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

CHAPTER 2: Mathematics for Microeconomics 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 f ( x, y)  4 x2  3 y 2 . a. f x  8x, f y  6 y. b. Constraining f ( x, y)  16 creates an implicit function between the variables.The f dy 8 x slope of this function is given by for combinations of x and y  x  dx fy 6y that satisfy the constraint. c. Since f (1, 2)  16 , we know that at this point d. The f ( x, y)  16 contour line is an ellipse centered at the origin.The slope of the line at any point is given by dy dx   8x 6 y . Notice that this slope becomes more negative as x increases and y decreases. dy 8 1 2   . dx 62 3 2.2 a. Profits are given by   R  C  2q 2  40q  100. The maximum value is found by setting the derivative equal to 0: d =  4q + 40  0 , dq implies q*  10 and  *  100. b. Since d 2 dq =  4  0, this is a global maximum. c. MR = dR dq = 70  2q. MC = dC dq = 2q  30. So, q*  10 obeys MR  MC  50. 2 2 2.3 Chapter 2: Mathematics for Microeconomics First, use the substitution method.Substituting y  1  x yields f ( x)  f ( x,1  x)  x(1  x)  x  x 2 . Taking the first-order condition, f  ( x) = 1  2 x  0, and solving yields x* = 0.5, y* = 0.5 , and f ( x* )  f ( x* , y* ) = 0.25. Since f ( x* )  2  0, this is a local and global maximum. Next, use the Lagrangemethod.The Lagrangian is L  xy   (1  x  y). The firstorder conditions are Lx = y   = 0, Ly = x   = 0, L = 1  x  y = 0. Solving simultaneously, x  y. Using the constraint gives x*  y*  0.5,  = 0.5, and x* y*  0.25. 2.4 Setting up the Lagrangian, L  x  y   (0.25  xy). The first-order conditions are Lx  1   y , L y  1   x, L  0.25  xy  0. So x  y. Using the constraint ( xy  x 2  0.25) gives x*  y*  0.5 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 f (t )  0.5gt 2  40t. The value of t for which height is maximized is found by using the first-order condition: df dt =  gt  40  0, implying t *  40 g . b. Substituting for t * , 2  40   40  800 f (t )  0.5 g    40    . g  g   g  * Hence, df (t * ) 800  2 . dg g c. Differentiation of the original function at its optimal value yields df (t * )  0.5(t * )2 . dg Because the optimal value of t depends on g , Chapter 2: Mathematics for Microeconomics 3 2  40  800 df (t * ) =  0.5(t * )2  0.5    2 , dg g  g  as was also shown in part (c). d. If g  32, t *  5 4. Maximum height is 800 32  25. If g  32.1, maximum height is 800 32.1  24.92, a reduction of 0.08.This could have been predicted from the envelope theorem, since  800   25  df (t * )   dg    (0.01)  0.08. 2   32   32  2.6 a. This is the volume of a rectangular solid made from a piece of metal, which is x by 3x with the defined corner squares removed. b. The first-order condition for maximum volume is given by V  3x 2  16 xt  12t 2  0. t Applying the quadratic formula to this expression yields 16 x  256 x 2  144 x 2 16 x  10.6 x   0.225 x. 24 24 The second value given by the quadratic (1.11x) is obviously extraneous. t c. If t  0.225 x, V  0.67 x3  0.04 x3  0.05×3  0.68×3. Sovolume increases without limit. d. This would require a solution using the Lagrangian method.The optimal solution requires solving three nonlinear simultaneous equations, a task not undertaken here.Butit seems clear that the solution would involve adifferent relationship between t and x than in parts (a–c). 2.7 a. Set up the Lagrangian: L  x1  5ln x2   (k  x1  x2 ). The firstorder conditions are Lx1 = 1   = 0, 5    0, x2 L  k  x1  x2  0. Lx2  Hence,   1  5 x2 . With k  10, the optimal solution is x1*  x2*  5. b. With k  4, solving the first-order conditions yields x1*  1 and x2*  5. c. If all variables must be nonnegative, it is clear that any positive value for x1 reduces y 4 Chapter 2: Mathematics for Microeconomics d. If k  20, optimal solution is x1*  15, x2*  5, and y*  15  5ln5. Because x2 provides a diminishing marginal increment to y as its value increases,whereas x1 does not, all optimal solutions require that once x2 reaches 5, any extra amounts be devoted entirely to x1. 