Problem 33 - A Graphical Approach to College Algebra

Question

Modeling I Exercises \(33-36\) (a) Express the cost \(C\) as a function of \(x,\) where \(x\) represents the number of items as described. (b) Express the revenue \(R\) as a function of \(x .\) (c) Determine analytically the value of \(x\) for which revenue equals cost. (d) Graph \(y_{1}=C(x)\) and \(y_{2}=R(x)\) on the same \(x y\) -axes and interpret the graphs. Stuffing Envelopes\(\quad\) A student stuffs envelopes for extra income during her spare time. Her initial cost to obtain the necessary information for the job was \(\$ 200.00 .\) Each envelope costs \(\$ 0.02,\) and she gets paid \(\$ 0.04\) per envelope stuffed. Let \(x\) represent the number of envelopes stuffed.

Step-by-Step Solution

Answer

She needs to stuff 10,000 envelopes to break even.

Step 1: Express the Cost Function

The cost of stuffing envelopes includes a fixed initial cost and a variable cost per envelope stuffed. - Initial cost: \(200.00 - Cost per envelope: \)0.02The total cost function can be expressed as:\[ C(x) = 200 + 0.02x \] where \( x \) is the number of envelopes stuffed.

Step 2: Express the Revenue Function

The revenue from stuffing envelopes is based on the amount she gets paid per envelope stuffed. - Revenue per envelope: $0.04The total revenue function can be expressed as:\[ R(x) = 0.04x \] where \( x \) is the number of envelopes stuffed.

Step 3: Determine Break-even Point

The break-even point occurs when revenue equals cost. Set the cost function equal to the revenue function and solve for \( x \):\[ 200 + 0.02x = 0.04x \] Subtract \( 0.02x \) from both sides:\[ 200 = 0.02x \] Divide both sides by \( 0.02 \):\[ x = \frac{200}{0.02} = 10000 \] Thus, she needs to stuff 10,000 envelopes to break even.

Step 4: Graph and Interpret the Functions

To graph \( C(x) = 200 + 0.02x \) and \( R(x) = 0.04x \), plot both functions on the \( x-y \) axes:- Cost function \( C(x) \) starts at 200 and increases with a slope of 0.02.- Revenue function \( R(x) \) starts at 0 and increases with a slope of 0.04.The intersection of these two lines, which occurs at \( x = 10000 \), shows the break-even point. For \( x < 10000 \), cost exceeds revenue. For \( x > 10000 \), revenue exceeds cost, indicating profit.

Key Concepts

Cost Function

The cost function is a mathematical way of expressing the total cost incurred by a business or individual in producing a specific quantity of goods. For our student, the cost function includes both fixed and variable costs.

Fixed costs are those that remain constant regardless of the level of production or, in this context, the number of envelopes stuffed. Here, the student has a fixed cost of \(200, which covers initial expenses.

Variable costs, however, change with the number of items produced. In this case, each envelope stuffed adds a cost of \)0.02. Therefore, the cost increases proportionately with the number of envelopes.

The total Cost Function, combining both fixed and variable components, is given by:
\[ C(x) = 200 + 0.02x \]

where \(x\) represents the number of envelopes stuffed.

Revenue Function

The revenue function illustrates how much money is earned from selling or, in this situation, stuffing a specific number of goods. Revenue, in its simplest form, is the product of the number of units sold and the price per unit.

For the student, each envelope stuffed earns her $0.04.

Therefore, the revenue earned from stuffing \(x\) envelopes is modeled by the equation:
\[ R(x) = 0.04x \]

This indicates that her revenue is directly proportional to the number of envelopes she stuffs.

Unlike costs, there are no fixed components here. Every envelope contributes equally and directly to the revenue.

Break-even Point

The break-even point is a critical financial concept, especially for anyone operating a business or managing a project. It’s the point at which total revenue equals total costs. Beyond this point, one begins to generate profit.

For our envelope-stuffing student, determining the break-even point requires setting the cost function equal to the revenue function:
\[ 200 + 0.02x = 0.04x \]

Solving this equation isolates the value of \(x\) where the two functions are equal:
Subtract \(0.02x\) from both sides:
\[ 200 = 0.02x \]
Divide both sides by \(0.02\):
\[ x = \frac{200}{0.02} = 10000 \]

At this intersection, the costs and revenues balance out, indicating that stuffing 10,000 envelopes covers her expenses exactly.

Beyond this number, every additional envelope stuffed provides a surplus, translating into profit.

Graphing Functions

Graphing functions provides a visual representation of how different values interact and can help in understanding concepts like the break-even point.

For the student, graphing the Cost Function \(C(x) = 200 + 0.02x\) and the Revenue Function \(R(x) = 0.04x\) transforms the numerical data into an intuitive picture.

- The cost graph starts at $200 when \(x = 0\), reflecting the fixed costs, with a slope of 0.02, depicting the small incremental cost per envelope.
- The revenue graph starts at the origin (0,0) because if no envelopes are stuffed, zero revenue is produced, and it has a steeper slope of 0.04.

Where these two lines intersect on the graph is key:

• At \(x = 10000\), they cross, showing the break-even point.
• For \(x < 10000\), the cost line lies above the revenue line, indicating losses.
• For \(x > 10000\), the revenue line surpasses the cost line, marking profit.

Visualizing these functions helps to quickly see how many envelopes to stuff to start making money.