Example - bicycle factories: A math problem involving two factories producing 3-speed and 10-speed bicycles with cost and production constraints.

Let's solve this linear programming problem step by step.

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Problem Summary:

We are given:

- Factory A produces:
- 16 three-speed bikes per day
- 20 ten-speed bikes per day
- Cost to operate: $100/day

- Factory B produces:
- 12 three-speed bikes per day
- 20 ten-speed bikes per day
- Cost to operate: $80/day

An order requires:
- 96 three-speed bikes
- 140 ten-speed bikes

We need to find:
- How many days each factory should operate (x = days for A, y = days for B)
- To minimize total cost, while meeting the order.

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Step 1: Define Variables

Let:
- $ x $ = number of days Factory A operates
- $ y $ = number of days Factory B operates

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Step 2: Objective Function (Minimize Cost)

Cost = $100x + 80y$

We want to minimize:
$$
\text{Cost} = 100x + 80y
$$

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Step 3: Constraints (Production Requirements)

#### Three-speed bikes:
- Factory A: 16 per day → $16x$
- Factory B: 12 per day → $12y$
- Total needed: 96
$$
16x + 12y \geq 96
$$

#### Ten-speed bikes:
- Factory A: 20 per day → $20x$
- Factory B: 20 per day → $20y$
- Total needed: 140
$$
20x + 20y \geq 140
$$

Also, non-negativity:
$$
x \geq 0, \quad y \geq 0
$$

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Step 4: Simplify Constraints

1. $16x + 12y \geq 96$
Divide entire inequality by 4:
$$
4x + 3y \geq 24 \quad \text{(Constraint 1)}
$$

2. $20x + 20y \geq 140$
Divide by 20:
$$
x + y \geq 7 \quad \text{(Constraint 2)}
$$

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Step 5: Graphical or Algebraic Solution

We'll solve this using the corner point method.

We need to find the feasible region defined by:
- $4x + 3y \geq 24$
- $x + y \geq 7$
- $x \geq 0$, $y \geq 0$

Then evaluate the cost function at the corner points.

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Step 6: Find Intersection Points (Corner Points)

We find where the constraint lines intersect.

#### Line 1: $4x + 3y = 24$
#### Line 2: $x + y = 7$

Solve simultaneously:

From equation 2: $y = 7 - x$

Substitute into equation 1:
$$
4x + 3(7 - x) = 24 \\
4x + 21 - 3x = 24 \\
x + 21 = 24 \\
x = 3
$$
Then $y = 7 - 3 = 4$

So intersection point: $(3, 4)$

Now find other boundary points (intercepts):

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#### Constraint 1: $4x + 3y = 24$

- When $x = 0$: $3y = 24 \Rightarrow y = 8$ → Point: $(0, 8)$
- When $y = 0$: $4x = 24 \Rightarrow x = 6$ → Point: $(6, 0)$

But we must also satisfy $x + y \geq 7$. So check if these points are feasible.

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#### Constraint 2: $x + y = 7$

- When $x = 0$: $y = 7$ → $(0, 7)$
- When $y = 0$: $x = 7$ → $(7, 0)$

Now, find feasible corner points of the region defined by both inequalities.

The feasible region is bounded by:
- $4x + 3y \geq 24$
- $x + y \geq 7$
- $x \geq 0$, $y \geq 0$

We look for vertices of this region.

Possible corner points:
1. Intersection of $4x + 3y = 24$ and $x + y = 7$: $(3, 4)$ ✔
2. Intersection of $4x + 3y = 24$ with $x = 0$: $(0, 8)$
- Check if satisfies $x + y \geq 7$: $0 + 8 = 8 \geq 7$ → Yes ✔
3. Intersection of $x + y = 7$ with $y = 0$: $(7, 0)$
- Check $4x + 3y = 4(7) = 28 \geq 24$ → Yes ✔
4. Intersection of $x + y = 7$ with $x = 0$: $(0, 7)$
- Check $4(0) + 3(7) = 21 < 24$ → Not feasible ✘
5. Intersection of $4x + 3y = 24$ with $y = 0$: $(6, 0)$
- Check $x + y = 6 < 7$ → Not feasible ✘

So the feasible corner points are:
- $(3, 4)$ — intersection of two constraints
- $(0, 8)$ — from $4x + 3y = 24$, $x=0$
- $(7, 0)$ — from $x + y = 7$, $y=0$

Wait — let’s double-check feasibility:

- At $(0,8)$: $x + y = 8 \geq 7$ → OK
- At $(7,0)$: $4x + 3y = 28 \geq 24$ → OK

But is there a point where $x + y = 7$ intersects $4x + 3y = 24$? We already found that: $(3,4)$

Now, what about the point where $x + y = 7$ meets $y$-axis? $(0,7)$: $4(0)+3(7)=21<24$ → violates first constraint → not feasible

Similarly, $(6,0)$: $x+y=6<7$ → not feasible

So the three corner points of the feasible region are:
1. $(3, 4)$
2. $(0, 8)$
3. $(7, 0)$

Let’s evaluate cost at each.

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Step 7: Evaluate Cost at Each Corner Point

Cost = $100x + 80y$

1. At $(3, 4)$:
$$
100(3) + 80(4) = 300 + 320 = \$620
$$

2. At $(0, 8)$:
$$
100(0) + 80(8) = 0 + 640 = \$640
$$

3. At $(7, 0)$:
$$
100(7) + 80(0) = 700 + 0 = \$700
$$

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Step 8: Minimum Cost

Minimum cost is \$620 at $(x, y) = (3, 4)$

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Step 9: Verify Production

Check if $(3, 4)$ meets demand:

- Three-speed bikes:
- A: $16 \times 3 = 48$
- B: $12 \times 4 = 48$
- Total: $48 + 48 = 96$ ✔

- Ten-speed bikes:
- A: $20 \times 3 = 60$
- B: $20 \times 4 = 80$
- Total: $60 + 80 = 140$ ✔

Perfect!

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✔ Final Answer:

- Factory A should operate for 3 days
- Factory B should operate for 4 days
- Minimum cost = \$620

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📌 Summary:

| Factory | Days | Cost |
|--------|------|------|
| A | 3 | $300 |
| B | 4 | $320 |
| Total | 7 | $620 |

This is the minimum cost solution that fills the order.