Question

The daily costs for a hamburger vendor are \(\$ 135\) per day plus \(\$ 1.25\) per hamburger sold. He sells each burger for \(\$ 3.50,\) and the maximum number of hamburgers he can sell in a day is 300. a) Write equations to represent the total cost, \(C,\) and the total revenue, \(R,\) as functions of the number, \(n,\) of hamburgers sold. b) Graph \(C(n)\) and \(R(n)\) on the same set of axes. c) The break-even point is where \(C(n)=R(n) .\) Identify this point. d) Develop an algebraic and a graphical model for the profit function. e) What is the maximum daily profit the vendor can earn?

Short answer

a) \(C(n) = 135 + 1.25n\), \(R(n) = 3.50n\). b) Graph both functions. c) Break-even at \(n = 60\). d) Profit function \(P(n) = 2.25n - 135\). e) Maximum profit is \(\$540\).

Step-by-step solution

Define the Cost Equation

The total cost, \(C(n)\), is the sum of the fixed daily cost and the variable cost per hamburger sold. The fixed cost is \(\$135\) per day, and the variable cost is \(\$1.25\) per hamburger. Therefore, the cost function can be written as: \[ C(n) = 135 + 1.25n \]

Define the Revenue Equation

The total revenue, \(R(n)\), is the product of the number of hamburgers sold and the selling price per hamburger. The vendor sells each hamburger for \(\$3.50\). Therefore, the revenue function can be written as: \[ R(n) = 3.50n \]

Graph the Cost and Revenue Equations

Graph the cost function \(C(n) = 135 + 1.25n\) and the revenue function \(R(n) = 3.50n\) on the same set of axes. The x-axis represents the number of hamburgers sold, \(n\), and the y-axis represents the amount in dollars.

Identify the Break-even Point

The break-even point occurs where \(C(n) = R(n)\). To find this, set the equations equal to each other and solve for \(n\): \[ 135 + 1.25n = 3.50n \] Solve for \(n\): \[ 135 = 2.25n \] \[ n = 60 \] At the break-even point, the vendor sells 60 hamburgers.

Define the Profit Equation

Profit, \(P(n)\), is the difference between revenue and cost. Therefore, the profit function is: \[ P(n) = R(n) - C(n) \] Substitute the expressions for \(R(n)\) and \(C(n)\): \[ P(n) = 3.50n - (135 + 1.25n) \] Simplify: \[ P(n) = 2.25n - 135 \]

Determine the Maximum Daily Profit

The maximum number of hamburgers that can be sold in a day is 300. Substitute \(n = 300\) into the profit function to find the maximum daily profit: \[ P(300) = 2.25 \times 300 - 135 \] \[ P(300) = 675 - 135 \] \[ P(300) = 540 \] Therefore, the maximum daily profit is \(\$540\).

Key concepts

Cost Equation

The cost equation is crucial for any business because it helps determine how much money is spent to produce a certain number of products. For the hamburger vendor, the total cost each day includes a fixed cost and a variable cost. The fixed cost is the amount of money that has to be spent regardless of how many hamburgers are sold, which is \$135\$ per day. On top of this, the vendor incurs a variable cost of \$1.25\$ for each hamburger sold. Therefore, the total cost function, \( C(n) \), where \( n \) represents the number of hamburgers sold, can be expressed as: \[ C(n) = 135 + 1.25n \] Breaking it down, the \$135\$ is the fixed cost, and \$1.25n\$ represents the variable cost. By understanding this equation, the vendor can predict how costs increase with the number of hamburgers sold.

Revenue Equation

Revenue is the income generated from selling goods or services. For our hamburger vendor, the revenue equation helps track the money earned from selling hamburgers. The vendor sells each hamburger for \$3.50\$. Therefore, if \( n \) hamburgers are sold in a day, the revenue function, \( R(n) \), can be expressed as: \[ R(n) = 3.50n \] Simply put, this equation multiplies the selling price by the number of hamburgers sold. By using this equation, the vendor can estimate daily earnings based on the number of hamburgers sold.

Break-even Point

The break-even point is a key concept in profit calculation. It represents the point where total costs equal total revenue, meaning there is no profit or loss. To find the break-even point for our vendor, we set the cost function equal to the revenue function: \[ C(n) = R(n) \] Substituting the expressions, we have: \[ 135 + 1.25n = 3.50n \] Solving this for \( n \), we get: \[ 135 = 2.25n \] \[ n = 60 \] Thus, the vendor must sell 60 hamburgers to break even. At this point, the revenue generated exactly covers the costs.

Profit Function

Profit is what a business has left after all costs are subtracted from total revenue. For our vendor, the profit function, \( P(n) \), is derived by subtracting the total cost function from the total revenue function: \[ P(n) = R(n) - C(n) \] Plugging in the expressions for \( R(n) \) and \( C(n) \), we get: \[ P(n) = 3.50n - (135 + 1.25n) \] Simplifying this, we have: \[ P(n) = 2.25n - 135 \] This function helps the vendor understand how profit varies with the number of hamburgers sold. It’s an essential tool for maximizing profit.

Graphing Equations

Graphing the cost and revenue equations provides a visual understanding of how they relate. Plotting \( C(n) = 135 + 1.25n \) and \( R(n) = 3.50n \) on the same axes can show where they intersect, pointing out the break-even point.

• The x-axis represents the number of hamburgers sold, \( n \).
• The y-axis represents the amount in dollars.

The point where the two lines intersect is where \( C(n) \) equals \( R(n) \), which we've calculated to be when \( n = 60 \). This graphical method not only highlights the break-even point but also shows how profit (or loss) changes with sales volume.