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ECN 505 HANDOUT SPRING 2009 EUCLIDEAN SPACE AND FUNCTIONS OF MORE THAN ONE VARIABLE Continuity Partial Derivatives Di erentiability Chain Rules Implicit thctions EUCLJDEAN SPACE De nition 1 nadimensional Euclidean space denoted R is the set of vectors with n realiualued compo nents that is R xlxn 1 ERfor alliln Also let R R denote the set of nonnegatiue strictly positive real numbers Then RS 111 E11xnxi 6 Eur R for all i ln We may use at to denote the vector x1 7 for example We Write as 2 y if at 2 pi i ln WeWriteacgtyifac7Eyand at 2 pi i ln We Write at gt y if acigt pi i ln De nition 2 The Euclidean norm associated with at E R denoted is x gt 2 Ex iZI De nition 3 Operations in Euclidean space Let 3531 6 R and t E R l 2 a Vector addition at y 211 311 95 b Scalar multiplication tap E tam tacn c Inner Product ac y E 21 any De nition 4 The open neighborhood nbd ofac E R of radius 6 where e gt 0 denoted News is yeanwinlte De nition 5 A set S g R is open R iffor each at E S there is an e gt 0 such that N605 g S De nition 6 A set T g R is closed if R 7 T is open REALVALUED FUNCTIONS DEFINED ON EUCLIDEAN SPACE Continuity Throughout the remainder of this handout unless otherwise indicated f S A R represents a function With nonempty domain S g R and range in R De nition 7 f has a limit L at a i1 in and we write 7213 f at L iffor each positive 6 no matter how small there is a corresponding positive 6 such that ifac 6 N50 OS and at i mmimlte De nition 8 The function f is continuous at a E S if a liming exists and b liming fac The function f is continuous on S if it is continuous at all at E S Partial Derivatives De nition 9 If the limit Em fa1ahan 7f222 has h exists we call it the ith partial derivative If f at x where i 1 n We deuute it by flit 07 fag 0r Dfx 0r 31W 3x 3f 3x 5555 In the following diagram We illustrate a geometric interpretation of fag GEOMETRIC INTERPRETATION oE PARTIAL DERIVATIVE Zf ny slopeZpar lal derivatlve of f mm at r i9 f i De nition 10 If all If the rst partial derivatives If f exist at x then we de ne the gradith If f at x deuuted Vfx 0r Dfx as 185 425 Math Fact As you might suspect giVen the results of the last handout continuity of f does not guarantee that the gradient of f exists The result that might surprise you however is that if the gradient of f at 9 exists f is not necessarily continuous there Derivatives De nition 11 Let h E Rquot The functiuu f is differentiable at x E S if there exists a vectur b E Rquot such that hm fxh7fx7b h 0 hgao h If such a vectur 1 exists it is called the derivative If f at x If f is di ere atiaMe at all x E S then it is differentiable Theorem 12 If f is differentiable at x E S then it is cautiuuuus there Theorem 13 Assume that f is cautiuuuus at x E S If 1 the derivative If f at 5 exists then s0 does Vfx and 1 wet Theorem 14 Assume that f S gt R where S is upeu If all If the rst partial derivatives 0quot f exist and are cautiuuuus an S then the derivative 0quot the functiuu exists and is cautiuuuus uh OPTIMZATION De nition 15 A function has a local or relative maximum minimum at x E S iffor some Nex it is the case that 2 fx for all x E Nex S De nition 16 A function has a global or absolute maximum minimum at x if 2 ltfx for all xES lf strict inequality holds in either of the last two de nitions for all x 7E 55 then strong or strict precedes the appropriate terms It should be clear that if f has a global maximum minimum at x then it has a local maximum minimum there De nition 17 A critical stationary point for the function f is a point x E S where 0 for all i 1 n Example Determine all critical points for the function f R2 A R Where 101 100 7 as 7 as Since f1 QJEL f2 23027 we nd that the only critical point is x 0 Theorem 18 Assume that S is an open set and that all of the rst partial derivatives of f exist and are continuous