Linear programming solution examples - Free Printable

The image you've provided is a graphical representation of a linear programming (LP) problem, showing constraints, the feasible region, and an iso-profit line. Let's analyze and solve this step by step.

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🔍 Step 1: Understand the Problem

We are given:

- Two decision variables:
$ x_a $ and $ x_b $

- Constraints:
1. $ 12x_a + 25x_b = 1800 $ → This is a resource constraint (likely material or budget).
2. $ x_b - 0.4x_a = 0 $ → This is a proportional relationship between $ x_b $ and $ x_a $.
3. Non-negativity: $ x_a \geq 0, x_b \geq 0 $ (implied by the graph).

- Objective function:
- The dashed line labeled $ 3x_a + 5x_b = 180 $ is an iso-profit line.
- This means the objective is to maximize $ Z = 3x_a + 5x_b $.

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📐 Step 2: Identify the Feasible Region

From the graph:

- The feasible region is the area bounded by the constraints and in the first quadrant ($ x_a \geq 0, x_b \geq 0 $).
- The two solid lines represent:
- $ 12x_a + 25x_b = 1800 $
- $ x_b = 0.4x_a $

These two lines intersect at a point, and the feasible region is the intersection of the half-planes defined by these inequalities.

But since both lines are drawn as equalities, we assume the inequalities are:
- $ 12x_a + 25x_b \leq 1800 $ (below the line)
- $ x_b \geq 0.4x_a $ (above the line)

Also, $ x_a \geq 0 $, $ x_b \geq 0 $

So the feasible region is the bounded area where all constraints are satisfied.

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✔ Step 3: Find Corner Points (Vertices)

In linear programming, the optimal solution occurs at a corner point of the feasible region.

Let’s find the intersection points.

#### 🔹 Intersection of $ x_b = 0.4x_a $ and $ 12x_a + 25x_b = 1800 $

Substitute $ x_b = 0.4x_a $ into the first equation:

$$
12x_a + 25(0.4x_a) = 1800 \\
12x_a + 10x_a = 1800 \\
22x_a = 1800 \\
x_a = \frac{1800}{22} = 81.818... \approx 81.82 \\
x_b = 0.4 \times 81.818 = 32.727...
$$

So, one vertex: $ (81.82, 32.73) $

#### 🔹 Intersection of $ x_b = 0.4x_a $ with $ x_a = 0 $

At $ x_a = 0 $, $ x_b = 0 $. So point: $ (0, 0) $

But check if it satisfies $ 12x_a + 25x_b \leq 1800 $? Yes. But is it in feasible region?

Wait — look at the graph: the feasible region starts from a point on the $ x_b $-axis around $ (0, 40) $? Let’s double-check.

Wait! There’s a dashed line from $ (0, 40) $ to $ (60, 0) $, but that's not a constraint. Actually, let’s re-express.

Wait — the constraint $ x_b - 0.4x_a = 0 $ passes through origin and goes up.

But the feasible region is bounded by:

- $ x_b \geq 0.4x_a $ (above the line)
- $ 12x_a + 25x_b \leq 1800 $ (below the line)
- $ x_a \geq 0 $, $ x_b \geq 0 $

So the feasible region is bounded by three lines:

1. $ x_b = 0.4x_a $
2. $ 12x_a + 25x_b = 1800 $
3. $ x_a = 0 $ or $ x_b = 0 $? Not necessarily.

But looking at the graph:

- The feasible region is a triangle bounded by:
- $ x_b = 0.4x_a $
- $ 12x_a + 25x_b = 1800 $
- And possibly the $ x_b $-axis? No — because at $ x_a = 0 $, $ x_b $ must be ≥ 0, but also $ x_b \geq 0.4(0) = 0 $, so okay.

Wait — the only vertices are:

1. Origin: $ (0, 0) $
- But does it satisfy $ x_b \geq 0.4x_a $? Yes.
- Does it satisfy $ 12x_a + 25x_b \leq 1800 $? Yes.
- So it's a corner point.

2. Intersection of $ x_b = 0.4x_a $ and $ 12x_a + 25x_b = 1800 $: $ (81.82, 32.73) $

3. Where $ 12x_a + 25x_b = 1800 $ intersects $ x_b $-axis?
- Set $ x_a = 0 $: $ 25x_b = 1800 $ → $ x_b = 72 $
- Point: $ (0, 72) $
- But does it satisfy $ x_b \geq 0.4x_a $? Yes, since $ x_a = 0 $, $ x_b = 72 \geq 0 $
- So yes, this point is in feasible region.

