Problem 43 - Calculus Early Transcendentals

Question

(a) A chemical manufacturer sells sulfuric acid in bulk at a price of \(\$ 100\) per unit. If the daily total production cost in dollars for \(x\) units is $$ C(x)=100,000+50 x+0.0025 x^{2} $$ and if the daily production capacity is at most 7000 units, how many units of sulfuric acid must be manu- factured and sold daily to maximize the profit? (b) Would it benefit the manufacturer to expand the daily production capacity? (c) Use marginal analysis to approximate the effect on profit if daily production could be increased from 7000 to 7001 units.

Step-by-Step Solution

Answer

(a) 7000 units. (b) No benefit. (c) Decreased profit.

Step 1: Define the Profit Function

Profit is the difference between the revenue and the cost. The revenue function is given by the price per unit times the number of units: \( R(x) = 100x \). The cost function is given as \( C(x) = 100,000 + 50x + 0.0025x^2 \). Therefore, the profit function \( P(x) \) is:\[ P(x) = R(x) - C(x) = 100x - (100,000 + 50x + 0.0025x^2). \] Simplifying, we get \[ P(x) = 50x - 0.0025x^2 - 100,000. \]

Step 2: Find Critical Points of Profit Function

To maximize profit, we first find the derivative of \( P(x) \) and set it to zero.\[ P'(x) = 50 - 0.005x. \] Set \( P'(x) = 0 \) to find \[ 50 - 0.005x = 0 \Rightarrow x = 10,000. \]

Step 3: Evaluate Production Capacity Constraint

Since the daily production capacity is at most 7000 units, we need to check \( P(x) \) at 7000 units and possible end values. Although 10,000 units maximize the profit theoretically, it exceeds the given capacity, so we evaluate \( P \) at 7000 units.

Step 4: Calculate and Compare Profits at Boundaries

Calculate \( P(7000) \):\[ P(7000) = 50(7000) - 0.0025(7000^2) - 100,000. \] Doing the math gives \( P(7000) = 125,000. \)

Step 5: Decision on Production Expansion

To consider expanding the production capacity, we calculate \( P(7001) \) to approximate the change in profit. From this, we again use \( P(x) = 50x - 0.0025x^2 - 100,000 \) to find \( P(7001) \) and take the difference \( P(7001) - P(7000) \) to perform marginal analysis.

Step 6: Marginal Analysis for Additional Unit

Calculate \( P(7001) \):\[ P(7001) = 50(7001) - 0.0025(7001)^2 - 100,000. \] The result shows \( P(7001) \approx 125,000, \) indicating a decrease from \( P(7000) \), thus showing diminishing returns above full capacity.

Key Concepts

Revenue Function

In business, understanding the revenue function is crucial because it tells us how much money is coming in from selling products or services. The revenue function is calculated by multiplying the price per unit by the number of units sold. In our exercise, the chemical manufacturer sells sulfuric acid at $100 per unit. This makes the revenue function simple to express:

• Revenue, \( R(x) \), is equal to \( 100x \), where \( x \) is the number of units sold.

As we increase the number of units sold, the revenue also increases. However, this does not take into account the costs involved in producing these units, which is why we must also consider the cost function to assess profitability.

Cost Function

The cost function represents the total cost of producing a certain number of units. It's important to understand how costs increase with production, especially with nonlinear components like quadratic terms.

• In the exercise, the cost function \( C(x) \) is given as: \( 100,000 + 50x + 0.0025x^2 \).

This function shows that there is a fixed cost of \( 100,000 \) dollars, present regardless of the production level. It also includes variable costs:

• \( 50x \), a linear term representing the direct cost per unit of production.
• \( 0.0025x^2 \), a quadratic term representing increasing marginal costs, where costs grow at an increasing rate as production volume rises.

Understanding the cost function helps in setting the production level where costs are optimized against revenues for maximum profit.

Marginal Analysis

Marginal analysis is a powerful tool used in economics to examine the additional benefits or costs from producing one more unit. It helps in making informed production decisions.

• To perform marginal analysis, we derive the profit function, \( P(x) \), which is calculated as the revenue minus the cost: \( 50x - 0.0025x^2 - 100,000 \).

By taking the derivative of this profit function, \( P'(x) = 50 - 0.005x \), we set it to zero to find the point where additional production will no longer increase profit; this is critical for profit maximization.

• Initially, we found that \( x = 10,000 \) would be optimal, but the production capacity limits this to 7000 units.
• Checking with marginal analysis, we evaluated the profit change from 7000 to 7001 units and found diminishing returns, cementing that \( x = 7000 \) is optimal given the current capacity.

Therefore, marginal analysis confirms the optimal production level, emphasizing the importance of evaluating marginal benefits and costs before making production adjustments.