Problem 42 - Calculus Early Transcendentals

Question

A fertilizer producer finds that it can sell its product at a price of \(p=300-0.1 x\) dollars per unit when it produces \(x\) units of fertilizer. The total production cost (in dollars) for \(x\) units is $$ C(x)=15,000+125 x+0.025 x^{2} $$ If the production capacity of the firm is at most 1000 units of fertilizer in a specified time, how many units must be manufactured and sold in that time to maximize the profit?

Step-by-Step Solution

Answer

Manufacture and sell 700 units to maximize profit while staying within capacity.

Step 1: Define Profit Function

Profit is defined as total revenue minus total cost. Here, the revenue function is given by \( R(x) = x \, p = x(300 - 0.1x) = 300x - 0.1x^2 \). The cost function is given as \( C(x) = 15,000 + 125x + 0.025x^2 \). The profit function \( P(x) \) is \( P(x) = R(x) - C(x) = (300x - 0.1x^2) - (15,000 + 125x + 0.025x^2) \). Simplifying, we find \( P(x) = -0.125x^2 + 175x - 15,000 \).

Step 2: Determine the Vertex of Profit Function

Since the profit function \( P(x) = -0.125x^2 + 175x - 15,000 \) is a downward-opening parabola, the maximum profit is at the vertex. The vertex formula for \( ax^2 + bx + c \) is \( x = -\frac{b}{2a} \). Here, \( a = -0.125 \) and \( b = 175 \), so \( x = -\frac{175}{2(-0.125)} = 700 \).

Step 3: Confirm Feasibility within Constraints

The production capacity constraint is \( x \leq 1000 \). Since \( x = 700 \) falls within this constraint, 700 units is feasible. Therefore, manufacturing and selling 700 units is possible within the given capacity.

Step 4: Conclusion

Therefore, to maximize profit, the firm should manufacture and sell 700 units of fertilizer within the given production capacity.

Key Concepts

Revenue Function

Understanding the revenue function is crucial for profit maximization. In this scenario, the revenue function is derived from both the quantity sold and the price per unit. When the fertilizer producer sells its product at a price of \(p = 300 - 0.1x\), it means the price decreases as more units \(x\) are produced and sold.
The revenue function \(R(x)\) represents the total income generated from selling the product. It is calculated by multiplying the number of units sold, \(x\), by the price per unit, \(p\).

• Mathematically: \( R(x) = x(300 - 0.1x) \)

By expanding this expression, we get \( R(x) = 300x - 0.1x^2 \). Here, \(300x\) indicates the potential revenue if prices did not decrease with quantity. However, \(0.1x^2\) represents the lost revenue due to price reductions, showcasing the law of diminishing returns.
The revenue function is typically a quadratic function, meaning it may have a maximum or minimum point, depending on the parabola's direction. In our case, it opens downwards, implying there's a point where revenue is maximized.

Cost Function

The cost function is essential in determining profit since it outlines the expenses associated with producing a certain number of units. In this exercise, the cost function, \(C(x)\), is given as:

• \( C(x) = 15,000 + 125x + 0.025x^2 \).

This equation considers fixed and variable costs.

• The \(15,000\) signifies fixed costs—expenses that do not change with the level of output, like rent or salaries.
• The term \(125x\) reflects variable costs, which increase linearly with each additional unit produced, such as raw materials.
• The term \(0.025x^2\) represents increasing marginal costs, showing how costs swell more substantially as production ramps up, perhaps due to overuse of machinery or overtime hours.

Understanding this function allows businesses to comprehend how total costs scale with production and to decide efficiently how many units to produce to achieve profitability.

Quadratic Optimization

Quadratic optimization involves finding the maximum or minimum values of a quadratic function. In the context of this problem, it's used to maximize profit, which can be expressed as a quadratic formula:

• \( P(x) = -0.125x^2 + 175x - 15,000 \).

The expression indicates a parabola that opens downward due to the negative coefficient of \(x^2\). To determine the point of maximum profit, we need to find the vertex of this parabola.
Using the vertex formula \( x = -\frac{b}{2a} \), where \(a\) is the coefficient of \(x^2\) and \(b\) is the coefficient of \(x\), helps locate this point.

• Here, \( a = -0.125 \) and \( b = 175 \), resulting in \( x = -\frac{175}{2(-0.125)} = 700 \).

This result reveals that producing 700 units will yield maximum profit, within the constraints provided, thus illustrating the power of quadratic optimization in practical business scenarios. This method provides a precise way to determine the optimal production quantity for maximum financial outcomes.