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. Bythe fundamental theorem of calculus, q  MC ( x) dx  TC (q)  TC (0), 0 where TC (0) is the fixed cost, which we will denote TC (0)  K for short.Rearranging, q TC (q )   MC ( x) dx  K 0 q   ( x  1)dx  K 0 xq  x2     x  K  2  x 0  b. q2  q  K. 2 For profit maximization, p  MC (q)  q  1, implying q  p  1. But p  15 implies q  14. Profit are TR  TC  pq  TC (q)  142   15 14    14  K   2   98  K . If the firm is just breaking even, profit equals 0, implying fixed cost is K  98. c. When p  20 and q  19, follow the same steps as in part (b), substituting fixed cost K  98. Profit are Chapter 2: Mathematics for Microeconomics 5 TR  TC  pq  TC (q )  192   20 19    19  K   2   180.5  98  82.5. d. Assuming profit maximization, we have  ( p)  pq  TC (q)  ( p  1)2   p( p  1)    ( p  1)  98  2  2 ( p  1)   98. 2 e. i. Using the above equation,  ( p  20)   ( p  15)  82.5  0  82.5. ii. The envelope theorem states that d dp  q* ( p). That is, thederivative of the profit function yields this firm’s supply function.Integrating over p shows the change in profits by the fundamentaltheorem of calculus: 20 d  (20)   (15)   dp dp 15 20   ( p  1)dp 15 p  20  p2    p  2  p 15  180  97.5  82.5. Analytical Problems 2.9 Concave and quasi-concave functions The proof is most easily accomplished through the use of the matrix algebra of quadratic forms.See, for example, Mas Colell et al.,1995, pp. 937–939.Intuitively, because concave functions lie below any tangent plane, their level curves must also be convex.But the converse is not 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. 6 Chapter 2: Mathematics for Microeconomics A counter example would be the Cobb–Douglas function, which is always quasiconcave, but convex when     1. 2.10 The Cobb–Douglas function a. f1   x1 1 x2  0, f 2   x1 x2 1  0, f11   (  1) x1  2 x1  0, f 22   (   1) x1 x2  2  0, f12  f 21   x1 1 x2 1  0. Clearly, all the terms in Equation 2.114 are negative. b. A contour line is found by setting the function equal to a constant: y  c  x1 x2 , implying x2  c1  x1  . Hence, dx2  0. dx1 Further, d 2 x2  0, dx12 implying the countour line is convex. c. 2.11 Using Equation 2.98, f11 f 22  f122   (1     ) x12 2 x22  2 , which is negative for     1. The power function a. Since y  0 and y  0, the function is concave. b. Because f11 , f 22  0 and f12  f 21  0, Equation 2.98 is satisfied, and the function is concave.Because f1 , f 2  0, Equation 2.114 is also satisfied, so 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 easily by f (2 x1 , 2 x2 )  2 f ( x1 , x2 ). 2.12 Proof of envelope theorem Chapter 2: Mathematics for Microeconomics a. 7 The Lagrangian for this problem is L ( x1 , x2 , a)  f ( x1 , x2 , a)   g ( x1, x2 , a). The first-order conditions are L1  f1   g1  0, L2  f 2   g 2  0, L  g  0. b.,c. Multiplication of each first-order condition by the appropriate deriviative yields dx dx dx   dx f1 1  f 2 2    g1 1  g 2 2   0. da da da   da d. The optimal value of f is given by f  x1 (a), x2 (a), a  . Differentiation of this with respect to a shows how this optimal value changes with a : dx dx df *  f1 1  f 2 2  f a . da da da e. Differentiation of the constraint g  x1 (a), x2 (a), a   0 yields dx dx dg  0  g1 1  g 2 2  g a . da da da 2.13 f. Multiplying the results from part (e) by  and using parts (b) and (c) yields df *  f a   g a  La . da This proves the envelope theorem. g. In Example 2.8, we showed that   P 8. This shows how much an extra unit of perimeter would raise the enclosed area.Direct differentiation of the original Lagrangian shows also that dA*  LP  . dP This shows that the Lagrange multiplier does indeed show thisincremental gain in this problem. Taylor approximations a. A function in one variable is concave if f ( x)  0. Using the quadratic Taylor formula to approximate this function at point a : f ( x)  f (a)  f (a)( x  a)  0.5 f (a)( x  a) 2  f (a)  f (a)( x  a). The inequality