Suppose that a firm has a marginal cost function given by \(M C(q)=q+1\) What is this firm's total cost function? Explain why total costs are known only up to a constant of integration, which represents fixed costs. b. As you may know from an earlier economics course, if a firm takes price ( \(p\) ) as given in its decisions then it will produce that output for which \(p=M C(q)\). If the firm follows this profit-maximizing rule, how much will it produce when \(p=15 ?\) Assuming that the firm is just breaking even at this price, what are fixed costs? c. How much will profits for this firm increase if price increases to \(20 ?\) d. Show that, if we continue to assume profit maximization, then this firm's profits can be expressed solely as a function of the price it receives for its output. e. Show that the increase in profits from \(p=15\) to \(p=20\) can be calculated in two ways: (i) directly from the equation derived in part (d); and (ii) by integrating the inverse marginal cost function \(\left[M C^{-1}(p)=p-1\right]\) from \(p=15\) to \(p=20\) Explain this result intuitively using the envelope theorem.

Short Answer

Expert verified
Short Answer: To find the total cost function, we integrated the given marginal cost function MC(q) = q + 1 to get C(q) = (1/2)q^2 + q + F, where F represents fixed costs. At p=15 (break-even point), we found the profit-maximizing output level, q=14, and calculated the fixed costs, F=42. When the price increased to $20, the new profit-maximizing output level became q=19. The increase in profits from p=15 to p=20 can be calculated using the profit equation, Profit(p) = p(q) - C(q), where q = p - 1, and by integrating the inverse marginal cost function. The result can be explained intuitively through the envelope theorem, which states that the gradient of the profit function concerning the changing variable (price) will be equivalent to the difference between the new and old variable values.

Step by step solution

01

Find the total cost function

Given MC(q) = q + 1, integrate the marginal cost function to find the total cost function C(q): C(q) = ∫(MC(q)dq) = ∫(q + 1)dq
02

Calculate the total cost function

After integrating MC(q) with respect to q, we get: C(q) = (1/2)q^2 + q + F where F represents fixed costs.
03

Determine the profit-maximizing output level at p=15

When the firm is profit maximizing, p = MC(q). At p=15: 15 = MC(q) = q + 1 Solving for q, we get: q = 14
04

Calculate fixed costs

The question states that the firm is just breaking even at p=15. In this case, total revenues equal total costs: Total Revenue = Total Cost 15 * 14 = C(14) 210 = (1/2)(14)^2 + 14 + F Solve for F: F = 42
05

Calculate the increase in profits when price increases to $20

First, find the new profit-maximizing output level when price increases to $20: 20 = MC(q) = q + 1 Solving for q, we get: q = 19 Now, calculate the increase in profits: ∆Profit = (20 * 19) - C(19) - ((15 * 14) - C(14))
06

Show that the firm's profits can be expressed solely as a function of price

Profit = Total Revenue - Total Cost Profit(p) = p*q - C(q) We know that p = MC(q) = q + 1. Thus, q = p - 1 Substitute this back into the profit equation: Profit(p) = p(p - 1) - C(p - 1)
07

Calculate the increase in profit from p=15 to p=20 using the inverse marginal cost function

Using the inverse marginal cost function MC^(-1)(p) = p - 1, integrate from p=15 to p=20: ∆Profit = ∫[(20-1) - (15-1)]dp The result should be the same as the previous method.
08

Explain the result intuitively using the envelope theorem

The envelope theorem states that if a function (in this case, profit) is dependent on changing variables (here, price), the gradient of the function with respect to the changing variable will be equivalent to the difference between the new and the old values of the variable. This means that the increase in profits from p=15 to p=20 can be found by evaluating the integral of the inverse marginal cost function.

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Key Concepts

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

Total Cost Function
Understanding the total cost function is crucial for businesses trying to comprehend how their expenses behave at different production levels. In economics, the total cost function represents the combined cost of production, including both fixed and variable costs. Fixed costs are expenses that do not change with the quantity of output produced, such as rent or salaries, while variable costs change with production volume.

Given the marginal cost function, which is the cost of producing an additional unit of output, the total cost function can be derived by integration. For example, for a marginal cost function given by \( MC(q) = q + 1 \), the total cost function \( C(q) \) can be found by integrating the marginal cost function with respect to quantity \( q \). The integration gives \( C(q) = \frac{1}{2}q^2 + q + F \), where \( F \) is the integration constant representing the fixed costs.

