https://file.aiquickdraw.com/mathbot/file/1730049923181glnqprc3.jpg

To solve this optimization problem, let's define the variables and constraints based on the details provided.

### Problem Setup

1. Let:
   - $ x $ = acres of crop A
   - $ y $ = acres of crop B

2. **Constraints**:
   - Total acreage: $ x + y = 30 $
   - Cost constraint for seeds: $ 10x + 20y \leq 400 $
   
3. **Objective Function**:
   - The goal is to maximize profit, which is given by:
     \[
     \text{Profit} = 150x + 130y
     \]

### Solution Steps

#### Step 1: Write the constraints in terms of inequalities

From the constraints:
1. $ x + y = 30 $ (total acreage constraint)
   - This can be rearranged as $ y = 30 - x $.
2. $ 10x + 20y \leq 400 $ (cost constraint)
   - Simplify by dividing everything by 10:
     \[
     x + 2y \leq 40
     \]
   - Rearrange for $ y $:
     \[
     y \leq 20 - \frac{x}{2}
     \]

#### Step 2: Determine the feasible region

The feasible region is defined by the intersection of:
- $ y = 30 - x $
- $ y \leq 20 - \frac{x}{2} $

#### Step 3: Evaluate the objective function at corner points of the feasible region

To maximize profit, we’ll evaluate the profit function $ 150x + 130y $ at the intersection points of the constraints:

1. **Point (0, 0)**: If $ x = 0 $ and $ y = 0 $, then profit = $ 150(0) + 130(0) = 0 $.

2. **Point (0, 20)**: If $ x = 0 $ and $ y = 20 $, then profit = $ 150(0) + 130(20) = 2600 $.

3. **Point (20, 10)**: If $ x = 20 $ and $ y = 10 $, then profit = $ 150(20) + 130(10) = 3000 + 1300 = 4300 $.

4. **Point (30, 0)**: If $ x = 30 $ and $ y = 0 $, then profit = $ 150(30) + 130(0) = 4500 $.

#### Conclusion

The maximum profit of **$4500** is achieved when the farmer plants:
- **30 acres of crop A**
- **0 acres of crop B**

### Final Answer
- Acres of crop A: 30
- Acres of crop B: 0

Would you like further details on any step, or have any other questions?

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Here are five additional questions related to this problem:

1. What would be the optimal solution if the cost constraint was increased to $500?
2. How would the solution change if crop B had a higher profit per acre than crop A?
3. Can you calculate the break-even point for each crop individually?
4. How does the feasible region change if the farmer had 40 acres instead of 30?
5. What would happen if the cost of seeds for crop A increased to $15 per acre?

**Tip:** When solving linear programming problems graphically, always identify the constraints and feasible region before calculating corner points.

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 $400 on seed. If crop B brings in a profit of $130 per acre, and crop A brings in a profit of $150 per acre, how many acres of each crop should the farmer plant to maximize her profit?

Learn how to solve a farm optimization problem using linear programming. This example covers the allocation of 30 acres between two crops with different seed costs and profits, subject to budget constraints. Suitable for students in Grades 9-12.

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