CBSE Class 12 Mathematics Linear Programming Worksheet Set A - Free Printable

Let's solve the given Linear Programming Problem (LPP) graphically and explain each step carefully.

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Problem Statement:

Minimize and Maximize
$$
z = 3x + 9y
$$

Subject to the constraints:
1. $ x + 3y \leq 60 $
2. $ x + y \geq 10 $
3. $ x \leq y $
4. $ x \geq 0, y \geq 0 $ *(Implicit non-negativity constraints in LPP)*

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## 🔍 Step-by-Step Solution:

We will solve this graphically, so we need to:
1. Graph all constraints.
2. Find the feasible region.
3. Identify the corner points (vertices) of the feasible region.
4. Evaluate the objective function $ z = 3x + 9y $ at each corner point.
5. Determine the minimum and maximum values of $ z $.

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✔ Step 1: Graph the Constraints

We'll convert each inequality into an equation and plot the lines. Then determine which side satisfies the inequality.

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#### Constraint 1: $ x + 3y \leq 60 $

- Line: $ x + 3y = 60 $
- Intercepts:
- When $ x = 0 $: $ 3y = 60 $ → $ y = 20 $
- When $ y = 0 $: $ x = 60 $
- So, line passes through $ (0, 20) $ and $ (60, 0) $
- Shade below the line (since it's ≤)

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#### Constraint 2: $ x + y \geq 10 $

- Line: $ x + y = 10 $
- Intercepts:
- $ x = 0 $: $ y = 10 $
- $ y = 0 $: $ x = 10 $
- Passes through $ (0, 10) $ and $ (10, 0) $
- Shade above the line (since ≥)

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#### Constraint 3: $ x \leq y $

- Line: $ x = y $
- This is a diagonal line passing through origin with slope 1
- Shade below the line (since $ x \leq y $), i.e., where $ y \geq x $

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#### Non-negativity: $ x \geq 0, y \geq 0 $
- We are only considering the first quadrant

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✔ Step 2: Plot and Find Feasible Region

Now, let’s find the common region satisfying all constraints.

We will find the intersection points of the boundary lines to get the vertices of the feasible region.

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✔ Step 3: Find Corner Points (Vertices)

We find intersection points of pairs of constraint lines that lie within the feasible region.

#### Intersection A: $ x + 3y = 60 $ and $ x + y = 10 $

Solve:
$$
\begin{cases}
x + 3y = 60 \\
x + y = 10
\end{cases}
$$

Subtract second from first:
$$
(x + 3y) - (x + y) = 60 - 10 \Rightarrow 2y = 50 \Rightarrow y = 25
$$
Then $ x = 10 - y = 10 - 25 = -15 $

→ $ (-15, 25) $ → Not feasible (x < 0)

So, no intersection in first quadrant for these two.

Wait — maybe check other intersections.

Let’s try all possible pairwise intersections and see which are feasible.

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#### Intersection 1: $ x + 3y = 60 $ and $ x = y $

Set $ x = y $ in $ x + 3y = 60 $:
$$
x + 3x = 60 \Rightarrow 4x = 60 \Rightarrow x = 15, y = 15
$$
Point: $ (15, 15) $

Check if it satisfies other constraints:
- $ x + y = 15 + 15 = 30 \geq 10 $ → Yes
- $ x \leq y $ → $ 15 \leq 15 $ → Yes
- $ x \geq 0, y \geq 0 $ → Yes

✔ Valid point: P1 = (15, 15)

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#### Intersection 2: $ x + y = 10 $ and $ x = y $

Set $ x = y $ in $ x + y = 10 $:
$$
x + x = 10 \Rightarrow 2x = 10 \Rightarrow x = 5, y = 5
$$
Point: $ (5, 5) $

Check:
- $ x + 3y = 5 + 15 = 20 \leq 60 $ → Yes
- $ x \leq y $ → Yes
- Non-negative → Yes

✔ Valid point: P2 = (5, 5)

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#### Intersection 3: $ x + 3y = 60 $ and $ x + y = 10 $

We already tried: gives $ (-15, 25) $ → Not feasible (x < 0)

So skip.

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#### Intersection 4: $ x + 3y = 60 $ with y-axis ($ x = 0 $)

When $ x = 0 $: $ 3y = 60 \Rightarrow y = 20 $ → Point: $ (0, 20) $

Check constraints:
- $ x + y = 0 + 20 = 20 \geq 10 $ → Yes
- $ x \leq y $ → $ 0 \leq 20 $ → Yes
- $ x = 0 \geq 0 $ → Yes

✔ Valid point: P3 = (0, 20)

But wait: Is $ (0, 20) $ satisfying $ x \leq y $? Yes. But does it satisfy $ x + y \geq 10 $? Yes.

But is it on the boundary of feasible region?

Also, check if it lies within the region defined by all constraints.

Yes.

But now check if $ (0, 20) $ is connected via constraints.

