Assume a fixed cost of $900, a variable cost of $5.25, and a selling price of $6.00.

a. What is the break-even point? (Roundup your answer to the next whole number.)

b. How many units must be sold to make a profit of $500.00? (Roundup your answer to the next whole number.)

c. How many units must be sold to average $0.25 profit per unit? $0.50 profit per unit? (Roundup your answer to the next whole number. Loss should be indicated with a minus sign.)

Let's break down each part of the problem:

### Given:
- Fixed cost $ F = 900 $
- Variable cost per unit $ V = 5.25 $
- Selling price per unit $ P = 6.00 $

### Part (a): Break-Even Point
The break-even point occurs when the total revenue equals the total cost. This means:

\[
\text{Revenue} = \text{Total Cost}
\]

The revenue from selling $ x $ units is $ P \times x $, and the total cost is the sum of the fixed cost and the variable costs: $ F + V \times x $.

Thus, the break-even equation is:

\[
P \times x = F + V \times x
\]

Substituting the given values:

\[
6.00 \times x = 900 + 5.25 \times x
\]

Solving for $ x $:

\[
6.00x - 5.25x = 900
\]
\[
0.75x = 900
\]
\[
x = \frac{900}{0.75} = 1200
\]

So, the break-even point is **1200 units**.

### Part (b): Profit of $500.00
For a profit of $500, the equation becomes:

\[
\text{Profit} = \text{Revenue} - \text{Total Cost}
\]

Thus, the revenue needed to make a profit of $500 is:

\[
500 = P \times x - (F + V \times x)
\]

Substituting the values:

\[
500 = 6.00 \times x - (900 + 5.25 \times x)
\]
\[
500 = 6.00x - 900 - 5.25x
\]
\[
500 = 0.75x - 900
\]
\[
0.75x = 1400
\]
\[
x = \frac{1400}{0.75} = 1867
\]

Thus, to make a profit of $500, **1867 units** must be sold.

### Part (c): Profit Per Unit
1. **For $0.25 profit per unit:**
   
   The profit per unit equation is:

\[
\text{Profit per unit} = P - V - \frac{F}{x}
\]

Substitute $ P = 6.00 $, $ V = 5.25 $, and set the profit per unit to $0.25:

\[
0.25 = 6.00 - 5.25 - \frac{900}{x}
\]
\[
0.25 = 0.75 - \frac{900}{x}
\]
\[
\frac{900}{x} = 0.75 - 0.25 = 0.50
\]
\[
x = \frac{900}{0.50} = 1800
\]

So, to have a profit of $0.25 per unit, **1800 units** must be sold.

2. **For $0.50 profit per unit:**

Substitute the values and set the profit per unit to $0.50:

\[
0.50 = 6.00 - 5.25 - \frac{900}{x}
\]
\[
0.50 = 0.75 - \frac{900}{x}
\]
\[
\frac{900}{x} = 0.75 - 0.50 = 0.25
\]
\[
x = \frac{900}{0.25} = 3600
\]

So, to have a profit of $0.50 per unit, **3600 units** must be sold.

### Summary of Results:
- (a) Break-even point: **1200 units**
- (b) Units needed for $500 profit: **1867 units**
- (c) Units needed for $0.25 profit per unit: **1800 units**, for $0.50 profit per unit: **3600 units**

Would you like more details or have any questions?

### Related Questions:
1. What would be the break-even point if the variable cost increased to $5.50?
2. How would the break-even point change if the fixed cost decreased to $800?
3. What profit would be made by selling 2000 units?
4. How does increasing the selling price to $6.50 affect the break-even point?
5. What is the total cost for producing 2500 units?

### Tip:
The break-even point is highly sensitive to both fixed and variable costs, so keeping variable costs low can significantly improve profitability.

Assume a fixed cost of $900, a variable cost of $5.25, and a selling price of $6.00.

a. What is the break-even point? (Round up your answer to the next whole number.)

b. How many units must be sold to make a profit of $500.00? (Round up your answer to the next whole number.)

c. How many units must be sold to average $0.25 profit per unit? $0.50 profit per unit? (Round up your answer to the next whole number.)

Calculate the break-even point, profit of $500, and units required to achieve $0.25 and $0.50 profit per unit. This problem explores key business math concepts such as break-even analysis and profit calculation. Suitable for Grades 9-12.

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