https://file.aiquickdraw.com/mathbot/file/17216646668406lq8jcpo.png

Sure, I'll solve problems 1, 2, 7, 9, and 10 from the uploaded image.

### Problem 1: Profit
**Given:**
- Unit selling price: $20
- Unit cost: $15
- Fixed costs: $600,000

**Solution:**
Let $ n $ be the number of units that must be sold to make a profit.

Profit occurs when total revenue > total cost:
\[ 20n > 15n + 600,000 \]

Simplifying:
\[ 5n > 600,000 \]
\[ n > 120,000 \]

**Answer:**
The least number of units that must be sold for the company to have a profit is **120,001**.

### Problem 2: Profit
**Given:**
- Cost of material per unit: $2.50
- Cost of labor per unit: $4
- Fixed cost: $5000
- Wholesale price per unit: $7.40

**Solution:**
Let $ n $ be the number of units that must be sold to make a profit.

Total cost per unit = cost of material + cost of labor:
\[ \text{Total cost per unit} = 2.50 + 4 = 6.50 \]

Profit occurs when total revenue > total cost:
\[ 7.40n > 6.50n + 5000 \]

Simplifying:
\[ 0.90n > 5000 \]
\[ n > \frac{5000}{0.90} \]
\[ n > 5555.56 \]

**Answer:**
The least number of units that must be sold to make a profit is **5556**.

### Problem 7: Investment
**Given:**
- Total amount to invest: $30,000
- Annual interest rates: 5% and $ 6\frac{3}{4} $%
- Minimum yield: 6.5%

**Solution:**
Let $ x $ be the amount invested at 5%, and $ y $ be the amount invested at $ 6\frac{3}{4} $%.
\[ x + y = 30,000 \]
\[ 0.05x + 0.0675y = 0.065 \times 30,000 \]

Simplifying the second equation:
\[ 0.05x + 0.0675y = 1950 \]

From $ x + y = 30,000 $:
\[ x = 30,000 - y \]

Substituting into the second equation:
\[ 0.05(30,000 - y) + 0.0675y = 1950 \]
\[ 1500 - 0.05y + 0.0675y = 1950 \]
\[ 0.0175y = 450 \]
\[ y = \frac{450}{0.0175} \]
\[ y = 25,714.29 \]

Therefore:
\[ x = 30,000 - 25,714.29 \]
\[ x = 4285.71 \]

**Answer:**
The company must invest **$4285.71** at 5% and **$25,714.29** at $ 6\frac{3}{4} $% to achieve the minimum yield.

### Problem 9: Sales Allocation
**Given:**
- Initial product units: 2500
- Initial price: $4 per unit
- Price increase next month: $0.50 per unit
- Total revenue target: at least $10,750

**Solution:**
Let $ n $ be the number of units sold this month.

Revenue from sales this month:
\[ 4n \]

Revenue from remaining units sold next month:
\[ (2500 - n)(4.50) \]

Total revenue:
\[ 4n + (2500 - n)4.50 \geq 10,750 \]

Simplifying:
\[ 4n + 11250 - 4.50n \geq 10,750 \]
\[ -0.50n + 11250 \geq 10,750 \]
\[ -0.50n \geq -500 \]
\[ n \leq 1000 \]

**Answer:**
The maximum number of units that can be sold this month is **1000**.

### Problem 10: Revenue
**Given:**
- Price function: $ p = \frac{200}{q} + 3 $
- Revenue function: $ R = pq = \left( \frac{200}{q} + 3 \right)q $

**Solution:**
To find the minimum number of units $ q $ such that revenue is no less than $9000:
\[ R \geq 9000 \]
\[ \left( \frac{200}{q} + 3 \right)q \geq 9000 \]
\[ 200 + 3q \geq 9000 \]
\[ 3q \geq 8800 \]
\[ q \geq \frac{8800}{3} \]
\[ q \geq 2933.33 \]

**Answer:**
The minimum number of units that must be sold is **2934**.

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**Tip:** When solving word problems, carefully define your variables and set up equations based on the relationships described in the problem. This will help in systematically solving for the unknowns.

Explore detailed solutions to mathematical problems involving profit analysis, investment strategies, sales allocation, and revenue functions. Understand key formulas and methods used to solve complex business and financial scenarios. Suitable for advanced high school or college students.

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