on S If there is a local maximum or local minimum of f at x E S then it must be a critical point 0 Given this result it follows that for the previous example if there is a maximum or minimum for the function r 100 7 xi 7 9037 then it is at x 0 Theorem 19 Weierstrauss Assume that f is a realiualued continuous function on S where S g R is closed and bounded Then f assumes a global maximum and a global minimum on S Example Let f S A R be de ned by 90 3019027 and S x1x2 E R2 x1x2 2 0 and x1 x2 S 10 f is realevalued and continuous in addition S g R2 is closed and bounded Therefore f assumes a global maximum and global minimum on S In fact we convince ourselves that 00 is the unique strict global minimum and 5 5 is the unique strict global maximum De nition 20 We can de ne second order partial derivatives a Second order partial derivative with respect to xi where i 1 n is mm Elli fx1xhxnlifx1xx7 if the limit exists It is also denoted by 39 82fx 727 8x Duf t0 t A second order mned or cross pdmdz dermdtme 15 ltigthm Mil ijwt inzrfmis it No where w 7 1 m 7 gm zlmlt gusts 1t dzso denoted ty 3 8f gt as B i new T 37G Do 31 31 m Theorem 21 Young39s Theorem Let f f f dnd fr wst dnd te eontmoods m d ntd ofi e s Then flea fo nght now slnoe we have not yet covered mstnx algebra we gwe second order sul olent oondltlons only lor a loan msxtmum mlnlmum gwen s lunotlon ol two vsrlsbles only Theorem 22 Let f s 7 R where s g a dud suppose mdt f hds eontmuons rst dnd second pdmdz dermdmes m d neighborhood ofd crmmlpomt 2122 Let A 7 fni1izt B 7 roam C 7 fzziliz 1f d A lt u dnd AC 757 gt u there 15 d zoedz mallmnm ati1i t A gt a dud Ac 757 gt u there is d loedz mmmmm dt i1i e AC 757 lt u there 15 d sdddze pomt dt 2122 Example For the lunotlon z 7 100 7e 7e we determlned that there ls s orltlosl polnt st 7 7 a We nd that fni 2 lt 0 frri 72 821 f11ifzzi mi 4M4 therelore there ls s 10ch msxtmum st 7 7 a In fact slnoe fz 7 100 7 e mg gtn esquot en there ls a unlque strlot globsl msxlmum st 1 Example ora Saddle Point Let f s 7 R where s g R7 The polnt 24 e s ls a sdddze pomt ol f llor some may one ol the lollowlng holds 2 NM 5 Hi 5 Km for EU WNW E NAM ms or 00 Km 5 NW 5 My for all WNW E NAM NS ILLUSTRATION or A SADDLE POINT CHAIN RULES A formal statement of a chain rule follows Theorem 23 A Chain Rule Let y hacl7 12 where 11 ft7 and 2 gt Let Ft hftgt Assume that the rst derivatives of f and g exist and are continuous at t i Then the rst derivative ofF exists at t t is continuous at t t and Fm h1i17i2f 1207017 HEMDy where E17502 0790 Example Let y 9017902 xi 37 and 1 x10 i3 120 i2 Find the rate of change of y WRT ie7 with respect to t First we do this without using the preceding theorem Notice that Fm fltx1ltzx2lttz2 2 tz4 dy F i 1 43 dz Now we use the preceding theorem Another way of writing the result is dy 8y dam By 112 1 dt 811 dt 812 dt dam 1 7 1352 t 2t dt 2 1 dt 311 2 2 8x1 3017 a 12 In terms of t7 8 1 8y 2 2tquot 2t 811 17 812 Now substitute into the chain rule at Q f 2t2 2t 14t3 the same as we obtained before Problem for you There are many other chain rules For example7 assume that 2log3523127 actu7 yt7u Notice that 2 depends on t and u through at and y The appropriate chain rule with which to determine 2 1s 8t 82 82 81 82 By y a39EU yw a Find that g7 2 8t7t2u2 in two ways substitution and then differentiation and ii the chain rule and then substitution b Determ1ne the appropr1ate cha1n rule w1th wh1ch to determine Z Then use the cha1n rule to u 8 nd that 82 Zn E 22 u239 Now obtain the same result by substitution and then differentiation Problems for You In both of the following cases7 express appropriate chain rules which allow you to arrive at the indicated derivatives Verbally interpret your rules d a w Fm x m y gm Find d f dw d2w b w 1105721 21 Flnd E7 IMPLICIT FUNCTIONS A statement of an implicit function theorem follows Theorem 24 Let f be