Problem 38 - Single Variable Calculus

Question

The Advance Visual Systems Corporation realizes a total profit of $$P(x)=-0.000002 x^{3}+0.016 x^{2}+80 x-70,000$$ dollars per week from the manufacture and sale of \(x\) units of their 26-in. LCD HDTVs. a. Find the marginal profit function \(P^{\prime}\). b. Compute \(P^{\prime}(2000)\) and interpret your result.

Step-by-Step Solution

Answer

The marginal profit function is \(P'(x) = -0.000006x^2 + 0.032x + 80\). When the company manufactures and sells 2000 units, the marginal profit is 24.00. This means increasing production by one more unit will result in an additional profit of approximately $24.00.

Step 1: Identify the Profit Function

The given profit function, \(P(x)\), is: \(P(x) = -0.000002x^3 + 0.016x^2 + 80x - 70000\)

Step 2: Find the Derivative of the Profit Function

To find the marginal profit function \(P'(x)\), we need to differentiate each of the terms in \(P(x)\) with respect to x: \(P'(x) = \frac{d}{dx}(-0.000002x^3 + 0.016x^2 + 80x - 70000)\) Now, differentiating each term: - \(\frac{d}{dx}(-0.000002x^3) = -0.000006x^2\) - \(\frac{d}{dx}(0.016x^2) = 0.032x\) - \(\frac{d}{dx}(80x) = 80\) - \(\frac{d}{dx}(-70000) = 0\) Then, combining the terms, we have the marginal profit function: \(P'(x) = -0.000006x^2 + 0.032x + 80\) #b. Compute \(P'(2000)\) and interpret your result#

Step 3: Compute the Marginal Profit for 2000 Units

We need to calculate \(P'(2000)\) by plugging in the value x = 2000 in the marginal profit function: \(P'(2000) = -0.000006(2000)^2 + 0.032(2000) + 80\) Calculating this, we get: \(P'(2000) \approx 24.00\)

Step 4: Interpret the Result

The value of \(P'(2000)\) is 24.00, which means that when the company manufactures and sells 2000 units of their 26-inch LCD HDTVs, increasing the production by one more unit will result in an additional profit of approximately $24.00.

Key Concepts

Profit Function

A Profit Function in calculus represents the total profit a business makes from producing and selling a certain number of products. The function is usually expressed in terms of a mathematical equation that helps businesses understand the relationship between production quantity and profitability. In this specific case, the profit function is given by \[P(x) = -0.000002x^3 + 0.016x^2 + 80x - 70000\]where \(x\) stands for the number of units produced and sold.

This equation combines several components:

• The cubic term \(-0.000002x^3\) affects the shape of the graph, potentially representing diminishing returns at higher production levels.
• The quadratic term \(0.016x^2\) could relate to efficiencies or cost savings.
• The linear term \(80x\) shows the base profit gain per unit.
• The constant \(-70000\) indicates fixed costs or starting capital investment.

The goal is to understand how changing \(x\) influences the overall profit, which is where the derivative comes into play.

Marginal Profit

The concept of Marginal Profit relates to the additional profit gained from producing one more unit of a product. It is a way to measure the sensitivity of total profit to a small change in the production quantity. To find the marginal profit, you need to calculate the derivative of the profit function, often noted as \(P'(x)\).

In the original problem, the marginal profit function was derived:\[P'(x) = -0.000006x^2 + 0.032x + 80\]This expression helps businesses make decision-making processes easier by analyzing the effect of change in the production level on profit.

• The marginal profit provides insights into whether increasing production will be beneficial or not.
• It helps in determining the optimal level of production where profit is maximized.

Derivative

The Derivative is a fundamental tool in calculus used to determine the rate at which one quantity changes with respect to another. In the context of the profit function, taking the derivative of \(P(x)\)—denoting it as \(P'(x)\)—helps to calculate the marginal profit.

The process of differentiating involves applying rules of calculus:

• The power rule is used to differentiate terms of the form \(x^n\).
• The constant rule implies the derivative of a constant is zero.

Using these rules, every term in the profit function is reduced to form the marginal profit function. Computing derivatives allows analysts to effectively predict changes in profitability under small quantity adjustments.

Interpretation of Derivative

Understanding the Interpretation of Derivative is crucial when analyzing profit changes. For a business, \(P'(x)\) indicates how sensitive the profit is to an increase in production by one additional unit. In practical terms:

In the problem, computing \(P'(2000)\) yields approximately 24.00, suggesting that producing one additional HDTV unit when \(x = 2000\) results in an increased profit of $24.00.

• This value informs stakeholders about the effects of slight adjustments in production size.
• Businesses use this result to decide whether to increase production beyond a specific target level.

Thus, interpreting derivatives helps companies optimize production levels to achieve maximum profitability.