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Microeconomic Theory: Basic Principles and Extensions , 12th Edition Solution Manual

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    Microeconomic Theory: Basic Principles and Extensions 12 th Edition Solutions Manual Walter Nicholson & Christopher Snyder

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    The problems in this chapter are primarily mathematical. They are intended to give students some practice with the concepts introduced in Chapter 2, but the problems in themselves offer few economic insights. Consequently, no commentary is provided. Results from some of the analytical problems are used in later chapters, however, and in those cases the student will be directed back to this chapter . Solutions 2.1 2 2 ( , ) 4 3 . f x y x y   a. 8 , x f x  6 . y f y  b. Constraining ( , ) 16 f x y  creates an implicit function between the variables. The slope of this function is given by 8 6 x y f dy x dx f y     for combinations of x and y that satisfy the constraint. c. Since (1, 2) 16 f  , we know that at this point 8 1 2 6 2 3 dy dx       . d . The ( , ) 16 f x y  contour line is an e llipse centered at the origin. The slope of the line at any point is given by 8 6 . dy dx x y   Notice that this slope becomes more negative as x increases and y decreases. 2.2 a. Profits are given by 2 2 40 100. R C q q        The maximum value is found by setting the derivative equal to 0 : 4 40 0 d = q + dq    , i mplies * 10 q  and * 100.   b. Since 2 4 0, 2 d = dq    this is a global maximum. c. 70 2 . MR = dR dq = q  2 30. MC = dC dq = q  S o , * 10 q  obeys 50. MR MC   CHAPTER 2 : Mathematics for Microeconomics

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    Chapter 2: Mathematics for Microeconomics 2 2.3 First, use the s ubstitution method. Substituting 1 y x   yields 2 ( ) ( ,1 ) (1 ) . f x f x x x x x x       Taking the first - order condition, ( ) 1 2 0, f x = x    and solving yields * 0.5, x = * 0.5 y = , and * * * ( ) ( , ) 0.25. f x f x y =  S ince * ( ) 2 0, f x     this is a local and global maximum. Next, use the Lagrange m ethod . The Lagrangian is (1 ). xy x y      L The first - order conditions are 0, 0, 1 0. x y = y = = x = = x y =        L L L So lving simultaneously, . x y  U sing the constraint gives * * 0.5, x y   0.5, =  and * * 0.25. x y  2.4 Setting up the Lagrangian , (0.25 ). x y xy      L The first - order conditions are 1 , 1 , 0.25 0. x y y x xy           L L L So . x y  Using the constraint 2 ( 0.25) xy x   gives * * 0.5 x y   and 2. =  Note that the solution is the same here as in Problem 2.3, but here the value for the Lagrangian multiplier is the reciprocal of the value in Problem 2.3. 2.5 a. The height of the ball is given by 2 ( ) 0.5 40 . f t gt t    The value of t for which height is maximized is found by using the first - order condition : 40 0, df dt = gt    implying * 40 . t g  b. Substituting for * , t 2 * 40 40 800 ( ) 0.5 40 . f t g g g g                 Hence, * 2 ( ) 800 . df t dg g   c. Differentiation of the original function at its optimal value yield s * * 2 ( ) 0.5( ) . df t t dg   B ecause the optimal value of t d epends on , g

