Math Problem Statement

A firm produces a product that has the production cost function

​C(x)equals=340x+9520

and the revenue function

​R(x)=425x.

No more than

261

units can be sold. Find and analyze the​ break-even quantity, then find the profit function.

Solution

Let's go through the steps to find and analyze the break-even quantity, as well as the profit function.

Step 1: Understand the Given Functions

  • Cost Function: C(x)=340x+9520C(x) = 340x + 9520
  • Revenue Function: R(x)=425xR(x) = 425x

Here, xx represents the number of units produced and sold.

Step 2: Find the Break-Even Quantity

The break-even point occurs when the revenue equals the cost, meaning R(x)=C(x)R(x) = C(x).

Set the revenue function equal to the cost function: 425x=340x+9520425x = 340x + 9520

Step 3: Solve for xx

Subtract 340x340x from both sides to isolate the variable: 425x340x=9520425x - 340x = 9520 85x=952085x = 9520 x=952085=112x = \frac{9520}{85} = 112

So, the break-even quantity is x=112x = 112 units.

Step 4: Analyze the Break-Even Quantity

At x=112x = 112, the firm neither makes a profit nor a loss. If the firm produces and sells fewer than 112 units, it incurs a loss. If it sells more than 112 units, it makes a profit.

Step 5: Find the Profit Function

The profit function P(x)P(x) is the difference between the revenue and the cost: P(x)=R(x)C(x)P(x) = R(x) - C(x)

Substitute the given functions: P(x)=425x(340x+9520)P(x) = 425x - (340x + 9520)

Distribute and simplify: P(x)=425x340x9520P(x) = 425x - 340x - 9520 P(x)=85x9520P(x) = 85x - 9520

So, the profit function is: P(x)=85x9520P(x) = 85x - 9520

Step 6: Determine the Profit at Maximum Sales

Given that no more than 261 units can be sold, we can evaluate the profit at this maximum quantity:

P(261)=85(261)9520P(261) = 85(261) - 9520 P(261)=221859520=12665P(261) = 22185 - 9520 = 12665

Thus, if the firm sells the maximum of 261 units, the profit will be $12,665.

Summary

  • Break-even quantity: x=112x = 112 units.
  • Profit function: P(x)=85x9520P(x) = 85x - 9520.
  • Maximum possible profit (at 261 units): $12,665.

Would you like more details or have any questions?

Related Questions:

  1. What happens to the profit if the cost per unit increases by $10?
  2. How would a change in fixed costs affect the break-even quantity?
  3. What is the revenue at the break-even point?
  4. How does the break-even analysis change if the price per unit is reduced?
  5. How would changes in demand (e.g., selling fewer than 261 units) impact profit?

Tip:

Understanding the break-even point helps businesses decide on production levels. Beyond this point, each additional unit sold contributes to profit.

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Math Problem Analysis

Mathematical Concepts

Cost Functions
Revenue Functions
Break-even Analysis
Profit Functions

Formulas

Cost Function: C(x) = 340x + 9520
Revenue Function: R(x) = 425x
Profit Function: P(x) = R(x) - C(x)

Theorems

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Suitable Grade Level

Advanced High School