A farmer is going to divide her 30 acre farm between two crops. Seed for crop A costs $10 per acre. Seed for crop B costs $20 per acre. The farmer can spend at most $500 on seed.

If crop B brings in a profit of $200 per acre, and crop A brings in a profit of $80 per acre, how many acres of each crop should the farmer plant to maximize her profit?

This problem can be solved using linear programming, where the objective is to maximize the profit subject to certain constraints.

### Define the Variables:
- Let $ x $ be the number of acres of crop A.
- Let $ y $ be the number of acres of crop B.

### Objective Function:
The profit for crop A is $80 per acre and for crop B is $200 per acre. Therefore, the objective function, which represents the total profit, is:
\[
\text{Maximize } P = 80x + 200y
\]

### Constraints:
1. The total amount of land is 30 acres, so:
   \[
   x + y \leq 30
   \]
2. The cost constraint is based on the budget of $500, with crop A costing $10 per acre and crop B costing $20 per acre:
   \[
   10x + 20y \leq 500
   \]
3. Non-negativity constraints:
   \[
   x \geq 0, \quad y \geq 0
   \]

### Solve the System:
Now, we need to find the values of $ x $ and $ y $ that maximize the profit function while satisfying the constraints.

1. **Constraint 1**: $ x + y = 30 $
   - Solve for $ y $:
   \[
   y = 30 - x
   \]

2. **Constraint 2**: $ 10x + 20y = 500 $
   - Solve for $ y $:
   \[
   20y = 500 - 10x
   \]
   \[
   y = 25 - 0.5x
   \]

Next, we'll find the intersection points of these constraint lines and then evaluate the objective function at each corner point of the feasible region. 

### Corner Points:
1. At $ x = 0 $: $ y = 25 $
2. At $ x = 20 $: $ y = 10 $
3. At $ x = 30 $: $ y = 0 $

### Calculate the Profit at Each Corner Point:
1. At $ (x, y) = (0, 25) $:
   \[
   P = 80(0) + 200(25) = 5000
   \]
2. At $ (x, y) = (20, 10) $:
   \[
   P = 80(20) + 200(10) = 1600 + 2000 = 3600
   \]
3. At $ (x, y) = (30, 0) $:
   \[
   P = 80(30) + 200(0) = 2400
   \]

### Conclusion:
The maximum profit occurs when the farmer plants 0 acres of crop A and 25 acres of crop B, resulting in a profit of $5000.

### Final Answer:
- $ x = 0 $ acres of crop A
- $ y = 25 $ acres of crop B

Would you like details on any of the steps, or do you have further questions?

Here are 5 related questions to consider:
1. How would the solution change if the budget for seeds was increased?
2. What happens if crop A's profit increases to $120 per acre?
3. How would the optimal planting strategy change if the total land area was reduced to 20 acres?
4. What is the impact on the profit if crop B's cost per acre increases to $25?
5. How would you graphically represent the constraints and objective function?

**Tip:** In linear programming, always ensure you evaluate the objective function at all the corner points of the feasible region to find the optimal solution.

A farmer is going to divide her 30 acre farm between two crops. Seed for crop A costs $10 per acre. Seed for crop B costs $20 per acre. The farmer can spend at most $500 on seed. If crop B brings in a profit of $200 per acre, and crop A brings in a profit of $80 per acre, how many acres of each crop should the farmer plant to maximize her profit?

Explore how to maximize profit for a farmer dividing 30 acres between two crops with a $500 budget for seed. This problem involves linear programming concepts, with the objective of optimizing profit based on crop costs and land constraints. Suitable for high school students learning optimization techniques.

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