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Bernheim calculusfinal.1.Calculus Supplement to Accompany B. Douglas Bernheim and Michael D. Whinston Microeconomics, 1st edition Anita Alves Pena† Colorado State University October 3, 2008 † Email: anita.pena@colostate.edu. Address: 1771 Campus Delivery, Department of Economics, ColoradoState University, Fort Collins, CO.Part I: Introduction 1.Chapter 1: Preliminaries Note to the Student Your textbook is written to promote “thinking like an economist” in order to solve quan-titative problems, relevant to real world economic decision making, whether or not you arefluent in advanced mathematical tools such as calculus. This supplement, however, extendswhat you have learned in your text by showing how you can solve economic problems usingcalculus. A theme of your book is the separability of the economics from the mathematics.This supplement continues that theme but presents the relationships between economics andcalculus, instead of between economics and algebra. Although you might find the idea of ap-plying calculus concepts to economic problems intimidating at first, be forewarned that youmay find the calculus approaches outlined here to be easier than the algebraic approachesof your textbook in some cases.

In addition, recognizing relationships between economicsand calculus will be useful if you decide to study even more advanced microeconomics in thefuture. Before jumping into economic problem solving, however, let’s introduce one calculusconcept that you may not have studied in your calculus class, but which will be useful forthis supplement. Partial Derivatives One of the central themes of microeconomics, as presented in your textbook, is that op-timal choices are made at the margin. In other words, in microeconomics, we study whetheror not small adjustments (marginal changes) in consumption or production make the agentsin our models better off. We can use calculus to formalize these margins mathematically. You should remember standard derivatives from your calculus class. However, if youtook a course focusing on single variable calculus, you might not have learned about partialderivatives.

Since partial derivatives are a mathematical tool commonly used in economicsand will be used throughout this supplement, let’s define a partial derivative now. When 2.you studied standard one variable derivatives, you considered functions such as f (X). Youtook the derivative of this function f with respect to its single argument X.

For example,if f (X) = X 2, then you were able to calculate the derivative of f with respect to X asdf d(f (X)) d(X 2 ) d(X n )dX = dX = dX = 2X using the common derivative rule that dX = nX n−1. Partial derivatives are useful when you have a function of two or more variables and wantto take a derivative with respect to only one variable while holding the other variable or othervariables constant. Instead of dealing with f (X), you might be dealing with f (X, Y ).

Orbetter yet, you might be dealing with f (X, Y, Z). For the time being though, let’s considerthe two variable case where the function is f (X, Y ). Your objective might be to take thepartial derivative of f with respect to X, or alternately, to take the partial derivative of fwith respect to Y. As you will see in a moment, these calculations are not equivalent. ∂f The partial derivative of f with respect to X is written ∂X. Similarly, the partial deriva- ∂ftive of f with respect to Y is written ∂Y. An easy way to think about calculating the partialderivative of f (X, Y ) with respect to X is to pretend that Y is a constant, not a variable.

D(cf (X)) (X)Recall the constant rule for derivatives that dX = c dfdX. Using this fact, you can takethe derivative of f (X, Y ) with respect to X just like you would take the derivative of f (X)with respect to X, remembering that Y is just a constant. Likewise, if you wanted to takethe partial derivative of f (X, Y ) with respect to Y, just pretend that X is a constant andtake the derivative of f (X, Y ) with respect to Y just as you would solve for the derivativeof f (Y ) with respect to Y. This time, just remember that for your purposes, X is treatedas a constant. Let’s look at a couple of examples.WORKED-OUT PROBLEM The Problem Consider the function f (X, Y ) = X 2 + Y 2. What is ∂f ∂X? What is ∂f ∂Y?

The Solution The problem asks for both the partial derivative of f with respect to X,∂f ∂f ∂f ∂(f (X,Y ))∂X, and for the partial derivative of f with respect to Y, ∂Y. In this case, ∂X = ∂X =∂(X 2 +Y 2 ) d(X 2 ) 2 ∂X = dX + d(Y ) = 2X + 0 = 2X. Since we are treating Y as a constant, we also can dXtreat Y 2 as a constant since a constant to a power is still a constant.

Since the derivative 3.d(Y 2 )of a constant is zero (here dX = 0), this term drops out. Similarly, we can calculate the ∂f ∂(f (X,Y )) ∂(X 2 +Y 2 )partial derivative of f (X, Y ) with respect to Y. In that case, ∂Y = ∂Y = ∂Y =d(X 2 ) 2) dY + d(Y = 0+2Y = 2Y.

Notice that we are treating X as a constant for this calculation. DYWORKED-OUT PROBLEM The Problem Let f (X, Y ) = X 2 Y 2. What is ∂f ∂X? What is ∂f ∂Y? The Solution In this example, the X and Y parts of the function are multiplied byeach other. This is in contrast to the previous worked-out problem where they were added.The general procedure to calculate partial derivatives, however, stays the same.