Wait — but earlier we thought the feasible region was bounded by $ x_b \geq 0.4x_a $, so the lower boundary is $ x_b = 0.4x_a $, and the upper boundary is $ 12x_a + 25x_b = 1800 $

So the feasible region has two vertices:

- Where $ x_b = 0.4x_a $ meets $ 12x_a + 25x_b = 1800 $: $ (81.82, 32.73) $
- Where $ 12x_a + 25x_b = 1800 $ meets $ x_b $-axis: $ (0, 72) $
- And where $ x_b = 0.4x_a $ meets $ x_a = 0 $: $ (0, 0) $

But wait — at $ (0, 0) $, is it feasible?

Yes.

But now, is $ (0, 72) $ feasible? Let's check $ x_b \geq 0.4x_a $: $ 72 \geq 0 $, yes.

But is $ (0, 72) $ connected to $ (81.82, 32.73) $? Only if the constraint $ x_b \geq 0.4x_a $ is active.

Actually, the feasible region is bounded by:

- From $ (0, 0) $ along $ x_b = 0.4x_a $ to $ (81.82, 32.73) $
- Then from $ (81.82, 32.73) $ along $ 12x_a + 25x_b = 1800 $ back to $ (0, 72) $
- Then from $ (0, 72) $ down to $ (0, 0) $? But no — that would violate $ x_b \geq 0.4x_a $ only if $ x_a > 0 $, but at $ x_a = 0 $, it's okay.

Wait — actually, the feasible region is not including $ (0, 72) $ unless $ x_b \geq 0.4x_a $ is satisfied.

But at $ x_a = 0 $, $ x_b \geq 0 $, so $ (0, 72) $ is feasible.

But what about $ (0, 40) $? There's a mark there.

Wait — look at the graph again.

There is a dashed line from $ (0, 72) $? No — actually, the graph shows:

- A line from $ (0, 72) $? Wait, the line $ 12x_a + 25x_b = 1800 $ goes from $ (0, 72) $ to $ (150, 0) $? Let's check:

At $ x_a = 0 $: $ 25x_b = 1800 $ → $ x_b = 72 $

At $ x_b = 0 $: $ 12x_a = 1800 $ → $ x_a = 150 $

So the line goes from $ (0, 72) $ to $ (150, 0) $

But the other line: $ x_b = 0.4x_a $, which goes from $ (0, 0) $ to $ (150, 60) $

Now, the feasible region is where:
- $ x_b \geq 0.4x_a $
- $ 12x_a + 25x_b \leq 1800 $
- $ x_a \geq 0 $, $ x_b \geq 0 $

So the feasible region is bounded by:

1. $ (0, 0) $ — but is this included?
- At $ (0, 0) $: $ x_b = 0 \geq 0.4(0) = 0 $ → OK
- $ 12(0) + 25(0) = 0 \leq 1800 $ → OK
- So yes.

2. Along $ x_b = 0.4x_a $ until it hits $ 12x_a + 25x_b = 1800 $: $ (81.82, 32.73) $

3. Then along $ 12x_a + 25x_b = 1800 $ until it hits $ x_b = 0 $: $ (150, 0) $

But at $ (150, 0) $: $ x_b = 0 $, $ x_a = 150 $, then $ x_b \geq 0.4x_a $? $ 0 \geq 60 $? ✘ NO!

So $ (150, 0) $ is not feasible.

Similarly, $ (0, 72) $: $ x_b = 72 $, $ x_a = 0 $, $ x_b \geq 0.4x_a $? $ 72 \geq 0 $ → YES

So $ (0, 72) $ is feasible.

But can we go from $ (0, 72) $ to $ (81.82, 32.73) $ along $ 12x_a + 25x_b = 1800 $? Yes, but only if $ x_b \geq 0.4x_a $

Let’s check whether $ 12x_a + 25x_b = 1800 $ lies above $ x_b = 0.4x_a $ in that segment.

We already found the intersection at $ (81.82, 32.73) $

Now, for $ x_a < 81.82 $, is $ x_b $ on the line $ 12x_a + 25x_b = 1800 $ greater than $ 0.4x_a $?

Try $ x_a = 0 $: $ x_b = 72 $, $ 0.4x_a = 0 $ → $ 72 > 0 $ → OK

At $ x_a = 60 $: $ 12(60) + 25x_b = 1800 $ → $ 720 + 25x_b = 1800 $ → $ 25x_b = 1080 $ → $ x_b = 43.2 $

$ 0.4x_a = 0.4×60 = 24 $, $ 43.2 > 24 $ → OK

So yes, the entire segment from $ (0, 72) $ to $ (81.82, 32.73) $ is above $ x_b = 0.4x_a $

So the feasible region is a triangle with vertices:

1. $ A = (0, 72) $
2. $ B = (81.82, 32.73) $
3. $ C = (0, 0) $? Wait — no.

