Chapter 1: Problem 32
It costs the \(\mathrm{ABC}\) Company \(400+5 \sqrt{x(x-4)}\) dollars to make \(x(x \geq 4)\) toy stoves that sell for \(\$ 6\) each. (a) Find a formula for \(P(x)\), the total profit in making \(x\) stoves. (b) Evaluate \(P(200)\) and \(P(1000)\). (c) How many stoves does \(\mathrm{ABC}\) have to make to just break even?
Short Answer
Step by step solution
Understanding Cost and Revenue Functions
Establishing the Profit Function
Calculating P(200)
Calculating P(1000)
Determining Break-Even Quantity
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cost Function
- The fixed cost (400 dollars) is an expense the company must pay regardless of how many stoves they produce. It's a one-time cost not influenced by production volume.
- The variable cost is represented by the term \( 5\sqrt{x(x-4)} \). This part of the function can be seen as increasing with the number of stoves, meaning the more stoves produced, the higher the variable costs.
Revenue Function
- The revenue is directly proportional to the number of stoves sold (\( x \)). This means that more stoves sold lead to more revenue.
- This linear function allows businesses to easily calculate expected revenue based on projected sales figures.
Break-Even Analysis
- Firstly, rearrange the profit function to \( 6x - 400 = 5\sqrt{x(x-4)} \).
- The algebraic manipulation involves isolating \( x \) and can be complex due to the square root and quadratic terms.
- Usually, the solution involves solving numerically or graphically to find that the break-even point is approximately 361 stoves.
Square Root in Functions
- Square roots are often used to represent diminishing returns or growth, where the rate of increase slows as the input grows larger.
- In the function \( 5\sqrt{x(x-4)} \), the expression \( x(x-4) \) inside the square root indicates that the costs increase at a diminishing rate as production expands, reflecting perhaps increased efficiency.
- These kinds of functions can be challenging to deal with algebraically, especially when solving for variables, as they may involve additional steps like squaring both sides of an equation.