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Suppose a firm's total revenues depend on the amount produced ( \(q\) ) according to the function \\[ R=70 q-q^{2} \\] Total costs also depend on \(q\) \\[ C=q^{2}+30 q+100 \\] a. What level of output should the firm produce to maximize profits \((R-C) ?\) What will profits be? b. Show that the second-order conditions for a maximum are satisfied at the output level found in part (a). c. Does the solution calculated here obey the "marginal revenue equals marginal cost" rule? Explain.

Short Answer

Expert verified
Answer: a. The output level that maximizes profit is 10 units, and the profit at that level is $100. b. Yes, the second-order conditions for a maximum are satisfied at the output level of 10 units. c. Yes, the solution follows the "marginal revenue equals marginal cost" rule as both marginal revenue and marginal cost are equal at the output level of 10 units.

Step by step solution

01

Calculate Profit Function

First, we need to find the profit function, which is the difference between total revenues and total costs. The profit function is given by: \\[ P(q) = R(q) - C(q) = (70q - q^2) - (q^2 + 30q + 100) \\] Simplify the profit function: \\[ P(q) = 70q - q^2 - q^2 - 30q - 100 = -2q^2 + 40q - 100 \\]
02

Find the First Derivative of the Profit Function

In order to find the maximum profit, we should find the first derivative of the profit function with respect to \(q\) and then set it to zero. The first derivative is given by: \\[ \frac{dP}{dq} = \frac{d}{dq}(-2q^2 + 40q - 100) \\] Differentiate with respect to \(q\): \\[ \frac{dP}{dq} = -4q + 40 \\]
03

Find the Output Level that Maximizes Profit

Now we'll set the first derivative equal to zero and solve for \(q\): \\[ -4q + 40 = 0 \Rightarrow q = 10 \\] At \(q=10\), the profit function has a critical point. Next, let's verify that it's a maximum.
04

Find the Second Derivative of the Profit Function

To check if the second-order conditions are satisfied at \(q=10\), find the second derivative of the profit function: \\[ \frac{d^2 P}{dq^2} = \frac{d^2}{dq^2}(-2q^2 + 40q - 100) \\] Differentiate with respect to \(q\) again: \\[ \frac{d^2 P}{dq^2} = -4 \\]
05

Verify the Second-Order Condition

Since the second derivative of the profit function is negative, i.e., \(\frac{d^2 P}{dq^2} = -4 < 0\), we conclude that the critical point at \(q=10\) represents a maximum. Now, we'll find the maximum profit value by substituting \(q=10\) into the profit function: \\[ P(10) = -2(10)^2 + 40(10) - 100 = 100 \\] Hence, the firm should produce 10 units of output to maximize its profit at $100.
06

