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.

---

Would you like details or have any questions about these solutions? Here are some additional questions you might be interested in:

1. How to solve quadratic equations?
2. What are the applications of derivatives in real life?
3. How do you integrate a function using substitution?
4. What is the geometric interpretation of eigenvalues and eigenvectors?
5. How do you find the inverse of a matrix?

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.