holds because f (a)  0. But the right-hand side of this equation is 8 Chapter 2: Mathematics for Microeconomics the equation for the tangent to the function at point a. So we 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 f11 f 22  f122  0. Hence, the quadratic form ( f11dx2  2 f12 dx dy  f 22 dy 2 ) 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 u  1  F ( x), du   f ( x), x  v, and dx  dv.   1  F ( x) dx  (1  F ( x)) x  0 x  x 0      f ( x)  xdx 0  0  E ( x)  E ( x). d. Use the hint to break up the integral defining expected value: t   E ( x) 1   t  xf ( x)dx  xf ( x)dx  t t 0    t 1 xf ( x)dx t   t 1 tf ( x)dx t    f ( x)dx t  P( x  t ). e. 1. Show that this function integrates to 1: Chapter 2: Mathematics for Microeconomics    1 3 2  f ( x)dx  2 x dx   x 2. x  x 1 9  1. Calculate the cumulative distribution function: x F ( x)   2t 3dt  t 2 tx  1  x 2 . t 1 1 3. Using the result from part (c):   1 1 E ( x)   1  F ( x) dx  x 2 dx   x 1 4. 1. x 1  1. To show Markov’s inequality use P( x  t )  1  F (t )  t 2  t 1  f. x  E ( x) . t Show that the PDF integrates to 1: x2 x2 x3 8  1 dx        1. 1 3 9 x 1 9  9  2 2. Calculate the expected value: x2 2 x3 x4 15 5 E ( x)   dx    . 3 12 x 1 12 4 1 3. Calculate P(1  x  0 ): x 0 0 x2 x3 1 dx   . 1 3 9 x 1 9 4. All we must do is adjust the PDF so that it now sums to 1 over the new, smaller interval.Since P( A)  8 9, f ( x) 3x 2 f ( x | A)   defined on 0  x  2. 89 8 5. The expected value is again found through integration: 2 6. 2.15 x2 3×3 3x 4 3 E ( x | A)   dx   . 8 32 x 0 2 0 Eliminating the lowest values ofx increases the expected value of the remaining values. More on variances a. This is just an application of the definition of variance: 10 Chapter 2: Mathematics for Microeconomics Var( x)  E  x  E ( x)  2  E  x 2  2 xE ( x)  [ E ( x)]2   E ( x 2 )  2[ E ( x)]2  [ E ( x)]2  E ( x 2 )  [ E ( x)]2 . b. Here, we let y  x   x and apply Markov’s inequality to y and remember that x can only take on positive values. E ( y 2 )  x2 P( y  k )  P( y 2  k 2 )   2. k2 k c. Let xi , i  1,, n be n independent random variables each with expected value  and variance  2 .  n  E  xi         n.  i1   n  Var  xi    2     2  n 2 .  i1  n Now, let x  i 1  xi n . n  . n n 2  2 Var( x )  2  . n n E( x )  d. Let X  kx1  (1  k ) x2 and E( X )  k   (1  k )  . Var( X )  k 2 2  (1  k )2  2  (2k 2  2k  1) 2 . dVar( X )  (4k  2) 2  0. dk Hence, variance is minimized for k  0.5. In this case, Var( X )  0.5 2 . If k  0.7, Var( X )  0.58 2 (not much of an increase). e. Suppose that Var( x1 )   2 and Var( x2 )  r 2 . Now Var( X )  k 2 2  (1  k )2 r 2  (1  r )k 2  2kr  r   2 . dVar( X )   2(1  r )k  2r  2  0. dk r k . 1 r For example, if r  2, then k  2 3 ,and optimal diversification requires that the lower risk asset constitute two-thirds of the portfolio.Note, however, that it is still 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 direct 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. Var(ax  by )  E[(ax  by ) 2 ]  [ E (ax  by)]2  a 2 E ( x 2 )  2abE ( xy )  b 2 E ( y 2 )  a 2 [ E ( x)]2  2abE( x)E( y)  b 2 [E( y)]2  a 2 Var( x)  b 2 Var( y )  2abCov( x, y). The 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 two-variable case, however, this is not necessarily the situation.For example, suppose that x and y are identically distributed and that Cov( x, y)  r 2 . Using the prior notation, Var( X )  k 2 2  (1  k )2  2  2k (1  k )r 2 . The first-order condition for a minimum is (4k  2  2r  4rk ) 2  0, implying 2  2r k*   0.5. 4  4r Regardless of the value of r. With more than two random variables, however, covariances may indeed affect optimal weightings. d. If x1  kx2 , 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 two variables. With less than a perfect linear relationship | Cov( x, y) |  Var( x)Var( y) . 0.5 e. If y     x, Cov( x, y )  E ( x  E ( x))( y  E ( y ))   E[( x  E ( x))(   x     E ( x))]   Var( x). 12 Chapter 2: Mathematics for Microeconomics Hence,  Cov( x, y ) . Var( x)