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    Chapter 2: Mathematics for Microeconomics 3 2 * * 2 ( ) 40 800 0.5( ) 0.5 , 2 df t = t dg g g            a s was also shown in part ( c ) . d. If 32, g  * 5 4. t  Maximum height is 800 32 25.  If 32.1, g  maximum height is 800 32.1 24.92,  a reduction of 0 .08. This coul d have been predicted from the envelope t heorem , since * 2 800 25 ( ) (0.01) 0.08. 32 32 df t dg                   2.6 a. This is the volume of a rectangular solid made from a piece of metal, w hich is x by 3 x with the defined corner squares removed. b. The first - order condition for maximum volume is given by 2 2 3 16 12 0. V x xt t t       Applying the quadratic formula to this expression yields 2 2 16 256 144 16 10.6 0.225 . 24 24 x x x x x t x       The second value given by the quadratic (1.11 ) x is obviously extraneou s. c. If 0.225 , t x  3 3 3 3 0.67 0.04 0.05 0.68 . V x x x x     S o volume increases without limit . d. This would require a solution using the Lagrangian method. The optimal solution requires solving three no nlinear simultaneous equations , a task not undertaken here. But it seems clear that the solution would involve a different relationship between t and x t han in p arts ( a – c ) . 2.7 a. Set up the Lagrangian : 1 2 1 2 5ln ( ). x x k x x       L The first - order conditions are 2 1 2 1 2 5 1 0 0, 0. , x x = x = k x x            L L L Hence, 2 1 5 . x    With 10, k  the optimal solution is * * 1 2 5. x x   b. With 4, k  solving the first - order conditions yields * 1 1 x   and * 2 5. x  c. If all variables must be nonnegative , it is clear that any positive value for 1 x reduces

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    Chapter 2: Mathematics for Microeconomics 4 d. If 20, k  optimal solution is * 1 15, x  * 2 5, x  and * 15 5ln5. y   Because 2 x provides a diminishing marginal increment to y as its value increases , whereas 1 x does not, all optimal solutions require tha t once 2 x reaches 5, any extra amounts be devoted entirely to 1 . x In consumer theory , this function can be used to illustrate how diminishing marginal usefulness can be modeled in a very simple setting. 2.8 a. Because MC is the derivative of , TC TC is an antiderivative of . MC By the fundamental theorem of calculus , 0 ( ) ( ) (0), q MC x dx TC q TC    where (0) TC is the fixed cost, which we will denote (0) TC K  for short. Rearranging, 0 0 2 0 2 ( ) ( ) ( 1) 2 . 2 q q x q x TC q MC x dx K x dx K x x K q q K                      b. For profit maximization, ( ) 1, p MC q q    implying 1. q p   But 15 p  implies 14. q  Profit are 2 ( ) 14 15 14 14 2 98 . TR TC pq TC q K K                 If the firm is just breaking even, profit equals 0, implying fixed cost is 98. K  c. When 20 p  and 19, q  follow the same steps as in part ( b ) , substituting fixed cost 98. K  Profit are

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    Chapter 2: Mathematics for Microeconomics 5 2 ( ) 19 20 19 19 2 180.5 98 82.5. TR TC pq TC q K                  d. Assuming profit maximization, we have 2 2 ( ) ( ) ( 1) ( 1) ( 1) 98 2 ( 1) 98. 2 p pq TC q p p p p p                    e. i . Using the above equation, ( 20) ( 15) 82.5 0 82.5. p p         ii. The envelope theorem states that * ( ). d dp q p   That i s, the derivative of the profit function yields this firm’s supply function. Integrating over p shows the change in profits by the fundamental theorem of calculus: 20 15 20 15 20 2 15 (20) (15) ( 1) 2 180 97.5 82.5. p p d dp dp p dp p p                       Analytical Problems 2.9 Concave and quasi - concave functions The proof is most easily accomplished through the use of the matr ix algebra of quadratic forms. See, for example, M as Colell et al ., 1995, pp. 937 – 939. Intuitively, because concave functions lie below any tangent plane, their lev el curves must also be convex. But the converse is n ot true. Quasi - concave functions may exhibit “ increasing returns to scale”; even though their level curves are convex, they may rise above the tangent plane when all variables are increased together.

