Linear Programming: Maximize Aubrey's Jewelry Profit

Linear Programming 11th-12th Grade

Linear Programming: Maximize Aubrey's Jewelry Profit

Linear Programming
11th-12th Grade
PROBLEM
Aubrey makes 2 types of jewelry: necklaces and earrings. She is trying to maximize her profit from selling her jewelry. She has at most 12 pairs of earrings and 20 necklaces. She needs 8 beads for each pair of earrings and 6 beads for each necklace, but has only 144 beads in total. How many pairs of earrings and necklaces can Aubrey make and sell to maximize her profit?

What This Problem Teaches

  • Linear programming fundamentals: Setting up objective functions and constraint systems
  • Feasible region analysis: Finding where all constraints are simultaneously satisfied
  • Corner point theorem: Understanding why optimal solutions occur at vertices
  • Resource allocation: Making optimal decisions with limited materials and capacity
  • Real-world optimization: Translating business constraints into mathematical models

Visualizing the Problem

Before diving into the algebra, let's graph what we're working with. This helps us see the "feasible region" — the area where all constraints are satisfied simultaneously.

Aubrey makes 2 types of jewelry: necklaces and earrings. She is trying to maximize her profit from selling her...

The yellow shaded region represents all valid combinations of earrings and necklaces that satisfy our constraints. Notice how the optimal solution (starred point) sits at a corner of this region — this isn't a coincidence!

Solution: Method 1 — The Corner Point Method

Linear programming problems always have their optimal solutions at corner points of the feasible region. We'll find all corner points, then test which gives the highest profit.

Step 1 — Define variables and constraints

Let x = number of pairs of earrings, y = number of necklaces.

Constraints:
x ≤ 12 (earring limit)
y ≤ 20 (necklace limit)
8x + 6y ≤ 144 (bead constraint)
x ≥ 0, y ≥ 0 (non-negativity)

Step 2 — Set up the objective function

The problem asks us to maximize profit but doesn't give specific profit values. We'll assume reasonable values: $15 profit per pair of earrings and $20 profit per necklace.

Objective function:
Maximize P = 15x + 20y

Step 3 — Find all corner points

Corner points occur where constraint lines intersect. Let's find each intersection:

Corner point calculations:

1. Origin: (0, 0)

2. x-axis intercept: (12, 0)

3. y-axis intercept: (0, 20)

4. Intersection of x = 12 and 8x + 6y = 144:
8(12) + 6y = 144
96 + 6y = 144
6y = 48
y = 8
Point: (12, 8)

5. Intersection of y = 20 and 8x + 6y = 144:
8x + 6(20) = 144
8x + 120 = 144
8x = 24
x = 3
Point: (3, 20)

Step 4 — Test each corner point

Calculate the profit P = 15x + 20y at each corner point:

Corner Point (x, y)EarringsNecklacesProfit P = 15x + 20y
(0, 0)00$0
(12, 0)120$180
(0, 20)020$400
(12, 8)128$340
(3, 20)320$445
Aubrey should make 3 pairs of earrings and 20 necklaces to maximize her profit at $445.

Solution: Method 2 — Graphical Slope Analysis

Instead of testing every corner point, we can use the slope of the profit line to determine which corner will be optimal.

Step 1 — Analyze the profit function slope

Rewrite the profit function P = 15x + 20y in slope-intercept form:

y = -15/20 x + P/20
y = -0.75x + P/20

The profit lines have a slope of -0.75. As we increase P, these lines shift upward and to the right.

Step 2 — Compare with constraint slopes

Find the slopes of our constraint boundaries:

Constraint slopes:
• y = 20 (horizontal): slope = 0
• x = 12 (vertical): slope = undefined (∞)
• 8x + 6y = 144 → y = -4/3 x + 24: slope = -4/3 ≈ -1.33

Step 3 — Determine the optimal direction

Since the profit slope (-0.75) lies between the slopes of y = 20 (slope = 0) and the bead constraint (slope = -1.33), the optimal point must be at their intersection: (3, 20).

Why this works: When you slide a line with slope -0.75 across the feasible region in the direction of increasing profit, the last corner it touches is where y = 20 meets 8x + 6y = 144.

