Problem 55 - College Algebra

Question

The cost function for a certain company is \(C=60 x+300\) and the revenue is given by \(R=100 x-0.5 x^{2} .\) Recall that profit is revenue minus cost. Set up a quadratic equation and find two values of \(x\) (production level) that will create a profit of \(\$ 300\) .

Step-by-Step Solution

Answer

The two production levels are 20 and 60 units.

Step 1: Define the Profit Function

The profit is calculated by subtracting the cost function from the revenue function. Write the profit function as follows:\[ P(x) = R(x) - C(x) = (100x - 0.5x^2) - (60x + 300). \]

Step 2: Simplify the Profit Function

Simplify the expression from Step 1 to obtain the quadratic equation:\[ P(x) = 100x - 0.5x^2 - 60x - 300 = -0.5x^2 + 40x - 300. \]

Step 3: Set the Profit Equation Equal to Desired Profit

Since we want the profit to be $300, set the profit equation equal to 300:\[ -0.5x^2 + 40x - 300 = 300. \]

Step 4: Rearrange the Equation

Rearrange the equation to set it to zero for solving:\[ -0.5x^2 + 40x - 300 - 300 = 0, \]which simplifies to:\[ -0.5x^2 + 40x - 600 = 0. \]

Step 5: Multiply Through by -2

To eliminate the decimal, multiply the entire equation by -2:\[ x^2 - 80x + 1200 = 0. \]

Step 6: Solve for x using the Quadratic Formula

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, \) \( b = -80, \) and \( c = 1200.\) Substitute these values in:\[ x = \frac{80 \pm \sqrt{(-80)^2 - 4 \cdot 1 \cdot 1200}}{2 \cdot 1} = \frac{80 \pm \sqrt{6400 - 4800}}{2}. \]

Step 7: Continue Solving the Quadratic Formula

Calculate the values under the square root and solve for \( x \):\[ x = \frac{80 \pm \sqrt{1600}}{2} = \frac{80 \pm 40}{2}. \]

Step 8: Find the Two Values of x

Calculate the two solutions:- \( x = \frac{80 + 40}{2} = 60 \).- \( x = \frac{80 - 40}{2} = 20 \).

Key Concepts

Profit Calculation

Understanding profit calculation is fundamental to analyzing a company's financial health. Profit is how much money a company makes after taking out expenses. In mathematical terms, profit is described as {}

• Revenue Function (R): This shows the total income generated from selling goods or services.
• Cost Function (C): This includes all expenses made to produce the goods or services, such as materials, labor, and overheads.

To find profit, simply subtract the total costs from total revenues: \[ P(x) = R(x) - C(x) \]. Profit calculation isn't just about knowing earnings; it's crucial for planning, decision-making, and determining a company's viability. For this exercise, we set the profit equation to $300, meaning we're looking for the values of production level \(x\) that generate exactly this amount.

Cost Function

The cost function represents the total expenses incurred by a company for producing a specific number of units. In the given exercise, the cost function is expressed as \[ C(x) = 60x + 300 \], where \(60x\) is the variable cost that changes with the number of units produced, and \(300\) is the fixed cost that remains constant irrespective of production levels.

• Variable Costs: These fluctuate with production volume and can include costs like materials and labor directly used in production.
• Fixed Costs: These remain constant and include expenses such as rent, salaries, and utilities that don't change with production levels.

Understanding the cost function is essential for managers to gauge how efficiently resources are being used. Identifying components within the cost function can help businesses find areas to cut costs and thus increase profit.

Revenue Function

The revenue function is pivotal in understanding a business's ability to generate income from sales. Defined by a mathematical equation, it calculates the total amount of money made from selling \(x\) units of a product.In our exercise, it is given as:\[ R(x) = 100x - 0.5x^2 \].The term \(100x\) represents the revenue made per unit, showing a direct linear relationship between the number of units sold and sales revenue. The quadratic term \(-0.5x^2\) implies diminishing returns as production increases, possibly due to market saturation or inefficiencies setting in as production scales up.

• Linear Term (100x): Represents initial sales growth with each additional unit sold.
• Quadratic Term (-0.5x²): Shows how revenue growth slows with higher production levels.

The revenue function helps businesses estimate potential income and plan their production goals accordingly, considering both direct and adaptable market conditions.

Quadratic Formula

The quadratic formula is a solution to any quadratic equation of the form \(ax^2 + bx + c = 0\). It's a critical tool in mathematics and especially useful in business scenarios like this one.For this problem, we derive the quadratic equation: \[ x^2 - 80x + 1200 = 0 \].The quadratic formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].In our equation, \(a = 1\), \(b = -80\), and \(c = 1200\). Inserting these values gives:\[ x = \frac{80 \pm \sqrt{6400 - 4800}}{2} = \frac{80 \pm \sqrt{1600}}{2} \].Solving this, we get two values for \(x\): 60 and 20.

• \(\pm\) Symbol: Determines two possible x-values, meaning the quadratic equation yields two solutions.
• Discriminant (\(b^2 - 4ac\)): Indicates the nature of roots (real and distinct, real and repeating, or complex).

This formula is invaluable for pinpointing exact production levels where a company meets specific profit goals, informing production, and financial strategy.