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    Chapter 2: Mathematics for Microeconomics 6 A counter example would be the Cobb – Douglas function , which is always quasi - c oncave, but convex when 1.     2.10 The Cobb – Douglas function a. 1 2 1 2 2 11 1 1 2 22 1 2 1 1 12 21 1 1 1 2 1 2 0, ( 1) 0, ( 1) 0, 0. 0, f x x f x x f x x x f f x x f x                                      Clearly, all the terms in Equation 2.114 are negative. b. A contour line is found by setting the function equal to a constant : 1 2 , y c x x     implying 1 2 1 . x c x      Hence, 2 1 0. dx dx  Further, 2 2 2 1 0, d x dx  implying t he countour line is convex. c. Using E quation 2.98, 2 2 2 2 2 11 22 12 1 2 (1 ) , f f f x x            which is negative for 1.     2.11 The power function a. Since 0 y   and 0, y   the function is concave. b. Because 11 22 , 0 f f  and 12 21 0, f f   Equation 2.98 is satisfie d, and the function is concave. Because 1 2 , 0, f f  Equation 2.114 is also satisfie d, s o the function is quasi - concave. c. y is quasi - concave as is . y  However, y is not concave for 1.   This can be shown most ea sily by 1 2 1 2 (2 , 2 ) 2 ( , ). f x x f x x   2.12 Proof of envelope theorem

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    Chapter 2: Mathematics for Microeconomics 7 a. The Lagrangian for this problem is 1 2 1 2 1 2 ( , , ) ( , , ) ( , , ). x x a f x x a g x x a    L The first - order conditions are 1 1 1 2 2 2 0, 0, 0. f g f g g            L L L b. , c. Multiplication of each first - order condition by the appropriate deriviative yields 2 2 1 1 1 2 1 2 0. dx dx dx dx f f g g da da da da            d. The optimal value of f is given by   1 2 ( ), ( ), . f x a x a a Differentiation of this with respect to a shows how this optimal value changes with a : * 1 2 1 2 . a da dx dx d df f f f da a    e. Di fferentiation of the constraint   1 2 ( ), ( ), 0 g x a x a a  yields 1 1 2 2 0 . a da dx dx dg g g g da da     f. Multiplying the results from part ( e ) by  and using parts ( b ) and ( c ) yields * . a a a df f g da     L This proves the envelope theorem. g . In Example 2.8 , we showed that 8. P   Th is sh ows how much an extra unit of perimeter would raise the enclosed area. Direct differentiation of the original Lagrangian shows also that * . P dA dP    L This shows that the Lagrange multiplier does indeed show this incremental gain in this problem. 2.13 Taylor ap proximations a. A function in one variable is concave if ( ) 0. f x   Using the quadratic Taylor formula to approximate this function at point a : 2 ( ) ( ) ( )( ) 0.5 ( )( ) ( ) ( )( ). f x f a f a x a f a x a f a f a x a            The in equality holds because ( ) 0. f a   But the right - hand side of this equation is

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    Chapter 2: Mathematics for Microeconomics 8 the equation for the tangent to the function at point . a S o w e have shown that any concave function must lie on or below the tangent to the function at that point. b. A function in two variables is concave if 2 11 22 12 0. f f f   Hence , the quadratic form 2 2 11 12 22 ) ( 2 f dx f dx dy f dy   will also be negative. But this says that the final portion of the Taylor expansion will be negative (by setting dx x a   and dy y b   ) , and hence the function will be below its tangent plane. 2 .14 More on expected value a. The tangent to ( ) g x at the point ( ) E x will have the form ( ) c dx g x   for all values of x and ( ) ( ( )). c dE x g E x   But, because the line c dx  is above the function ( ) g x , we know ( ( )) ( ) ( ) ( ( )). E g x E c dx c dE x g E x      This proves Jensen’s inequality. b. Use the same procedure as in part ( a ) , but reverse the inequalities. c. Let 1 ( ), u F x   ( ), du f x   , x v  and . dx dv        0 0 0 1 ( ) (1 ( )) ( ) 0 ( ) ( ). x x F x dx F x x f x xdx E x E x               d. Use the hint to break up the integral defining expected value: 1 0 1 1 ( ) ( ) ( ) ( ) ( ) ( ) ( ). t t t t t E x t xf x dx xf x dx t t xf x dx t tf x dx f x dx P x t                          e. 1. Show that this function integrates to 1 :