dfunction which is continuously di erentidble on some open neighborhood S g R3 of 551772 6 R3 such that fi172 C and f3i172 7E 0 Then there exists a unique continuously di erentidble function 2 de ned on a neighborhood of 5517 7 such that 2 i717 2 and f 131 2 C Example Consider 23 27231 7 7123122 2 0 Near l7 127 is it permissible to view 2 as a function of ac and y Determine7 if possible7 e 817 83139 Example Assume that Uac7 y is continuously differentiable on 113 and that Uva 31 i 0 Can we determine dd ECN 505 001 HANDOUT SPRING 2009 Constrained Optimization Interior Solutions One Equality Constraint Assume that fg R1 A R Let ac x1 The form of the problem in this section is max objective function as st 91 c constraint at 2 0 constraint The Lagrangian function is 1317 A x A C 7 9M where A is the Lagrangian multiplier Assume that f and g are differentiable Then under certain conditions made more precise in class the following equalities are necessary and suf cient to de ne a solution 30 to the constrained optimization problem 1ac f1ac 7 X g xquot 0 Lunar fnx4gnxo Ax 0791quot Interior or Boundary Solutions One or More Inequality Constraints We look at a constrained optimization technique appropriate when there are one or more constraints when the constraints are weak inequality not necessarily equality constraints and when the solutions may be on the boundary they are not necessarily interior solutions Assume that fgj R1 A Rj 1 in Let ac x1 The form of the problem in this section 1s max f as objective function f st 971 3 07 j 1 m constraints at 2 0 constraint The Lagrangian function is ac A m 2A 07 7 91m j1 Assume that f and 97 j 1 m are differentiable Notice that m 13mm may 7 ZAJggm 239 1 n j1 LljltxAc7gtgjltx j1m The following two results apply to the problem in this section The following theorem shows inequality conditions that solutions must satisfy Theorem 1 Necessity Assume that the following hold 1 f and 9735 j 1 m are di erentiable where at 2 0 2 30 is a local solution to the problem in this section 5 There exists a 2 0 such that 9732 lt c7 j 1 m 4 971j 1m is quasiconvex where at 2 0 Also for allj 1m at least one of the rst partial derivatives of 97 when evaluated at 1 is nonzero Then there exists V such that m am Aquot mar 7 ZAggg39m g 0 x 2 0 15mm x 0 239 1 n jil 1579mm 0739 797M 2 0 A 2 0 minty A 0 j The next theorem tells us about second order suf cient conditions Theorem 2 Su iciency Assume that the following hold 1 f is di erentiable and quasiconcaue where x 2 0 gjx j 1 m is di erentiable and quasiconuex where x 2 0 2 There exists x satisfying the KuhneTucker conditions that is there exists x such that m Quay my 7 ZAgggw g 0 as 2 0 1xAx 0 239 j1 1570mm 0739 797M 2 0 A 2 0 Away A 0 j 5 is twice continuously di erentiable in some neighborhood of xquot7 and fi 7E 0 for at least one i 1 n Then xquot is a global solution to the problem in this section The rst theorem gives under speci c conditions the necessary conditions ie the KuhneTucker conditions for xquot to be a solution to the constrained maximization problem The second theorem states conditions under Which these inequality conditions are suf cient to guarantee that xquot is a solution to the constrained maximization problem Note that there are more general statements of these results Problems for You Consider an individual s constrained utility maximization problem in each of the fole lowing parts assume that the price of both goods is 1 and that the consumer s income is 10 Use the theorems in this section to derive hisher Marshallian demands and prove that your answers are correct a WWI 796 i 1V i y 7 1V b U067 12y c Uxy log x l log y 1 Problem for You An individual s utility function is Ux y log at log y Derive hisher optimal consumption When heshe faces both the usual budget constraint and a time constraint Let T be the total amount of consumption time available