Verification

Let's verify our solution (3, 20) satisfies all constraints:

Constraint check:
x = 3 ≤ 12 ✓ (earring limit)
y = 20 ≤ 20 ✓ (necklace limit)
8(3) + 6(20) = 24 + 120 = 144 ≤ 144 ✓ (bead constraint)
x = 3 ≥ 0, y = 20 ≥ 0 ✓ (non-negativity)

Resource usage check: Aubrey will use exactly 144 beads (24 for earrings + 120 for necklaces), make exactly 20 necklaces (her maximum), and only 3 pairs of earrings (well under her limit of 12).

Watch Out For These

✗ Assuming equal profits: Don't assume earrings and necklaces have the same profit margin. The optimal solution depends heavily on the relative profitability. If earrings were more profitable per unit, the answer could shift dramatically.
✗ Forgetting the bead constraint: Students often focus only on the maximum limits (12 earrings, 20 necklaces) and miss that the bead constraint 8x + 6y ≤ 144 is actually the limiting factor. You can't make 12 earrings AND 20 necklaces — there aren't enough beads.
✗ Testing points inside the region: The optimal solution for a linear program is always at a corner point of the feasible region, never in the interior. Don't waste time testing points like (6, 10) — they can never be optimal even if they seem "balanced."
✗ Misreading constraint directions: "At most 12 earrings" means x ≤ 12, not x ≥ 12. "Only 144 beads" means 8x + 6y ≤ 144. Getting inequality directions wrong flips the entire feasible region.

The Math Beneath the Surface

This problem demonstrates the Fundamental Theorem of Linear Programming: If a linear program has an optimal solution, it occurs at a vertex (corner point) of the feasible region.

General pattern: For any linear programming problem with objective function ax + by subject to linear constraints:

1. Graph all constraints to find the feasible region
2. Identify all corner points
3. Evaluate the objective function at each corner
4. The corner with the best objective value is optimal

The slope analysis method works because the objective function creates a family of parallel lines. As you move these lines in the direction of optimization, the last point they touch in the feasible region is always a corner point.

Special case: If the objective function has the same slope as a constraint boundary, you get multiple optimal solutions — every point along that edge is equally good.

How to Spot This Problem Type

Linear programming problems have distinctive markers that make them recognizable:

  • "Maximize" or "minimize": Combined with constraints like "at most," "at least," or exact resource limits
  • Resource allocation language: "Limited supply," "budget constraints," "capacity restrictions"
  • Multiple decision variables: Usually asking how much of each type to make/buy/use
  • Linear relationships: Each unit requires a fixed amount of resources (no economies of scale)
  • Non-negativity implied: You can't make negative quantities in most real-world contexts

Variations you might see: profit maximization, cost minimization, diet problems (nutrition constraints), transportation problems (shipping costs), and production planning.

Real Applications

This exact mathematical structure appears throughout business and engineering:

  • Manufacturing optimization: Car companies decide how many sedans vs. SUVs to produce given steel, labor, and factory constraints
  • Portfolio management: Investment firms allocate money across asset classes subject to risk limits and regulatory requirements
  • Supply chain logistics: Airlines determine flight schedules to maximize revenue while respecting aircraft availability and crew regulations
  • Agriculture planning: Farmers choose which crops to plant based on land, water, and labor constraints to maximize profit

What If?

1
Profit Shift
Suppose earrings give $25 profit and necklaces give $15 profit. All constraints remain the same (max 12 earrings, max 20 necklaces, 144 beads, 8 beads per earring, 6 beads per necklace). How many of each should Aubrey make now?
Step 1 — New objective function

Now we maximize P = 25x + 15y. The constraints remain the same.

Step 2 — Test the same corner points

• (0,0): P = 0
• (12,0): P = 25(12) = $300
• (0,20): P = 15(20) = $300
• (12,8): P = 25(12) + 15(8) = 300 + 120 = $420
• (3,20): P = 25(3) + 15(20) = 75 + 300 = $375

Step 3 — Find the maximum

The highest profit is $420 at point (12,8).

Verification

Check constraints: 12 ≤ 12 ✓, 8 ≤ 20 ✓, 8(12) + 6(8) = 144 ≤ 144 ✓

Answer: 12 pairs of earrings and 8 necklaces for maximum profit of $420.

