Problem 79 - Brief Applied Calculus

Question

BUSINESS: Break-Even Points and Maximum Profit A company that produces tracking devices for computer disk drives finds that if it produces \(x\) devices per week, its costs will be \(C(x)=180 x+16,000\) and its revenue will be \(R(x)=-2 x^{2}+660 x\) (both in dollars). a. Find the company's break-even points. b. Find the number of devices that will maximize profit, and the maximum profit.

Step-by-Step Solution

Answer

Break-even points are at 40 and 200 devices; maximum profit is \$12,800 at 120 devices.

Step 1: Define the Break-Even Points

The break-even point occurs when the cost equals the revenue. This means we need to set the cost function \(C(x) = 180x + 16,000\) equal to the revenue function \(R(x) = -2x^2 + 660x\). The equation becomes:\[-2x^2 + 660x = 180x + 16,000\]

Step 2: Simplify the Equation

To simplify, move all the terms to one side of the equation:\[-2x^2 + 660x - 180x - 16,000 = 0\]This simplifies to:\[-2x^2 + 480x - 16,000 = 0\].

Step 3: Solve the Quadratic Equation for x

We will use the quadratic formula to solve for \(x\). The quadratic equation is \(-2x^2 + 480x - 16,000 = 0\), with \(a = -2\), \(b = 480\), and \(c = -16,000\). The quadratic formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Substitute the values:\[x = \frac{-480 \pm \sqrt{480^2 - 4(-2)(-16,000)}}{2(-2)}\]Calculate \(b^2 - 4ac\):\[b^2 - 4ac = 480^2 - 4(2)(16,000) = 230,400 - 128,000 = 102,400\].

Step 4: Calculate the Break-Even Points

Continuing from the equation:\[x = \frac{-480 \pm \sqrt{102,400}}{-4}\]\[x = \frac{-480 \pm 320}{-4}\]This gives us two solutions:- \(x = \frac{-480 + 320}{-4} = 40\)- \(x = \frac{-480 - 320}{-4} = 200\)So, the break-even points are at 40 devices and 200 devices.

Step 5: Define Profit Function

Profit is calculated as revenue minus costs. So, the profit function \(P(x)\) is:\[P(x) = R(x) - C(x) = (-2x^2 + 660x) - (180x + 16,000)\]Simplify the profit function:\[P(x) = -2x^2 + 660x - 180x - 16,000\]\[P(x) = -2x^2 + 480x - 16,000\].

Step 6: Find Number of Devices for Maximum Profit

The maximum profit occurs at the vertex of the parabola described by the profit function \(P(x) = -2x^2 + 480x - 16,000\). The vertex \(x\) of a parabola \(ax^2 + bx + c\) is given by \(x = -\frac{b}{2a}\).Here, \(a = -2\) and \(b = 480\), so:\[x = -\frac{480}{2(-2)} = \frac{480}{4} = 120\]Thus, 120 devices will maximize the profit.

Step 7: Calculate Maximum Profit

Substitute \(x = 120\) into the profit function to find the maximum profit:\[P(120) = -2(120)^2 + 480(120) - 16,000\]\[-2(14,400) + 57,600 - 16,000 = -28,800 + 57,600 - 16,000 = 12,800\]The maximum profit is \$12,800.

Key Concepts

Quadratic Equations

Quadratic equations are essential when dealing with break-even analysis and profit maximization in business. A quadratic equation is one of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. These equations often arise when the relationship between variables is non-linear, particularly in cases involving squared terms, as in our cost and revenue functions.

To solve a quadratic equation, one common approach is to use the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

• The term inside the square root, \(b^2 - 4ac\), is called the discriminant. It helps determine the nature of the roots. If it is positive, there are two distinct real roots. If zero, there is exactly one real root, often called a repeated or double root. If negative, there are no real roots.
• In a business context, solving these equations can help identify critical points such as break-even points or maximum profit conditions.

This method was employed to find the break-even points for the production of tracking devices, ensuring that the company knows at which production levels its costs are exactly offset by revenues.

Profit Maximization

Profit maximization is at the core of any business operation. It involves identifying the level of production or sales that yields the highest profit. Profit itself is the difference between total revenue and total costs, often expressed through a mathematical function.

For instance, in our case, the profit function \(P(x)\) was derived from the revenue function \(R(x) = -2x^2 + 660x\) and the cost function \(C(x) = 180x + 16,000\). This led to:

• \(P(x) = -2x^2 + 480x - 16,000\)

We identify the point of maximum profit by finding the vertex of the parabolic profit function. Because the parabola opens downward (as the \(a\) coefficient, \(-2\), is negative), the vertex corresponds to the maximum point. This vertex can be calculated using:

• \(x = -\frac{b}{2a}\)

For our example, solving for \(x\) resulted in a production level of 120 devices, which was determined to maximize profits at $12,800. This calculation is vital as it helps businesses focus their resources effectively to achieve the best financial outcomes.

Revenue and Cost Functions

Understanding revenue and cost functions is key for effective financial management and strategic decision-making in businesses. These functions quantify how a company's revenue and costs vary with changes in production volume.

The revenue function \(R(x)\) and cost function \(C(x)\) are usually dependent on variables such as the number of items produced and sold.

• In our example, \(R(x) = -2x^2 + 660x\) represents the revenue, showing a downward-opening parabola. The negative coefficient in front of \(x^2\) implies diminishing returns at higher production levels.
• The cost function \(C(x) = 180x + 16,000\) represents the linear growth of production costs with a fixed overhead cost of \$16,000.

The intersection points of these functions, called break-even points, are critical in determining at which production levels the company neither profits nor incurs losses. This business insight allows the company to manage production levels simply by balancing revenue and cost to ensure profitability.