and ti and ty the time prices of goods x and 31 respectively ECN 505 HANDOUT SPRING 2009 Integration INDEFINITE INTEGRALS De nition 1 If F W f r then F is called a primitive or an inde nite integral or an antiderivative of f Proposition 2 Assume that F is an inde nite integral of f Then any other inde nite integral of f is of the form F 07 where c is a constant DEFINITE INTEGRALS Proposition 3 thdamental Theorem of Calculus ff is continuous on 21 and F is an inde nite integral of f7 then facdac Fb 7 Fox 2 Fm INTEGRATION RULES AND EXAMPLES Integration Rules Following is a brief list of integration rules please see a calculus text for a more complete set 1 f du u C 2 fcfudu cffudu 3 f u gudu ffudu fgudu 4 f undu u n 1 n1 CWhere n 71 l 5 f du log C Where log natural logarithm u 6 feudue C 7 fudvuvifvdu Example Find 2x 1dx Solution f2 1dx 1321 1 2m f2x1 gig byRule2 u du 211 T 2 C by Rule 4 Oh A 211C Example Probability density functions appear in probability and statistics these are functions f R A R suc l 2 0 for every at 2 f facdac 1 In this example7 we want to determine the constant c so that the function 7 012 where 0 S at S 2 0 otherwise is a probability density function pdf Solution We determine 0 so that that is7 2 chdx l 0 2 A012d11ltgt 013 lltgt0701ltgtc Then notice that x gm 2 0 where 0 S at S 2 Example An individual has the von Neumann Morgenstern utility function Uw wi where w wealth and 0 S w 3 2 Assume that w is determined according to the pdf of the previous example Determine the individual s expected utility Solution In general the expected utility associated with w is co ultwfltwdw 0 where is the pdf associated with w In this example7 the pdf associated with w is 3 2 7 Eu where 0 S w 3 2 7 otherwise Expected utility of wealth is where w wealth The individual starts with a total wealth we to be allocated between money m with return 0 and bonds we 7 m with risky return T then the value of the portfolio or wealth7 w7 after the return is m 1 rwo 7 Assume that 7 has the probability density function Determine the optimal amount of m 771 What happens to mquot as we increases Solution Note that 771 1 Tw0 7m 1 Tw0 7 Tm then w 1 Tw0 7 Tm Note that 8w 77quot 87m The individuals problem is to maXEU bw 7 cw2frdr FOC We assume an interior solution 8EU7 ba w7cw8 wfrdr0ltgt 8m 00 1007117 cwrfrdr 0 ltgt 7bioo 7 f7 d7 C7 o 7 17 wo 7 77 L1f7 d7 0ltgt ET 7bF c rw0frdr 72mg 7 mfrdr 0 4 7bFc 10000 Tf7 d7 w0 7m x V I 2F EEW 00 72frdr 04 xiv z 71 ewe Cwo 7 mE7 2 0 ltgt 7 mpg 7 b 77 m 7 we CE 72 SOC If we assume that c gt 07 E7 2 gt 07 2 oo BBEZU 70 72frdr 7CE7 2 lt 0 m 00 In addition to assuming that E7 2 gt 07 assume that 77 gt 0 then 8771 77 1 gt 0 8100 E 72 Example A consumer s demand curve is pix 10 Determine the consumer s surplus associated With the purchase of 1 unit at pf 10 ECN 505 HANDOUT SPRING 2009 MATRIX ALGEBRA DETERMINANTS AND DIFFERENTIALS ECONOMIC APPLICATIONS THE MATHEMATICS MATRICES Basic De nitions and Results De nition 1 A realeualued matrix A is a rectangular array of numbers 111 112 aln m 121 122 1121 rows aml am2 amn n columns The matrix A is of order or dimension m X n If m n then A is a square matrix Any matrix A can be denoted by aij Where aij is a representative element in the ith row and jth column this short hand notation allows us to represent outcomes of operations on matrices Which follow in a compact way De nition 2 Operations on Matrices Let A aij 7B by 7 and C cij be matrices and let t e R 1 Addition Let A and B both be m X n matrices Then A B C where cij aij bij where i lmj 17n 2 Multiplication Let A be an m X n matrix and B be an n X q matrix In this case AB C where for i 17m j 17q cij Z1aikbkj 5 Scalar Multiplication The multiplication of the matrix A by the scalar t is always de ned tA C where cij taij andi 17m j 17n In general7 AB and BA are not equal that is7 their corresponding entries are not