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    Chapter 2: Mathematics for Microeconomics 9 3 2 1 1 ( ) 2 1. x x f x dx x dx x              2. Calculate the cumulative distribution function: 3 2 2 1 1 ( ) 2 1 . x t x t F x t dt t x            3. Using the result from p art ( c ) :   2 1 1 1 1 ( ) 1 ( ) 1. x x E x F x dx x dx x               4. T o show Markov’s inequality use 2 1 ( ) ( ) 1 ( ) . E x P x t F t t t t         f. 1. Show that the PDF integrates to 1: 2 2 2 3 1 1 8 1 1. 3 9 9 9 x x x x dx                2. Calculate the expected value: 2 2 3 4 1 1 15 5 ( ) . 3 12 12 4 x x x x E x dx         3. Calculate ( 1 0 P x    ): 0 0 2 3 1 1 1 . 3 9 9 x x x x dx       4. All we must do is adjust the PDF so that it now sums to 1 over the new, smaller interval. Since ( ) 8 9, P A  2 ( ) 3 ( | ) defined on 0 2. 8 9 8 f x x f x A x     5. The expected value is again found through integration: 2 2 3 4 0 0 3 3 3 ( | ) . 8 32 2 x x x x E x A dx       6. E liminating the lowest values of x increases the expected value of the remaining values. 2.15 More on variances a. This is just an application of the definition of variance:

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    Chapter 2: Mathematics for Microeconomics 10   2 2 2 2 2 2 2 2 Var( ) ( ) 2 ( ) [ ( )] ( ) . 2[ ( )] [ ( )] ( ) [ ( )] x E x E x E x xE x E x E x E x E x E x E x               b. Here , we let x y x    and apply Markov’s inequality to y and remember that x can only take on positive values. 2 2 2 2 2 2 ( ) ( ) ( ) . x E y P y k P y k k k       c. Let , i x 1, , i n   be n independent random variables each with expected value  and varianc e 2 .  1 . n i i E x n                2 2 2 1 Var . n i i x n                Now , let   1 . n i i x x n    ( ) . n E x n     2 2 2 Var( ) . n x n n     d. Let 1 2 (1 ) X kx k x    and ( ) (1 ) . E X k k        2 2 2 2 2 2 Var( ) (1 ) (2 2 1) . X k k k k          2 Var( ) (4 2) 0. d X k dk     Hence , variance is minimized for 0.5. k  In this case, 2 Var( ) 0 . .5 X   If 0.7, k  2 Var( ) 0.58 X   ( not much of an increase ) . e. Suppose that 2 1 Var( ) x   and 2 2 Var( ) . x r   Now 2 2 2 2 2 2 Var( ) (1 ) (1 ) 2 . X k k r r k kr r                 2 Var( ) 2(1 ) 2 0. d X r k r dk      . 1 r k r   For example, if 2, r  then 2 3 k  , and optimal diversification requires that the lower risk asset constitute two - thirds of the portfolio. Note, however, that it is still

