Question

A firm has the following total-cost and demand functions: \(C=\frac{1}{3} Q^{3}-7 Q^{2}+111 Q+50\) \(Q=100-P\) (a) Does the total-cost function satisfy the coefficient restrictions of (9.5)\(?\) (b) Write out the total-revenue function \(R\) in terms of \(Q\) (c) Formulate the total-profit function \(\pi\) in terms of \(Q\) (d) Find the profit-maximizing level of output \(Q^{*}\) (e) What is the maximum profit?

Short answer

(a) Insufficient data to confirm restrictions. (b) \( R(Q) = 100Q - Q^2 \). (c) \( \pi(Q) = -\frac{1}{3} Q^3 + 6 Q^2 - 11 Q - 50 \). (d) \( Q^* = 11 \). (e) Maximum profit is approximately \( 111.33 \).

Step-by-step solution

Verify Total-Cost Function Restrictions

The total-cost function is given by \[ C=\frac{1}{3} Q^{3}-7 Q^{2}+111 Q+50 \]To check if it satisfies certain coefficient restrictions (which might be stated as conditions like positive output, increasing marginal cost, etc.), we need specifics on what these conditions from (9.5) are. However, typically, the cost function should ensure production is feasible and realistic given context, such as ensuring positive and increasing costs with increases in output. Without explicit restrictions from (9.5), generally, this cost function appears reasonable for typical scenarios.

Write Total-Revenue Function

The demand function is:\[ Q = 100 - P \]Solving for price gives:\[ P = 100 - Q \]The total-revenue function, as a function of quantity \( Q \), is:\[ R = P \times Q = (100-Q) \times Q = 100Q - Q^2 \]

Formulate Total-Profit Function

Profit \( \pi \) is total revenue minus total cost. Therefore:\[ \pi(Q) = R(Q) - C(Q) = (100Q - Q^2) - \left( \frac{1}{3} Q^3 - 7 Q^2 + 111 Q + 50 \right) \]Simplifying gives:\[ \pi(Q) = - \frac{1}{3} Q^3 + 6 Q^2 - 11 Q - 50 \]

Find Profit-Maximizing Output

To find the output \( Q^* \) that maximizes profit, take the derivative of the profit function with respect to \( Q \) and set it equal to zero to find critical points.\[ \frac{d\pi}{dQ} = -Q^2 + 12Q - 11 \]Set this equal to zero:\[ -Q^2 + 12Q - 11 = 0 \]Solve this quadratic equation for \( Q \). Using the quadratic formula, \( Q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -1 \), \( b = 12 \), \( c = -11 \):\[ Q = \frac{-12 \pm \sqrt{12^2 - 4\cdot (-1)\cdot (-11)}}{2\cdot(-1)} \]\[ Q = \frac{-12 \pm \sqrt{144 - 44}}{-2} \]\[ Q = \frac{-12 \pm 10}{-2} \]Thus, \( Q = 1 \) or \( Q = 11 \). Evaluate the profit function at these points to find the maximum.

Determine Maximum Profit

Calculate \( \pi(1) \) and \( \pi(11) \):\[ \pi(1) = -\frac{1}{3}(1)^3 + 6(1)^2 - 11(1) - 50 = -\frac{1}{3} + 6 - 11 - 50 = -55.33 \]\[ \pi(11) = -\frac{1}{3}(11)^3 + 6(11)^2 - 11(11) - 50 \]Simplifying:\[ -\frac{1}{3}(11)^3 = -\frac{1}{3}\cdot1331 = -443.67 \]\[ 6(11)^2 = 6\cdot121 = 726 \]\[ 11\cdot11 = 121 \]So:\[ \pi(11) = -443.67 + 726 - 121 - 50 = 111.33 \]The profit-maximizing output \( Q^* = 11 \) yields a maximum profit of approximately \( 111.33 \).

Key concepts

Profit Maximization

Profit maximization is a key objective for businesses, aiming to achieve the highest possible difference between total revenue and total cost. In economics, reaching this goal involves finding the output level where the marginal cost of production equals the marginal revenue, making no further profit possible by altering the level of goods produced.
To determine the profit-maximizing level of output, economists often use calculus to analyze the behavior of profit based on the quantity of goods sold. By deriving the profit function, which is the difference between total revenue and total cost, and finding its critical points, businesses can identify the output level where profit is at its peak.
In our solution, we found that setting the derivative of the profit function \( \pi(Q) = - \frac{1}{3} Q^3 + 6 Q^2 - 11 Q - 50 \) equal to zero enabled us to determine the quantities at which profit could potentially be maximized. By further evaluating these critical points, we identified that producing 11 units yields the maximum profit.

Quadratic Equations in Economics

Quadratic equations frequently appear in economic analyses, especially when exploring relationships involving maximization or minimization, such as cost, revenue, or profit functions.
A standard quadratic equation follows the format \( ax^2 + bx + c = 0 \,\) where \( a, b, \) and \( c \) are constants. Solving such equations often utilizes the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \.\)
In our scenario, understanding how to solve a quadratic equation was crucial to determining the profit-maximizing quantity. After taking the derivative of the profit function and setting it to zero, we applied the quadratic formula to solve for \( Q \.\) The output levels calculated (1 and 11) were the potential candidates where changes in output could dramatically impact the firm's profit.

Derivative in Economics

The concept of derivatives plays a crucial role in economic analyses, especially when examining change rates, such as cost minimization or profit maximization.
Essentially, a derivative represents the slope of a function at a given point, providing insight into how a small change in a variable can influence the function's outcome. In profit analysis, the derivative of the profit function in terms of quantity, \( \frac{d\pi}{dQ} \,\) reveals information about profit changes concerning the quantity produced.
In our exercise, we used the derivative of the profit function to identify points where the profit was neither increasing nor decreasing, known as critical points. These help determine where profits peak and guide firms in deciding optimal production levels to achieve maximum profit. Thus, taking derivatives in economics offers practical insights into strategic decision-making for businesses.

Economic Demand Function

The economic demand function is an equation that describes the relationship between the price of a good and the quantity demanded by consumers. It provides insights into consumer behavior and market dynamics, forming a core part of predicting revenue outcomes.
A typical linear demand function can be expressed as \( Q = a - bP, \) where \( a \) and \( b \) are constants, \( P \) represents price, and \( Q \) represents quantity demanded. Adjustments in price level generally result in changes in the quantity demanded, suggesting how sensitive consumers are to price changes.
In the provided exercise, the demand function \( Q = 100 - P \) was used to derive the total revenue function \( R = PQ. \) By substituting this into the total revenue and subsequently into the profit functions, the demand function played a foundational role in analyzing how different quantity levels impact both pricing strategies and profit maximization.