Specifically,when we are calculating the partial derivative of f with respect to X, we treat Y as if it werea constant, and vice versa when we are calculating the partial derivative of f with respect ∂(X 2 Y 2 )to Y. The partial derivative with respect to X therefore is ∂f ∂X = ∂X = 2XY 2. (Recall d(cf (X)) (X)that our constant rule for derivatives is dX = c dfdX.) The partial derivative of f with ∂(X 2 Y 2 )respect to Y is ∂f ∂Y = ∂Y = 2X 2 Y.ADDITIONAL EXERCISES ∂f ∂f 1. Consider the function f (X, Y ) = 2X + 3Y. Consider the function f (X, Y ) = 2X 2 + 3Y 3. What is ∂f ∂X? What is ∂f ∂Y?

Consider the function f (X, Y ) = 6X 2 Y 3. What is ∂f ∂X? What is ∂f ∂Y?4. Consider the function f (X, Y ) = 9X 1/9 Y 1/3.

What is ∂f ∂X? What is ∂f ∂Y?5. Consider the function f (X, Y ) = X 1/3 Y 1/2. What is ∂f ∂X? What is ∂f ∂Y?6.

Consider the function f (X, Y ) = X 3 Y 2 + 3Y. What is ∂f ∂X?

What is ∂f ∂Y?7. Let f (X, Y ) = X 7 Y 9 + 10X. What is ∂f ∂X? What is ∂f ∂Y? Let f (X, Y ) = 9X 1/9 Y 1/3 + 6X 1/3 Y 1/2. What is ∂f ∂X?

What is ∂f ∂Y?9. Let f (X, Y ) = X 4 Y 2 + 8X 2 Y + 4XY 2. What is ∂f ∂X? What is ∂f ∂Y?10.

Let f (X, Y ) = X 3 Y 3 + 12X 2 Y 2 + 6XY + 10. What is ∂f ∂X? What is ∂f ∂Y?11. Let f (X, Y ) = X 5 Y 6 + X 4 Y 3 + X 1/2 Y 1/4 + 10Y. What is ∂f ∂X?

What is ∂f ∂Y?12. Let f (X, Y ) = 7X 4 Y 3 + 14X 2 Y + 10X + 12Y + 6. What is ∂f ∂X? What is ∂f ∂Y? 6.Chapter 2: Supply and Demand Supply and Demand Functions A demand function for a good describes mathematically how that good’s quantity de-manded changes for different combinations of its price and other related factors such asincome or the price of a related product. A simple demand function might take the linearform Qd = A − BP where A is a constant and B is a slope parameter.

In this simple case,Qd is a function of only one variable, P, which is the price of the product. Note that we ∂Qd dQdcan interpret B in terms of calculus by noting that ∂P = dP = −B.

Similarly, a supplyfunction is a mathematical representation of how quantity supplied of a product varies fordifferent price levels and levels of other relevant factors. A simple linear supply curve mighttake the form Qs = C + DP where C is a constant and D is a slope parameter. In this case,∂Qs dQs∂P = dP = D. Similarly, we can consider a more complicated demand curve which is afunction of additional parameters such as the price of another good (PO ) and income (M):Qd = α − βP + γPO + δM.

We can again characterize the slope parameters given calculus. ∂Qd ∂Qd ∂QdHere, ∂P = −β, ∂PO = γ, and ∂M = δ. Notice that you can extend this framework formore complicated forms of the demand and supply functions.

The general method to findthe slope of the demand or supply function in the direction of a specific variable is to takethe first partial derivative of the relevant function with respect to the variable of interest. Substitutes and Complements Two goods are substitutes if an increase in the price of one causes buyers to demandmore of the other product. Two goods are complements if an increase in the price of onecauses buyers to demand less of the other product. The relationships of substitutes and ofcomplements therefore are relationships between the price of one good and the quantity ofanother good.

Considering the demand curve Qd = α − βP + γPO + δM, we can easily tell ∂Qdif products are substitutes or complements by examining ∂PO, which equals the coefficient 7.γ. If γ 0, then the two goods are substitutes indicating that an increase in the price of arelated good, PO, is associated with an increase in the quantity demanded of the good itself.If γ.

Find more information about:ISBN:007290027X 262207720711101411OCLC Number:156902383Notes:Includes index.Description:xxiii, 833 pages: illustrations (some color); 26 cmContents:1. Introduction -2. Supply and Demand -3. Balancing Benefits and Costs -4. Principles and Preferences -5.

Constraints, Choices, and Demand -6. From Demand to Welfare -7. Technology and Production -8. Profit Maximization -10. Choices Involving Time -11. Choices involving Risk -12. Choices Involving Strategy -13.

Behavioral Economics -14. Equilibrium and Efficiency -15. Market Intervention -16.

Microeconomics Theory Pdf

General Equilibrium, Efficiency, and Equity -17. Monopoly -18. Pricing Policies -19. Oligopoly -20. Externalities and Public Goods -21.

Microeconomics Bernheim Winston 2008 Edition Hotels

Information Imperfections.Responsibility:B. Douglas Bernheim, Michael D. Whinston.More information:.Abstract.