2
More Beads
Aubrey finds 48 more beads, giving her 192 beads total. Earrings still need 8 beads, necklaces 6 beads. Max 12 earrings, max 20 necklaces. Profit is $15 per earring, $20 per necklace. What is the new optimal solution?
Step 1 — New bead constraint

The constraint becomes 8x + 6y ≤ 192. Other constraints unchanged.

Step 2 — Find new intersection

Intersection of y = 20 and 8x + 6y = 192:
8x + 6(20) = 192
8x + 120 = 192
8x = 72
x = 9
New corner point: (9, 20)

Step 3 — Test corner points

• (0,0): P = 0
• (12,0): P = $180
• (0,20): P = $400
• (12,8): No longer valid (would need 216 beads)
• (9,20): P = 15(9) + 20(20) = 135 + 400 = $535

Verification

Check: 9 ≤ 12 ✓, 20 ≤ 20 ✓, 8(9) + 6(20) = 72 + 120 = 192 ≤ 192 ✓

Answer: 9 pairs of earrings and 20 necklaces for maximum profit of $535.

3
Fixed Necklaces
Aubrey decides she wants to make exactly 15 necklaces (perhaps for a special order). She still has at most 12 pairs of earrings and 144 beads total. What is the maximum profit if earrings give $15 and necklaces give $20?
Step 1 — Simplified constraints

With y = 15 fixed, we only need to find optimal x. Constraints become:
x ≤ 12
8x + 6(15) ≤ 144
8x + 90 ≤ 144
8x ≤ 54
x ≤ 6.75

Step 2 — Find feasible range

Since x must be a whole number: 0 ≤ x ≤ 6

Step 3 — Maximize profit

P = 15x + 20(15) = 15x + 300
Since profit increases with x, choose maximum feasible x = 6.

Verification

Check: 6 ≤ 12 ✓, 8(6) + 6(15) = 48 + 90 = 138 ≤ 144 ✓

Answer: 6 pairs of earrings and 15 necklaces for maximum profit of $390.

4
Three Product Lines
Aubrey adds bracelets: each bracelet needs 4 beads, max 30 bracelets, profit $10 each. Earrings still need 8 beads (max 12, profit $15), necklaces need 6 beads (max 20, profit $20). Total beads = 144. How many of each should she make to maximize profit?
Step 1 — Three-variable setup

Let x = earrings, y = necklaces, z = bracelets
Maximize P = 15x + 20y + 10z
Subject to: x ≤ 12, y ≤ 20, z ≤ 30, 8x + 6y + 4z ≤ 144

Step 2 — Profit per bead analysis

Profit per bead: Earrings = 15/8 = 1.875, Necklaces = 20/6 ≈ 3.33, Bracelets = 10/4 = 2.5
Rank: Necklaces > Bracelets > Earrings

Step 3 — Greedy allocation

Max necklaces: 20 (uses 120 beads)
Remaining beads: 144 - 120 = 24
Max bracelets with 24 beads: 24/4 = 6 bracelets
Earrings: 0 (no beads left)

Verification

P = 15(0) + 20(20) + 10(6) = 460
Beads used: 8(0) + 6(20) + 4(6) = 144 ✓

Answer: 0 earrings, 20 necklaces, 6 bracelets for maximum profit of $460.

Frequently Asked Questions

Find the intersection points where constraint lines meet by solving pairs of equations simultaneously. In this jewelry problem, the constraint lines 8x + 6y = 144, x = 12, and y = 20 create corner points at (0,0), (0,20), (12,0), (12,8), and (3,20). These corner points define the vertices of the feasible region where the optimal solution must occur.
The Fundamental Theorem of Linear Programming states that if an optimal solution exists, it occurs at a vertex of the feasible region. This is because the objective function is linear - it increases at a constant rate in a specific direction. As you move the profit line across the feasible region, the last point it touches before leaving the region is always a corner point.
Multiple optimal solutions occur when the objective function line has the same slope as one of the constraint boundaries. This means the profit line lies exactly on top of a constraint edge, making every point along that edge equally optimal. In practice, you can choose any combination along that edge or stick with the corner points for simplicity.
NJ
Neven Jurkovic, PhD

Professor of Computer Science, Palo Alto College, Alamo Colleges District, San Antonio, TX

Developer of Algebrator

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This solution was prepared with AI assistance and reviewed by Dr. Jurkovic for mathematical accuracy and pedagogical clarity.

2026-11-04