Check if the Marginal Revenue Equals Marginal Cost Rule is Satisfied

Now let's check whether the marginal revenue (\(MR\)) equals marginal cost (\(MC\)) at \(q=10\). First, find the first derivatives of the total revenue and total cost functions: \\[ \frac{dR}{dq} = \frac{d}{dq}(70q - q^2) = 70 - 2q \\ \frac{dC}{dq} = \frac{d}{dq}(q^2 + 30q + 100) = 2q + 30 \\ \\] Next, find the marginal revenue and marginal cost at \(q=10\): \\[ MR(10) = 70 - 2(10) = 50 \\ MC(10) = 2(10) + 30 = 50 \\] At \(q=10\), \(MR=MC=50\). Therefore, the solution follows the "marginal revenue equals marginal cost" rule. To summarize: a. The firm should produce \(10\) units of output to maximize its profit at $100. b. The second-order conditions for a maximum are satisfied at the output level found in part (a) since the second derivative of the profit function is negative. c. The "marginal revenue equals marginal cost" rule is satisfied at the calculated output level (\(q=10\)), as the marginal revenue and marginal cost are equal at this point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Total Revenue
Total Revenue is the total income that a company can earn from selling its goods or services. It depends on the number of products sold and their price.
In this case, the total revenue function is given by \[ R = 70q - q^2 \] where \(q\) represents the quantity of goods produced and sold.
  • The term \(70q\) indicates the revenue generated per unit by selling \(q\) units at a price of $70 each.
  • The term \(q^2\) reflects the decrease in revenue due to factors such as increased competition or diminishing returns from overproduction.
Understanding Total Revenue helps businesses assess whether they are generating the expected returns from their sales efforts and guides them in adjusting strategies to optimize their earnings.
Total Cost
Total Cost encompasses all expenses a business incurs in the production of goods or services. This typically includes raw materials, labor, and overhead costs.
For our problem, the total cost function is \[ C = q^2 + 30q + 100 \] indicating how costs change with varying output levels \(q\).
  • The term \(q^2\) suggests costs related to scaling up production efforts, where expenses might accelerate as more units are produced.
  • \(30q\) covers additional variable costs like labor and materials required for each unit produced.
  • The constant term \(100\) represents fixed costs, unavoidable expenses regardless of output level, such as rent or salaries.
Total Cost analysis enables businesses to evaluate their spending and optimize their cost management strategies to achieve maximum profitability mix.
Marginal Revenue
Marginal Revenue refers to the additional income received from the sale of one more unit of a product. It is crucial for firms trying to determine the optimal production level to maximize profit.
In our exercise, the Marginal Revenue is calculated as the derivative of the Total Revenue function:\[\frac{dR}{dq} = 70 - 2q \]
  • This calculation shows how each additional unit affects overall revenue.
  • At the output level of \(q=10\), the marginal revenue is 50, indicating additional revenue will be $50 for each extra unit sold beyond 10 units.
Understanding Marginal Revenue helps businesses optimize their sales targets to ensure that any additional production contributes positively to overall revenue.
Marginal Cost
Marginal Cost is the expense of producing one extra unit and is critical in determining production efficiency. This metric guides firms on how much output to produce.
In the context of this problem, the Marginal Cost comes from the derivative of the Total Cost function:\[\frac{dC}{dq} = 2q + 30\]
  • This outcome reveals how costs increase with additional units of production.
  • At the optimal quantity \(q=10\), the marginal cost is 50, meaning it costs an additional $50 to produce each new unit beyond 10.
By comparing Marginal Revenue and Marginal Cost, businesses can identify the production level where profits are maximized, ensuring no resources are wasted in overproduction or underproduction.

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Most popular questions from this chapter

One of the most important functions we will encounter in this book is the Cobb-Douglas function: \\[ y=\left(x_{1}\right)^{\alpha}\left(x_{2}\right)^{\beta} \\] where \(\alpha\) and \(\beta\) are positive constants that are each less than 1 a. Show that this function is quasi-concave using a "brute force" method by applying Equation 2.114 b. Show that the Cobb-Douglas function is quasi-concave by showing that any contour line of the form \(y=c\) (where \(c\) is any positive constant is convex and therefore that the set of points for which \(y>c\) is a convex set. c. Show that if \(\alpha+\beta>1\) then the Cobb-Douglas function is not concave (thereby illustrating again that not all quasiconcave functions are concave). Note: The Cobb-Douglas function is discussed further in the Extensions to this chapter.

Here are a few useful relationships related to the covariance of two random variables, \(x_{1}\) and \(x_{2}\) a show that \(\operatorname{Cov}\left(x_{1}, x_{2}\right)=E\left(x_{1} x_{2}\right)-E\left(x_{1}\right) E\left(x_{2}\right) .\) An important implication of this is that if \(\operatorname{Cov}\left(x_{1}, x_{2}\right)=0, E\left(x_{1} x_{2}\right)\) \(=E\left(x_{1}\right) E\left(x_{2}\right) .\) That is, the expected value of a product of two random variables is the product of these variables' expected values. b. Show that \(\operatorname{Var}\left(a x_{1}+b x_{2}\right)=a^{2} \operatorname{Var}\left(x_{1}\right)+b^{2} \operatorname{Var}\left(x_{2}\right)+2 a b \operatorname{Cov}\left(x_{1}, x_{2}\right)\) c. In Problem \(2.15 \mathrm{d}\) we looked at the variance of \(X=k x_{1}+(1-k) x_{2} \quad 0 \leq k \leq 1 .\) Is the conclusion that this variance is minimized for \(k=0.5\) changed by considering cases where \(\operatorname{Cov}\left(x_{1}, x_{2}\right) \neq 0 ?\) d. The correlation coefficient between two random variables is defined as \\[ \operatorname{Corr}\left(x_{1}, x_{2}\right)=\frac{\operatorname{Cov}\left(x_{1}, x_{2}\right)}{\sqrt{\operatorname{Var}\left(x_{1}\right) \operatorname{Var}\left(x_{2}\right)}} \\] Explain why \(-1 \leq \operatorname{Corr}\left(x_{1}, x_{2}\right) \leq 1\) and provide some intuition for this result. e. Suppose that the random variable \(y\) is related to the random variable \(x\) by the linear equation \(y=\alpha+\beta x\). Show that \\[ \beta=\frac{\operatorname{Cov}(y, x)}{\operatorname{Var}(x)} \\] Here \(\beta\) is sometimes called the (theoretical) regression coefficient of \(y\) on \(x\). With actual data, the sample analog of this expression is the ordinary least squares (OLS) regression coefficient.