Let’s check another point.

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#### Intersection 5: $ x + y = 10 $ with x-axis ($ y = 0 $)

$ x + 0 = 10 \Rightarrow x = 10 $ → Point: $ (10, 0) $

Check:
- $ x + 3y = 10 + 0 = 10 \leq 60 $ → Yes
- $ x \leq y $? $ 10 \leq 0 $? No → ✘ Not feasible

So $ (10, 0) $ is not feasible

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#### Intersection 6: $ x = y $ and $ x = 0 $ → $ (0, 0) $

Check constraints:
- $ x + 3y = 0 \leq 60 $ → Yes
- $ x + y = 0 \geq 10 $? No → ✘ Not feasible

So $ (0, 0) $ not feasible

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#### Intersection 7: $ x + 3y = 60 $ and $ y = 0 $

$ x = 60 $, $ y = 0 $ → $ (60, 0) $

Check:
- $ x + y = 60 \geq 10 $? Yes
- $ x \leq y $? $ 60 \leq 0 $? No → ✘ Not feasible

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#### Intersection 8: $ x + y = 10 $ and $ y = 0 $ → (10, 0) → already checked → not feasible

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Now consider where the boundaries intersect and form the feasible region

We have three candidate points:
- $ (15, 15) $: intersection of $ x+3y=60 $ and $ x=y $
- $ (5, 5) $: intersection of $ x+y=10 $ and $ x=y $
- $ (0, 20) $: intersection of $ x+3y=60 $ and $ x=0 $

Is there a fourth vertex?

Let’s check: Does $ x + y = 10 $ intersect $ x + 3y = 60 $ somewhere in feasible region? Earlier we got $ (-15, 25) $, which is invalid.

But what about intersection of $ x + y = 10 $ and $ x = 0 $?

$ x = 0 $, $ y = 10 $ → $ (0, 10) $

Check constraints:
- $ x + 3y = 0 + 30 = 30 \leq 60 $ → Yes
- $ x + y = 10 \geq 10 $ → Yes
- $ x \leq y $? $ 0 \leq 10 $ → Yes
- Non-negative → Yes

✔ So $ (0, 10) $ is feasible

But is it a vertex?

Let’s verify: It lies on $ x = 0 $ and $ x + y = 10 $

So yes — it's a corner point.

Similarly, $ (0, 20) $ is on $ x = 0 $ and $ x + 3y = 60 $

Now, is $ (0, 10) $ connected to $ (0, 20) $? Only if both on same edge.

But between $ (0, 10) $ and $ (0, 20) $, $ x = 0 $, but $ x \leq y $ → always true

But also need to satisfy $ x + y \geq 10 $ → at $ (0, y) $, $ y \geq 10 $

And $ x + 3y \leq 60 $ → $ 3y \leq 60 $ → $ y \leq 20 $

So along $ x = 0 $, $ y $ ranges from $ 10 $ to $ 20 $

So the segment from $ (0, 10) $ to $ (0, 20) $ is part of the feasible region.

But is it bounded?

Now, check if $ (0, 10) $ connects to $ (5, 5) $?

Let’s list all corner points of the feasible region.

We need to find where the boundaries intersect and form a polygon.

Let’s go step by step.

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🔎 Let's find all vertices of the feasible region:

We look for intersections of constraint lines that lie in the feasible region.

We already found:

1. A: $ (0, 10) $ → Intersection of $ x = 0 $ and $ x + y = 10 $

2. B: $ (0, 20) $ → Intersection of $ x = 0 $ and $ x + 3y = 60 $

3. C: $ (15, 15) $ → Intersection of $ x + 3y = 60 $ and $ x = y $

4. D: $ (5, 5) $ → Intersection of $ x + y = 10 $ and $ x = y $

Now, check whether all these points are connected in order.

Let’s test feasibility and connectivity.

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#### Check if $ (5, 5) $ connects to $ (0, 10) $?

The line between them: $ x + y = 10 $, from $ (5, 5) $ to $ (0, 10) $

Along this line: $ x + y = 10 $, and $ x \leq y $? At $ (5,5) $: equal; at $ (0,10) $: $ 0 \leq 10 $ → yes

Also $ x + 3y \leq 60 $? At $ (0,10) $: $ 30 \leq 60 $; at $ (5,5) $: $ 5 + 15 = 20 \leq 60 $

So yes, this edge is feasible.

Now, from $ (5,5) $ to $ (15,15) $? Along $ x = y $, from $ x = 5 $ to $ x = 15 $

Check if $ x + y \geq 10 $: $ 2x \geq 10 $ → $ x \geq 5 $ → yes

$ x + 3y = x + 3x = 4x \leq 60 $ → $ x \leq 15 $ → yes

So yes, this edge is feasible.

