Worksheet 3.2 - Linear Programming | PDF | Linear Programming ... - Free Printable

Let's solve each problem step by step using the given feasible regions and objective functions. We'll evaluate the objective function at each corner point (vertex) of the feasible region, since in linear programming, the maximum and minimum values occur at the vertices.

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Problem 1 & 2: First Feasible Region (Diamond Shape)

From the first graph:

The feasible region is a diamond (a square rotated 45°) with vertices at:
- A = (3, 8)
- B = (7, 4)
- C = (3, 0)
- D = (-1, 4)

Wait — let’s carefully read the coordinates from the grid:

Looking at the graph:
- The top vertex is at (3, 8) → labeled as A
- Right vertex is at (7, 4) → labeled as B
- Bottom vertex is at (3, 0) → labeled as C
- Left vertex is at (-1, 4) → labeled as D

So the four vertices are:
- A: (3, 8)
- B: (7, 4)
- C: (3, 0)
- D: (-1, 4)

Now we will evaluate the objective functions at these points.

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#### Problem 1: $ z = 3x + 2y $

Evaluate $ z = 3x + 2y $ at each vertex:

- At A (3, 8):
$ z = 3(3) + 2(8) = 9 + 16 = 25 $

- At B (7, 4):
$ z = 3(7) + 2(4) = 21 + 8 = 29 $

- At C (3, 0):
$ z = 3(3) + 2(0) = 9 + 0 = 9 $

- At D (-1, 4):
$ z = 3(-1) + 2(4) = -3 + 8 = 5 $

Results:
- Maximum: 29 at (7, 4)
- Minimum: 5 at (-1, 4)

✔ Answer for #1:
- Max: 29
- Min: 5

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#### Problem 2: $ z = x - 4y $

Evaluate $ z = x - 4y $ at each vertex:

- At A (3, 8):
$ z = 3 - 4(8) = 3 - 32 = -29 $

- At B (7, 4):
$ z = 7 - 4(4) = 7 - 16 = -9 $

- At C (3, 0):
$ z = 3 - 4(0) = 3 $

- At D (-1, 4):
$ z = -1 - 4(4) = -1 - 16 = -17 $

Results:
- Maximum: 3 at (3, 0)
- Minimum: -29 at (3, 8)

✔ Answer for #2:
- Max: 3
- Min: -29

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Problem 3 & 4: Second Feasible Region (Trapezoid-like shape)

From the second graph:

Vertices are labeled:
- A: (0, 18)
- B: (0, 15)
- C: (6, 18)
- D: (10, 8)
- E: (0, 0)

Wait — let’s look closely:

- Point A is at (0, 18)
- Point B is at (0, 15)
- Point C is at (6, 18)
- Point D is at (10, 8)
- Point E is at (0, 0)

But looking at the shaded region, it appears to be bounded by:
- From (0, 18) down to (0, 15)? That doesn't make sense.

Wait — actually, the graph shows:
- Top-left: (0, 18) → labeled A
- Top-right: (6, 18) → labeled C
- Then down to (10, 8) → labeled D
- Then down to (0, 0) → labeled E
- And back up to (0, 18)? But there's a point at (0, 15) labeled B.

Wait — this seems inconsistent.

Let me re-express based on the plot:

Actually, looking at the graph again:

- The region starts at (0, 18) → A
- Goes to (6, 18) → C
- Then to (10, 8) → D
- Then to (0, 0) → E
- Then back to (0, 18)? But that would skip (0, 15).

Wait — but there's a point labeled B at (0, 15).

Ah! It seems like the feasible region is bounded between:
- Left boundary: from (0, 15) to (0, 18) — but that’s vertical?
- Actually, the shaded region appears to be a quadrilateral with vertices:
- A: (0, 18)
- C: (6, 18)
- D: (10, 8)
- E: (0, 0)

And the point B at (0, 15) might be a mislabel or just an intermediate point?

Wait — no. Looking more carefully:

There is a line from (0, 18) to (6, 18), then to (10, 8), then to (0, 0), then back to (0, 18)? But that would not be convex.

But wait — the shaded region includes:
- From (0, 0) up along x-axis?
- Then up to (10, 8)?
- Then to (6, 18)?
- Then to (0, 18)?
- Then down to (0, 15)? No.

