The problem states that a manufacturer sells belts for $15 per unit. The fixed costs are $3000 per month, and the variable cost per unit is $10. (a) We need to write the equations for the revenue $R(x)$ and cost $C(x)$ functions. (b) We need to find the break-even point.

AlgebraLinear EquationsCost FunctionRevenue FunctionBreak-Even Analysis
2025/5/19

1. Problem Description

The problem states that a manufacturer sells belts for 15perunit.Thefixedcostsare15 per unit. The fixed costs are 3000 per month, and the variable cost per unit is $
1

0. (a) We need to write the equations for the revenue $R(x)$ and cost $C(x)$ functions.

(b) We need to find the break-even point.

2. Solution Steps

(a)
The revenue function R(x)R(x) is the amount of money earned from selling xx units, which is the price per unit times the number of units sold. Since the price per unit is $15, the revenue function is
R(x)=15xR(x) = 15x
The cost function C(x)C(x) is the total cost of producing xx units, which is the sum of the fixed costs and the variable costs. The fixed costs are 3000,andthevariablecostperunitis3000, and the variable cost per unit is 10, so the total variable cost is 10x10x. Therefore, the cost function is
C(x)=3000+10xC(x) = 3000 + 10x
(b)
The break-even point is where the revenue equals the cost, so we set R(x)=C(x)R(x) = C(x) and solve for xx.
15x=3000+10x15x = 3000 + 10x
Subtract 10x10x from both sides:
15x10x=300015x - 10x = 3000
5x=30005x = 3000
Divide by 5:
x=30005=600x = \frac{3000}{5} = 600
It takes 600 units to break even.

3. Final Answer

(a) R(x)=15xR(x) = 15x
C(x)=3000+10xC(x) = 3000 + 10x
(b) It takes 600 units to break even.

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