Maximize: z = 10x + 60y

Subject to:
x + y ≤ 200
x ≥ 20
y ≥ 4x



The given problem is a linear programming problem, and we are tasked with maximizing the objective function \( z = 10x + 60y \) under the given constraints:

### Objective function:
\[
z = 10x + 60y
\]

### Constraints:
1. \( x + y \leq 200 \)
2. \( x \geq 20 \)
3. \( y \geq 4x \)

To solve this, we need to:

1. **Graph the constraints** to identify the feasible region.
2. **Find the vertices** of the feasible region.
3. **Evaluate the objective function** at each vertex to determine the maximum value.

### Step 1: Graph the constraints
- The inequality \( x + y \leq 200 \) represents a line with an intercept at \( (200, 0) \) and \( (0, 200) \).
- The constraint \( x \geq 20 \) is a vertical line at \( x = 20 \).
- The constraint \( y \geq 4x \) is a line with a slope of 4, passing through the origin.

### Step 2: Find the vertices of the feasible region
We will calculate the intersection points of the boundary lines.

- Intersection of \( x + y = 200 \) and \( y = 4x \):

\[
x + 4x = 200 \implies 5x = 200 \implies x = 40
\]
\[
y = 4(40) = 160
\]
So, one vertex is \( (40, 160) \).

- Intersection of \( x = 20 \) and \( y = 4x \):

\[
y = 4(20) = 80
\]
So, another vertex is \( (20, 80) \).

- Intersection of \( x = 20 \) and \( x + y = 200 \):

\[
20 + y = 200 \implies y = 180
\]
So, the third vertex is \( (20, 180) \).

### Step 3: Evaluate the objective function at each vertex
We now evaluate \( z = 10x + 60y \) at each of the vertices:

1. At \( (40, 160) \):
\[
z = 10(40) + 60(160) = 400 + 9600 = 10,000
\]

2. At \( (20, 80) \):
\[
z = 10(20) + 60(80) = 200 + 4800 = 5,000
\]

3. At \( (20, 180) \):
\[
z = 10(20) + 60(180) = 200 + 10,800 = 11,000
\]

### Conclusion:
The maximum value of \( z \) is \( 11,000 \), which occurs at the point \( (20, 180) \).

Let me know if you want more details on the steps or any clarifications!

---

Here are 5 related questions for further practice:
1. What happens if the objective function is changed to \( z = 20x + 40y \)?
2. How would the solution change if the constraint \( y \geq 4x \) becomes \( y \geq 3x \)?
3. What are the implications of removing the constraint \( x \geq 20 \)?
4. Can the feasible region be unbounded for certain values of \( x \) and \( y \)?
5. How do you solve this problem using the Simplex method?

**Tip:** Always check if the constraints form a closed and bounded region to ensure that a maximum or minimum exists.

Maximize: z = 10x + 60y

Subject to:
x + y ≤ 200
x ≥ 20
y ≥ 4x

Solve the linear programming problem of maximizing z = 10x + 60y under the constraints x + y ≤ 200, x ≥ 20, and y ≥ 4x. Learn how to graph constraints, find vertices, and evaluate the objective function. Ideal for Grades 11-12.

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