Problem 80 - Applied Calculus

Question

BUSINESS: Break-Even Points and Maximum Profit City and Country Cycles finds that if it sells \(x\) racing bicycles per month, its costs will be \(C(x)=420 x+72,000\) and its revenue will be \(R(x)=-3 x^{2}+1800 x\) (both in dollars). a. Find the store's break-even points. b. Find the number of bicycles that will maximize profit, and the maximum profit.

Step-by-Step Solution

Answer

Break-even at 54 bicycles; maximum profit of $86,700 at 230 bicycles.

Step 1: Understand Break-Even Points

Break-even points occur where cost equals revenue, i.e., \( C(x) = R(x) \). According to the given function formulas, set the cost function equal to the revenue function and solve for \( x \).

Step 2: Set Up the Equation for Break-Even

Substitute the cost and revenue functions: \( 420x + 72,000 = -3x^2 + 1800x \). Rearrange it to form a quadratic equation: \( 3x^2 + 1380x - 72,000 = 0 \).

Step 3: Solve the Quadratic Equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 3 \), \( b = 1380 \), and \( c = -72,000 \). Calculate the discriminant: \( 1380^2 - 4\times 3\times (-72,000) = 2,899,200 \). Solve for \( x \): \( x = \frac{-1380 \pm \sqrt{2,899,200}}{6} \).

Step 4: Calculate Break-Even Points

Calculate each root from the quadratic formula: \( x = \frac{-1380 + 1701}{6} = 53.5 \) and \( x = \frac{-1380 - 1701}{6} = -513.5 \). Since \( x \) cannot be negative, the break-even points are \( x = 53.5 \) bicycles.

Step 5: Determine Profit Function

Profit \( P(x) \) is given by revenue minus cost: \( P(x) = R(x) - C(x) = (-3x^2 + 1800x) - (420x + 72,000) \). Simplify to get the profit function: \( P(x) = -3x^2 + 1380x - 72,000 \).

Step 6: Maximize the Profit Function

Find the vertex of the parabola for \( P(x) = -3x^2 + 1380x - 72,000 \). The x-coordinate of the vertex is given by \( x = \frac{-b}{2a} \) where \( a = -3 \), \( b = 1380 \). Thus, \( x = \frac{-1380}{-6} = 230 \).

Step 7: Calculate Maximum Profit

Substitute \( x = 230 \) back into the profit function: \( P(230) = -3(230)^2 + 1380(230) - 72,000 \). Calculate to find \( P(230) = -158,700 + 317,400 - 72,000 = 86,700 \). Thus, the maximum profit is \$86,700.

Key Concepts

Quadratic Equation

A quadratic equation is a polynomial equation of degree two. It has the standard form of \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. Quadratic equations are crucial in modeling various real-world scenarios including projectile motion, business profit analysis, and optimization problems.
To find the solutions of a quadratic equation, we often use the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

• The term under the square root sign, \( b^2 - 4ac \), is called the discriminant. It determines the nature of the roots.

• If the discriminant is positive, there are two real and distinct solutions.
• If the discriminant is zero, there is exactly one real solution, also known as a repeated or double root.
• If the discriminant is negative, the solutions are complex and imaginary.

In the exercise related to break-even points, the pedal bicycle shop set the cost function equal to the revenue function, resulting in a quadratic equation. Solving this equation, we get the points where the business neither makes a profit nor a loss.

Profit Maximization

Profit maximization is a fundamental objective in business where enterprises aim to achieve the highest possible profit. This is accomplished by finding the optimal level of production that generates the maximum difference between revenue and cost.
To determine this optimal point, it's essential to analyze the profit function, defined as the difference between total revenue and total cost:

• \( P(x) = R(x) - C(x) \)

In many cases, particularly when dealing with quadratic revenue and cost functions, the profit function will also appear as a quadratic equation. The task then is to find the vertex of this parabola because, for a downward-opening curve (where \( a < 0 \)), the vertex represents the maximum point.
The x-coordinate of the vertex, and thus the optimal number of products to produce and sell, can be calculated using:

• \( x = \frac{-b}{2a} \)

In our exercise, the profit function calculated led to identifying 230 bicycles as the optimal number to produce for maximum profit, resulting in a substantial profit of $86,700.

Revenue and Cost Functions

Revenue and cost functions model the financial aspects of a business. They allow for analysis of production and sales activities.
**Revenue Function (\( R(x) \))**:

• This function details how much income is generated from selling a certain number of units. It can be affected by price and quantity factors, often leading to a quadratic form when considering diminishing returns at higher production levels.

In the example provided, \( R(x) = -3x^2 + 1800x \). It shows that revenue initially increases with more bicycles sold but eventually decreases due to the \(-3x^2\) term.
**Cost Function (\( C(x) \))**:

• This function shows the total expenses incurred to produce a given number of units, encompassing fixed and variable costs.

For the cycle store, the cost function \( C(x) = 420x + 72,000 \) represents initial fixed costs \(\(72,000\) and a variable cost of \( \)420 \) per bicycle.
When you set \( R(x) = C(x) \), you find the break-even point, indicating no profit or loss. Understanding these functions helps businesses make informed decisions regarding production levels to achieve desired profitability.