $ (8,3) $: $ Z = 8(8) + 12(3) = 64 + 36 = 100 $
4. $ (12,0) $: $ Z = 8(12) + 12(0) = 96 $

Maximum is 100 at $ (8,3) $

So buy 8 Cabinet X and 3 Cabinet Y

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✔ Answer for Problem 1:
> Buy 8 Cabinet X and 3 Cabinet Y to maximize storage volume (100 cubic feet).

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Problem 2: Snack Bar – Hamburgers and Hot Dogs

Goal: Maximize profit

Let:
- $ h $ = number of hamburgers
- $ d $ = number of hot dogs

Profit:
- Hamburger: 33 cents → $ 0.33h $
- Hot dog: 21 cents → $ 0.21d $

So objective:
$$
\text{Maximize } P = 0.33h + 0.21d
$$

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#### Constraints:

1. At least 10 hamburgers: $ h \geq 10 $
2. No more than 40 hamburgers: $ h \leq 40 $
3. At least 30 hot dogs: $ d \geq 30 $
4. No more than 70 hot dogs: $ d \leq 70 $
5. Total items ≤ 90: $ h + d \leq 90 $
6. $ h \geq 0, d \geq 0 $ (redundant)

So constraints:

$$
\begin{cases}
h \geq 10 \\
h \leq 40 \\
d \geq 30 \\
d \leq 70 \\
h + d \leq 90 \\
h, d \in \mathbb{Z}^+
\end{cases}
$$

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#### Find Feasible Region Corners

We look for vertices of the feasible region.

The feasible region is bounded by these inequalities.

List corner points:

1. Intersection of $ h = 10 $ and $ d = 30 $: $ (10,30) $
2. $ h = 10 $ and $ h + d = 90 $: $ d = 80 $, but $ d \leq 70 $ → not feasible
- So instead: $ d = 70 $, $ h = 10 $: $ (10,70) $ — check total: $ 10 + 70 = 80 \leq 90 $ → OK
3. $ h = 10 $, $ d = 70 $: $ (10,70) $
4. $ d = 70 $, $ h + d = 90 $: $ h = 20 $ → $ (20,70) $
5. $ h = 40 $, $ d = 30 $: $ (40,30) $
6. $ h = 40 $, $ h + d = 90 $: $ d = 50 $ → $ (40,50) $
7. $ d = 30 $, $ h + d = 90 $: $ h = 60 $ → but $ h \leq 40 $ → not feasible
8. So instead: $ h = 40 $, $ d = 50 $ → already listed
9. $ h = 40 $, $ d = 30 $: $ (40,30) $
10. $ h = 40 $, $ d = 50 $: $ (40,50) $
11. $ h = 20 $, $ d = 70 $: $ (20,70) $
12. $ h = 40 $, $ d = 50 $: $ (40,50) $
13. $ h = 40 $, $ d = 30 $: $ (40,30) $

Now list all feasible corner points:

- $ (10,30) $: min h, min d
- $ (10,70) $: min h, max d
- $ (20,70) $: from $ d = 70 $, $ h + d = 90 $ → $ h = 20 $
- $ (40,50) $: $ h = 40 $, $ h + d = 90 $ → $ d = 50 $
- $ (40,30) $: max h, min d

Is $ (40,30) $ feasible? Yes: $ h = 40 \leq 40 $, $ d = 30 \geq 30 $, $ h + d = 70 \leq 90 $

Are there others?

What about $ (40,70) $? $ h + d = 110 > 90 $ → no

$ (10,30) $, $ (10,70) $, $ (20,70) $, $ (40,50) $, $ (40,30) $

Now check if $ (40,30) $ connects directly to $ (40,50) $? Yes, along $ h = 40 $

And $ (20,70) $ to $ (40,50) $? Along $ h + d = 90 $

So the vertices are:

1. $ (10,30) $
2. $ (10,70) $
3. $ (20,70) $
4. $ (40,50) $
5. $ (40,30) $

Now evaluate profit $ P = 0.33h + 0.21d $ at each:

1. $ (10,30) $: $ 0.33(10) + 0.21(30) = 3.3 + 6.3 = 9.60 $
2. $ (10,70) $: $ 0.33(10) + 0.21(70) = 3.3 + 14.7 = 18.00 $
3. $ (20,70) $: $ 0.33(20) + 0.21(70) = 6.6 + 14.7 = 21.30 $
4. $ (40,50) $: $ 0.33(40) + 0.21(50) = 13.2 + 10.5 = 23.70 $
5. $ (40,30) $: $ 0.33(40) + 0.21(30) = 13.2 + 6.3 = 19.50 $

Maximum is $23.70 at $ (40,50) $

So sell 40 hamburgers and 50 hot dogs

Check constraints:
- $ h = 40 $: between 10 and 40 → OK
- $ d = 50 $: between 30 and 70 → OK
- Total: $ 40 + 50 = 90 \leq 90 $ → OK

Yes, feasible.

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✔ Answer for Problem 2:
> Sell 40 hamburgers and 50 hot dogs to maximize profit ($23.70).

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✔ Final Answers:

1) Buy 8 Cabinet X and 3 Cabinet Y to maximize storage volume.

2) Sell 40 hamburgers and 50 hot dogs to maximize profit.

Let me know if you'd like the graphs sketched!