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    Chapter 2: Mathematics for Microeconomics 11 optimal to have some of the higher risk asset because asset returns are independent. 2.16 More on covariances a. This is a dire ct result of the definition of covariance:   Cov( , ) ( ( ))( ( )) [ ( ) ( ) ( ) ( )] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ). x y E x E x y E y E xy xE y yE x E x E y E xy E x E y E y E x E x E y E xy E x E y              b. 2 2 2 2 2 2 2 2 2 2 2 2 Var( ) [( ) ] [ ( )] ( ) 2 ( ) ( ) [ ( )] 2 E( )E( ) [E( )] Var( ) Var( ) 2 Cov( , ). ax by E ax by E ax by a x abE xy b y a E x ab x y b y a x b y a E x E b y               T he final line is a result of Problems 2.15a and 2.16a. c. The presence of the covariance term in the result of Problem 2.16b suggests that the results would differ. In the tw o - v ariable case, however, this is not necessarily the situation. For example, suppose that x and y are identically distributed and that 2 Cov( , ) . x y r   Using the prior notation, 2 2 2 2 2 Var( ) (1 ) 2 . (1 ) X k k k k r         The first - or der condition for a minimum is 2 (4 2 2 4 ) 0, k r rk      implying * 2 2 0.5. 4 4 r k r     Regardless of the value of . r With more than two random variables, however, covariances may indeed affect optimal weightings. d. If 1 2 , x kx  the correlation coefficient will be either 1  ( if k is positive) or 1  ( if k is negative ), since k will factor out of the definition leaving only the ratio of the common variance of the tw o variables. With less than a perfect linear relationship   0.5 | Cov( , ) | Var( )Var( . ) x y x y  e. If , y x       Cov( , ) ( ( ))( ( )) [( ( ))( ( ))] Var( ). x y E x E x y E y E x E x x E x x              

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    Chapter 2: Mathematics for Microeconomics 12 Hence, Cov( , ) . Var( ) x y x  

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    13 13 These problems provide some practice in examining utility functions by looking at indifference curve maps and at a few functional forms. The primary focus is on illustrating the notion of quasi - concavity (a diminishing MRS) in various contexts. The concepts of the budget constraint and utility maximization are not used until the next chapter. Comments on Problems 3.1 This problem requires students to graph indifference curves for a variety of functions, some of which are not quasi - concave. 3.2 This problem introduces the formal definition of quasi - concavity (from Chapter 2) to be applied to the functions studied graphically in Problem 3.1. 3.3 This problem shows that diminishing marginal utility is not required to obtain a diminishing MRS . All of the functions are monotonic transformations of one another, so this problem illustrates that diminishing MRS is preserved by monotonic transformations but diminishing marginal utility is not. 3.4 This problem focuses on whether some simple utility functions exhibit convex indifference curves. 3.5 This problem is an exploration of the fixed - proportions utility function. The problem also shows how the goods in such problems can be treated as a composite commodity. 3.6 This problem asks students to use their imaginations to explain how advertising slogans might be captured in the form of a utility function . 3.7 This problem shows how utility functions can be inferred from MRS segments. It is a very simple example of “integrability . ” 3.8 This problem offers some practice in deriving utility functions from indifference curve specifications. CHAPTER 3 : Preferences and Utility

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    Chapter 3: Preferences and Utility 14 Analytical Problems 3.9 Initial endowments. This problem shows how initial endowments can be treated in simple indifference curve analysis. 3.10 Cobb – Douglas utility. This problem provides some exercises with the Cobb – Douglas function , including how to integrate subsistence levels of consumption into the functional form. 3.11 Independent marginal utilities. This problem shows how analysis can be simplified if the cro ss - partials of the utility function are zero. 3.12 CES utility. This problem shows how distributional weights can be incorporated into the CES form introduced in the chapter without changing the basic conclusions about the function. 3.13 The quasi - linear function. This problem provides a brief introduction to the quasi - linear form , which (in later chapters) will be used to illustrate a number of interesting outcomes. 3.14 Preference relations . This problem provides a very brief introduction to how preferences can be treated formally with set - theoretic concepts. 3.15 The benefit function . This problem introduces Luenberger’s notion of reducing preferences to a cardinal number of replications of a basic bundle of goods. Solutions 3.1 Her e w e calculate the MRS for each of these functions: a. 3 . 1 x y U MRS U   M RS is constant. b. 0.5 0.5 0.5( ) . 0.5( ) x y U y x y MRS U y x x     Convex; MRS is diminishing. c. 0.5 0.5 . 1 x y U x MRS U    MRS is diminishing . d. 2 2 0.5 2 2 0.5 2 0.5( ) 2 . 0.5( ) x y U x x MRS x y y U x y y          MRS is increasing.

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