Actually, the graph has a polygon with:
- Vertex A: (0, 18)
- Vertex C: (6, 18)
- Vertex D: (10, 8)
- Vertex E: (0, 0)

But there is also a point labeled B at (0, 15) — perhaps it's not a vertex?

Wait — maybe the feasible region is bounded by:
- From (0, 18) to (6, 18) — horizontal
- Then to (10, 8) — diagonal
- Then to (0, 0) — diagonal
- Then back to (0, 18) — but that skips (0, 15)

But the shaded region includes all points from y=0 to y=18 at x=0? But only up to (0, 15) is marked?

Wait — I think there's a mistake in interpretation.

Let me re-analyze the second graph.

Looking at the axes:
- Y-axis goes from 0 to 18
- X-axis from 0 to 10

Points:
- A: (0, 18)
- B: (0, 15) — this is directly below A on y-axis
- C: (6, 18)
- D: (10, 8)
- E: (0, 0)

The shaded region appears to be bounded by:
- From (0, 15) to (6, 18) — upward slant
- From (6, 18) to (10, 8) — downward slant
- From (10, 8) to (0, 0) — downward slant
- From (0, 0) to (0, 15) — vertical

So the vertices are:
- B: (0, 15)
- C: (6, 18)
- D: (10, 8)
- E: (0, 0)

And point A (0, 18) is outside the feasible region? But the line from (0,18) to (6,18) is drawn, but the shaded region doesn’t include it?

Wait — actually, the shaded region starts at (0,15), not (0,18). So (0,18) is not a vertex of the feasible region.

So the four vertices of the feasible region are:
- B: (0, 15)
- C: (6, 18)
- D: (10, 8)
- E: (0, 0)

Yes, that makes sense.

So vertices:
- B: (0, 15)
- C: (6, 18)
- D: (10, 8)
- E: (0, 0)

We’ll use these for problems 3 and 4.

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#### Problem 3: $ z = 0.35x + 1.25y $

Evaluate at each vertex:

- At B (0, 15):
$ z = 0.35(0) + 1.25(15) = 0 + 18.75 = 18.75 $

- At C (6, 18):
$ z = 0.35(6) + 1.25(18) = 2.1 + 22.5 = 24.6 $

- At D (10, 8):
$ z = 0.35(10) + 1.25(8) = 3.5 + 10 = 13.5 $

- At E (0, 0):
$ z = 0.35(0) + 1.25(0) = 0 $

Results:
- Maximum: 24.6 at (6, 18)
- Minimum: 0 at (0, 0)

✔ Answer for #3:
- Max: 24.6
- Min: 0

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#### Problem 4: $ z = 1.5x - 0.5y $

Evaluate at each vertex:

- At B (0, 15):
$ z = 1.5(0) - 0.5(15) = 0 - 7.5 = -7.5 $

- At C (6, 18):
$ z = 1.5(6) - 0.5(18) = 9 - 9 = 0 $

- At D (10, 8):
$ z = 1.5(10) - 0.5(8) = 15 - 4 = 11 $

- At E (0, 0):
$ z = 1.5(0) - 0.5(0) = 0 $

Results:
- Maximum: 11 at (10, 8)
- Minimum: -7.5 at (0, 15)

✔ Answer for #4:
- Max: 11
- Min: -7.5

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✔ Final Answers:

#### #1: $ z = 3x + 2y $
- Maximum: 29 at (7, 4)
- Minimum: 5 at (-1, 4)

#### #2: $ z = x - 4y $
- Maximum: 3 at (3, 0)
- Minimum: -29 at (3, 8)

#### #3: $ z = 0.35x + 1.25y $
- Maximum: 24.6 at (6, 18)
- Minimum: 0 at (0, 0)

#### #4: $ z = 1.5x - 0.5y $
- Maximum: 11 at (10, 8)
- Minimum: -7.5 at (0, 15)

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🔍 Explanation:

In linear programming, the optimal value (maximum or minimum) of the objective function occurs at one of the corner points (vertices) of the feasible region. Therefore, we:
1. Identify the coordinates of all vertices.
2. Plug each into the objective function.
3. Compare the results to find max and min.

This method works because the feasible region is a convex polygon, and the objective function is linear.

Let me know if you'd like the answers formatted neatly for submission!