Show that if \(f\left(x_{1}, x_{2}\right)\) is a concave function then it is also a quasi-concave function. Do this by comparing Equation 2.114 (defining quasi-concavity) with Equation 2.98 (defining concavity). Can you give an intuitive reason for this result? Is the converse of the statement true? Are quasi-concave functions necessarily concave? If not, give a counterexample.

Because we use the envelope theorem in constrained optimization problems often in the text, proving this theorem in a simple case may help develop some intuition. Thus, suppose we wish to maximize a function of two variables and that the value of this function also depends on a parameter, \(a: f\left(x_{1}, x_{2}, a\right) .\) This maximization problem is subject to a constraint that can be written as: \(g\left(x_{1}, x_{2}, a\right)=0\) a. Write out the Lagrangian expression and the first-order conditions for this problem. b. Sum the two first-order conditions involving the \(x^{\prime}\) s. c. Now differentiate the above sum with respect to \(a\) - this shows how the \(x\) 's must change as \(a\) changes while requiring that the first-order conditions continue to hold. A. As we showed in the chapter, both the objective function and the constraint in this problem can be stated as functions of \(a: f\left(x_{1}(a), x_{2}(a), a\right), g\left(x_{1}(a), x_{2}(a), a\right)=0 .\) Differentiate the first of these with respect to \(a\). This shows how the value of the objective changes as \(a\) changes while keeping the \(x^{\prime}\) s at their optimal values. You should have terms that involve the \(x^{\prime}\) s and a single term in \(\partial f / \partial a\) e. Now differentiate the constraint as formulated in part (d) with respect to \(a\). You should have terms in the \(x\) 's and a single term in \(\partial g / \partial a\) f. Multiply the results from part (e) by \(\lambda\) (the Lagrange multiplier), and use this together with the first-order conditions from part (c) to substitute into the derivative from part (d). You should be able to show that \\[ \frac{d f\left(x_{1}(a), x_{2}(a), a\right)}{d a}=\frac{\partial f}{\partial a}+\lambda \frac{\partial g}{\partial a} \\] which is just the partial derivative of the Lagrangian expression when all the \(x^{\prime}\) 's are at their optimal values. This proves the envelope theorem. Explain intuitively how the various parts of this proof impose the condition that the \(x\) 's are constantly being adjusted to be at their optimal values. g. Return to Example 2.8 and explain how the envelope theorem can be applied to changes in the fence perimeter \(P\) -that is, how do changes in \(P\) affect the size of the area that can be fenced? Show that in this case the envelope theorem illustrates how the Lagrange multiplier puts a value on the constraint.

Consider the following constrained maximization problem: \\[ \begin{array}{ll} \text { maximize } & y=x_{1}+5 \ln x_{2} \\ \text { subject to } & k-x_{1}-x_{2}=0 \end{array} \\] where \(k\) is a constant that can be assigned any specific value. a. Show that if \(k=10\), this problem can be solved as one involving only equality constraints. b. Show that solving this problem for \(k=4\) requires that \(x_{1}=-1\) c. If the \(x^{\prime}\) s in this problem must be non-negative, what is the optimal solution when \(k=4 ?\) (This problem may be solved either intuitively or using the methods outlined in the chapter.) d. What is the solution for this problem when \(k=20 ?\) What do you conclude by comparing this solution with the solution for part (a)? Note: This problem involves what is called a quasi-linear function. Such functions provide important examples of some types of behavior in consumer theory-as we shall see.

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