From $ (15,15) $ to $ (0,20) $? That would be along $ x + 3y = 60 $

At $ (15,15) $: $ 15 + 45 = 60 $ → on line

At $ (0,20) $: $ 0 + 60 = 60 $ → on line

So the line segment from $ (15,15) $ to $ (0,20) $ is on $ x + 3y = 60 $

Now, check if all points on this segment satisfy other constraints.

Take a point in between: say $ x = 7.5 $, then $ 7.5 + 3y = 60 $ → $ 3y = 52.5 $ → $ y = 17.5 $

So $ (7.5, 17.5) $

Check:
- $ x + y = 25 \geq 10 $ → Yes
- $ x \leq y $? $ 7.5 \leq 17.5 $ → Yes

So yes, entire segment is feasible.

Now, from $ (0,20) $ to $ (0,10) $? Along $ x = 0 $

As before, $ y $ from 10 to 20, satisfies:
- $ x + y = y \geq 10 $
- $ x + 3y = 3y \leq 60 $ → $ y \leq 20 $
- $ x \leq y $ → $ 0 \leq y $

So yes.

Thus, the feasible region is a quadrilateral with vertices:

1. $ A = (0, 10) $
2. $ B = (0, 20) $
3. $ C = (15, 15) $
4. $ D = (5, 5) $

But wait — is $ (0,10) $ connected directly to $ (5,5) $? Yes, along $ x + y = 10 $

And $ (5,5) $ to $ (15,15) $ along $ x = y $

And $ (15,15) $ to $ (0,20) $ along $ x + 3y = 60 $

And $ (0,20) $ to $ (0,10) $ along $ x = 0 $

So yes, the feasible region is a quadrilateral with vertices:

> (0,10), (0,20), (15,15), (5,5)

But is $ (5,5) $ connected to $ (0,10) $? Yes.

So the order around the region is:

- Start at $ (0,10) $
- → $ (0,20) $
- → $ (15,15) $
- → $ (5,5) $
- → back to $ (0,10) $

Wait — is $ (5,5) $ directly connected to $ (0,10) $? Yes, via $ x + y = 10 $

So the four vertices are:

1. $ P_1 = (0, 10) $
2. $ P_2 = (0, 20) $
3. $ P_3 = (15, 15) $
4. $ P_4 = (5, 5) $

✔ These are the corner points of the feasible region.

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✔ Step 4: Evaluate Objective Function $ z = 3x + 9y $ at Each Vertex

| Point | $ x $ | $ y $ | $ z = 3x + 9y $ |
|-------|--------|--------|------------------|
| $ (0,10) $ | 0 | 10 | $ 3(0) + 9(10) = 90 $ |
| $ (0,20) $ | 0 | 20 | $ 0 + 180 = 180 $ |
| $ (15,15) $ | 15 | 15 | $ 45 + 135 = 180 $ |
| $ (5,5) $ | 5 | 5 | $ 15 + 45 = 60 $ |

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✔ Step 5: Find Minimum and Maximum

- Minimum value of $ z $: $ \boxed{60} $ at $ (5, 5) $
- Maximum value of $ z $: $ \boxed{180} $ at both $ (0, 20) $ and $ (15, 15) $

Note: Since both $ (0,20) $ and $ (15,15) $ give $ z = 180 $, the maximum is achieved at two points.

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✔ Final Answer:

#### Minimize $ z = 3x + 9y $:
- Minimum value = 60
- Occurs at $ (x, y) = (5, 5) $

#### Maximize $ z = 3x + 9y $:
- Maximum value = 180
- Occurs at $ (x, y) = (0, 20) $ and $ (15, 15) $

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📌 Summary of Feasible Region and Optimal Solutions

- The feasible region is a quadrilateral bounded by:
- $ x = 0 $ from $ y = 10 $ to $ y = 20 $
- $ x + 3y = 60 $ from $ (0,20) $ to $ (15,15) $
- $ x = y $ from $ (15,15) $ to $ (5,5) $
- $ x + y = 10 $ from $ (5,5) $ to $ (0,10) $

- The optimal solutions occur at the corner points as computed.

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✔ Graphical Interpretation (Mentally or on Paper):

Plot the lines:
- $ x + 3y = 60 $: from $ (0,20) $ to $ (60,0) $
- $ x + y = 10 $: from $ (0,10) $ to $ (10,0) $
- $ x = y $: diagonal line

Shade the region that satisfies:
- Below $ x + 3y = 60 $
- Above $ x + y = 10 $
- Below $ x = y $
- In first quadrant

The shaded region is a quadrilateral with vertices at $ (0,10), (0,20), (15,15), (5,5) $

Evaluate $ z $ at each — done.

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✔ Final Boxed Answers:

$$
\boxed{
\begin{aligned}
&\text{Minimum value of } z = 60 \text{ at } (5, 5) \\
&\text{Maximum value of } z = 180 \text{ at } (0, 20) \text{ and } (15, 15)
\end{aligned}
}
$$