Wait — at $ x_a = 0 $, $ x_b $ can be from $ 0 $ to $ 72 $, but $ x_b \geq 0.4x_a = 0 $, so $ x_b \in [0, 72] $

But can we have $ x_a > 0 $ and $ x_b < 0.4x_a $? No — because of the constraint $ x_b \geq 0.4x_a $

So the feasible region is bounded by:

- From $ (0, 0) $ along $ x_b = 0.4x_a $ to $ (81.82, 32.73) $
- Then from $ (81.82, 32.73) $ along $ 12x_a + 25x_b = 1800 $ to $ (0, 72) $
- Then from $ (0, 72) $ down to $ (0, 0) $ along $ x_a = 0 $

But is $ (0, 0) $ connected directly to $ (0, 72) $? Only if $ x_b \geq 0.4x_a $ is satisfied — which it is at $ x_a = 0 $

So the feasible region is a triangle with vertices:

1. $ P_1 = (0, 0) $
2. $ P_2 = (81.82, 32.73) $
3. $ P_3 = (0, 72) $

But wait — is $ (0, 0) $ connected to $ (0, 72) $? Yes, along $ x_a = 0 $

And $ (0, 72) $ to $ (81.82, 32.73) $ along $ 12x_a + 25x_b = 1800 $

And $ (81.82, 32.73) $ to $ (0, 0) $ along $ x_b = 0.4x_a $

So yes, the feasible region is a triangle with those three corners.

But wait — is $ (0, 0) $ really part of the feasible region?

Yes.

But now, let’s evaluate the objective function $ Z = 3x_a + 5x_b $ at each vertex.

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✔ Step 4: Evaluate Objective Function at Each Vertex

#### 🔹 Vertex 1: $ (0, 0) $
$$
Z = 3(0) + 5(0) = 0
$$

#### 🔹 Vertex 2: $ (81.82, 32.73) $
$$
Z = 3(81.82) + 5(32.73) = 245.46 + 163.65 = 409.11
$$

#### 🔹 Vertex 3: $ (0, 72) $
$$
Z = 3(0) + 5(72) = 360
$$

So maximum is at $ (81.82, 32.73) $ with $ Z = 409.11 $

But wait — the iso-profit line shown is $ 3x_a + 5x_b = 180 $, which is much lower than 409.11.

That suggests that the optimal solution is not on that line.

But the graph shows a dashed line $ 3x_a + 5x_b = 180 $, and it's parallel to the direction of increasing profit.

So why is it shown?

Ah — perhaps the iso-profit line is just a reference, and the actual optimal solution is at $ (81.82, 32.73) $

But let’s check: is the optimal value higher than 180? Yes.

So the maximum profit is approximately $ \boxed{409.11} $

But maybe the numbers are meant to be exact.

Let’s compute exactly.

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✔ Exact Solution

We had:

- $ x_b = 0.4x_a = \frac{2}{5}x_a $
- Plug into $ 12x_a + 25x_b = 1800 $:

$$
12x_a + 25\left(\frac{2}{5}x_a\right) = 1800 \\
12x_a + 10x_a = 1800 \\
22x_a = 1800 \\
x_a = \frac{1800}{22} = \frac{900}{11} \approx 81.818 \\
x_b = \frac{2}{5} \cdot \frac{900}{11} = \frac{1800}{55} = \frac{360}{11} \approx 32.727
$$

Now compute $ Z = 3x_a + 5x_b $:

$$
Z = 3\left(\frac{900}{11}\right) + 5\left(\frac{360}{11}\right) = \frac{2700 + 1800}{11} = \frac{4500}{11} \approx 409.09
$$

So optimal value is $ \frac{4500}{11} \approx 409.09 $

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🔍 Final Answer

✔ Optimal Solution:

- $ x_a = \frac{900}{11} \approx 81.82 $
- $ x_b = \frac{360}{11} \approx 32.73 $
- Maximum profit: $ Z = \frac{4500}{11} \approx 409.09 $

This occurs at the intersection of the two constraints:
- $ 12x_a + 25x_b = 1800 $
- $ x_b = 0.4x_a $

The iso-profit line $ 3x_a + 5x_b = 180 $ is just a reference line; the actual optimal solution is much higher.

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🧠 Summary

- Objective: Maximize $ Z = 3x_a + 5x_b $
- Constraints:
- $ 12x_a + 25x_b \leq 1800 $
- $ x_b \geq 0.4x_a $
- $ x_a \geq 0, x_b \geq 0 $
- Feasible region: Triangle with vertices $ (0,0), (0,72), (81.82, 32.73) $
- Optimal solution: At $ (81.82, 32.73) $, $ Z = 409.09 $

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📌 Note on Graph

The dashed line $ 3x_a + 5x_b = 180 $ is not the optimal solution. It's just an example of an iso-profit line. To find the optimal, move it parallel until it touches the last point in the feasible region — which is the intersection point we found.

Thus, the solution is at the intersection of the two constraint lines.

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✔ Final Answer:
$$
\boxed{
x_a = \frac{900}{11},\quad x_b = \frac{360}{11},\quad \text{Max } Z = \frac{4500}{11} \approx 409.09
}
$$