It is important to underscore why total costs are known only up to a constant of integration. This simply means that we can determine the behavior of the variable costs, but to obtain a complete total cost function, we must also know the fixed costs, which remain constant irrespective of how much is produced. Since fixed costs do not change, they are represented by a constant in the cost function equation.
Profit Maximization
Profit maximization is a key goal for any business. It represents the process by which a firm determines the price and output level that returns the greatest profit. To achieve profit maximization, firms set production such that the marginal cost of producing an extra unit is equal to the price of that unit, which is where marginal revenue equals marginal cost.

For the firm described in our exercise, the rule for profit maximization is given by the price \( p \) equaling the marginal cost \( MC(q) \). Therefore, when the price, \( p \), is set at 15, the firm will produce the quantity where \( p = MC(q) \), which happens to be 14 units. This quantity maximizes the firm’s profit given the price.

The concept also extends to determining fixed costs when we know that the firm is just breaking even at a certain price, implying total revenue equals total costs. In this scenario, fixed costs can be solved by substituting the profit-maximizing output level and the price into the total cost function and setting the total cost equal to total revenue. This enables us to determine the constant \( F \), which stands for fixed costs in our total cost function.
Envelope Theorem
The envelope theorem is an elegant component of microeconomic theory that helps understand how a company's profits change when the conditions of its business environment alter, such as the selling price of its products. This theorem tells us that the change in profit resulting from a small change in a parameter, like the selling price, is equivalent to the partial derivative of the profit function with respect to that parameter.

Why does this matter? Imagine the firm raising its price from \( p=15 \) to \( p=20 \) and you want to find out how much profit will increase. Using the envelope theorem, we understand that this profit change can be accurately captured either by direct calculation using the profit function, or by integrating the inverse marginal cost function over the range of the price change.

The true beauty of the theorem lies in its ability to simplify complex economic situations. It allows us to bypass the need for detailed information about all variables and focus only on how a change in price impacts profit. When we integrate the inverse marginal cost function from the old price to the new price, we're applying a practical use of the envelope theorem to predict changes in the firm's profits.

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Most popular questions from this chapter

Another function we will encounter often in this book is the power function: \\[ y=x^{\delta} \\] the form \(y=x^{8} / 8\) to ensure that the derivatives have the proper sign). a. Show that this function is concave (and therefore also, by the result of Problem \(2.9,\) quasi-concave). Notice that the \(\delta=1\) is a special case and that the function is "strictly" concave only for \(\delta<1\) b. Show that the multivariate form of the power function \\[ y=f\left(x_{1}, x_{2}\right)=\left(x_{1}\right)^{8}+\left(x_{2}\right)^{8} \\] is also concave (and quasi-concave). Explain why, in this case, the fact that \(f_{12}=f_{21}=0\) makes the determination of concavity especially simple. c. One way to incorporate "scale" effects into the function described in part (b) is to use the monotonic transformation \\[ g\left(x_{1}, x_{2}\right)=y^{y}=\left[\left(x_{1}\right)^{8}+\left(x_{2}\right)^{8}\right]^{y} \\] where \(\gamma\) is a positive constant. Does this transformation preserve the concavity of the function? Is \(g\) quasi-concave?

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.

A simple way to model the construction of an oil tanker is to start with a large rectangular sheet of steel that is \(x\) feet wide and \(3 x\) feet long. Now cut a smaller square that is \(t\) feet on a side out of each corner of the larger sheet and fold up and weld the sides of the steel sheet to make a traylike structure with no top. a. Show that the volume of oil that can be held by this tray is given by \\[ V=t(x-2 t)(3 x-2 t)=3 t x^{2}-8 t^{2} x+4 t^{3} \\] b. How should \(t\) be chosen to maximize \(V\) for any given value of \(x ?\) c. Is there a value of \(x\) that maximizes the volume of oil that can be carried? d. Suppose that a shipbuilder is constrained to use only 1,000,000 square feet of steel sheet to construct an oil tanker. This constraint can be represented by the equation \(3 x^{2}-4 t^{2}=1,000,000\) (because the builder can return the cut-out squares for credit). How does the solution to this constrained maximum problem compare with the solutions described in parts and (c)?

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.

Suppose that \(f(x, y)=x y .\) Find the maximum value for \(f\) if \(x\) and \(y\) are constrained to sum to \(1 .\) Solve this problem in two ways: by substitution and by using the Lagrange multiplier method.

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