Solving Linear Programming Problems Graphically (movie 2.3A) - Free Printable

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 to meet the order at minimum cost.
- What is the minimum cost?

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

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Step 1: Formulate the Objective Function

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

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Step 2: Write the Constraints

From production requirements:

#### Three-speed bikes:
Factory A produces 16 per day, Factory B produces 12 per day.
$$
16x + 12y \geq 96
$$

#### Ten-speed bikes:
Both produce 20 per day.
$$
20x + 20y \geq 140
$$

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

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Step 3: Simplify the Constraints

1. $ 16x + 12y \geq 96 $ → divide 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 4: Graphical Solution or Corner Point Method

We'll find the feasible region and evaluate the cost at corner points.

Let’s find the intersection points of the constraints.

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#### Find Intersection of Two Lines:

Solve:
1. $ 4x + 3y = 24 $
2. $ x + y = 7 $

From (2): $ y = 7 - x $

Substitute into (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) $

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Now, find where each constraint intersects axes.

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

- If $ x = 0 $: $ 3y = 24 $ → $ y = 8 $
- If $ y = 0 $: $ 4x = 24 $ → $ x = 6 $

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

- If $ x = 0 $: $ y = 7 $
- If $ y = 0 $: $ x = 7 $

But we must consider feasible region: both inequalities are ≥, so we want area above both lines.

The feasible region is bounded by the two lines and the axes, but only where both constraints are satisfied.

Let’s find the corner points of the feasible region.

They occur at:
1. Intersection of $ 4x + 3y = 24 $ and $ x + y = 7 $ → $ (3, 4) $
2. Where $ 4x + 3y = 24 $ intersects $ y = 0 $ → $ x = 6 $, $ y = 0 $
- Check if it satisfies $ x + y \geq 7 $: $ 6 + 0 = 6 < 7 $ → Not feasible
3. Where $ x + y = 7 $ intersects $ x = 0 $: $ x = 0, y = 7 $
- Check $ 4(0) + 3(7) = 21 \geq 24 $? No → 21 < 24 → Not feasible
4. Try other intersections:

Let’s find where $ 4x + 3y = 24 $ intersects $ x = 0 $: $ y = 8 $
Check $ x + y = 0 + 8 = 8 \geq 7 $ → yes → feasible → point $ (0, 8) $

Where $ x + y = 7 $ intersects $ y = 0 $: $ x = 7 $
Check $ 4(7) + 3(0) = 28 \geq 24 $ → yes → feasible → point $ (7, 0) $

Now check if $ (0, 8) $ and $ (7, 0) $ are on boundary.

So possible corner points:
- $ (0, 8) $
- $ (3, 4) $
- $ (7, 0) $

Wait — is $ (7, 0) $ satisfying both constraints?

- $ 4(7) + 3(0) = 28 \geq 24 $ ✔
- $ 7 + 0 = 7 \geq 7 $ ✔ → Yes

Is $ (0, 8) $ valid?
- $ 4(0) + 3(8) = 24 \geq 24 $ ✔
- $ 0 + 8 = 8 \geq 7 $ ✔ → Yes

And $ (3, 4) $: already checked ✔

Are there any other corners?

Let’s sketch mentally:

- The feasible region is a polygon with vertices:
- $ (0, 8) $: from constraint 1 intercept
- $ (3, 4) $: intersection of two lines
- $ (7, 0) $: from constraint 2 intercept

Yes — these are the three corner points.

Let’s evaluate cost at each.

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Step 5: Evaluate Objective Function at Each Corner Point

1. At $ (0, 8) $:
$$
\text{Cost} = 100(0) + 80(8) = 0 + 640 = \$640
$$

2. At $ (3, 4) $:
$$
\text{Cost} = 100(3) + 80(4) = 300 + 320 = \$620
$$

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

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Step 6: Find Minimum Cost

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

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Step 7: Verify Production

Check if this meets the order:

- 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 is \$620

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🟩 Answer:

> Factory A: 3 days
> Factory B: 4 days
> Minimum cost: $620