equal Proposition 3 Properties of Operations Let A B and C be matrices and let t be a real number Assume that the matrices are such that the following operations are de ned Then 1 A B B A 2 ABCABC 3 tAtB tAB 4 ABC ABC 5 ABC AB AC 6 AtB tAB tAB Types of Matrices and Results De nition 4 An identity matrix is an n X n matrix A such that 1 1 i 1n and aij 0 ij 1n i 7E j That is an identity matrix is a square matrix with 1 s along the diagonal and 0 s everywhere else It is denoted by I when its dimension can be understood from the context in which it appears Proposition 5 Let A be an n X n matrix Then A A A De nition 6 Let A be an ngtlt n matrix If an n X n matrix B exists such that AB I then B is the inverse of A It is denoted by A l Proposition 7 Let A be an n X n matrix If A 1 exists then it is unique and AA 1 A lA 1 Notice that it follows from the last proposition that A 1 1 A De nition 8 The transpose of the m X n matrix A denoted AT or A is C cij where cij aji i 1 m j 1 n Notice that the transpose of A is an n X m matrix De nition 9 IfA is an n X n matrix and aij aji ij 1 n then A is a symmetric matrix That is T an n X n matrix A is symmetric ifA A Proposition 10 Let A be any matrix Then 1 ATA is symmetric 2 ATT A DETERMINANTS De nition 11 Consider an n X n matrix A We describe how to obtain the determinant associated with A written detA or 1 Choose any row or column of the determinant 2 Consider the minor associated with each element of the row or column It is the determinant which is obtained by blocking out the row and column in which the element appears If the element is aij label its minor mij 5 Determine the cofactor of aij Ci E 71 jmij 4 Then A 221 aijCij if the expansion is along the ith row 21 aiJCij if the expansion is along the jth column Proposition 12 Properties of Determinants 1 If the rows of one determinant are the same as the co umns of another determinant then the two determinants are equa 2 If two columns rows of a determinant are interchanged the value of the resulting determinant is equal to the negative of the value of the given determinant 5 If the corresponding elements of two columns rows of a determinant are identical the value of the determinant is 0 4 If the elements of a column row of a determinant are multiplied by k the value of the determinant is multiplied by k F If the elements of the jth column ith row of a determinant D are of the form aij bij then D is the sum of the determinants D and D in which all of the columns rows of D D and D are the same except the jth ith The jth column ith row of D is aij i 1 n39 j 1n and the jth column ith row ofD is bij i 1n39 j 1n 6 a If in a determinant D the elements of the kth column am i 1 n are replaced by aik taij where aij i n are the elements of the jth column the determinant obtained equals D b If in a determinant D the elements of the th row agj j 1 n are replaced by agj taij where aij j 1 n are the elements of the ith row the determinant obtained equals D Problems for You Go back to the section Tests for Some of the Special Functions on the handout on special functions Given those tests determine if the following functions are strictly concave convex strictly quasiconcaveconvex Assume that xy gt 1 fxy alogx logy where a gt 0 2 fx y xay consider the three cases 04 lt 1 a 1 and Oz gt 1 3 my 21 APPLICATIONS OF DETERMINANTS AND MATRICES Second Order Conditions For the next theorem consider the following constrained maximization problem and its associated Lae grangian max f st c 7 gx 0 131 WE A C 7 900 Consider the following matrix a bordered Hessian matrix derived from the Lagrangian and evaluated at x A where a gtgt 0 111 a 5 51115075 910 0 BH 5 131 X 12 X 791 a 791 gn 0 Given the matrix de ne the border preserving leading principle minor of order i as 11x 3x 51239 507 3 91 BHi EJ 7 2 Iii I 132 3 799 0 91i 9215 0 Theorem 13 Constrained Maximization Let f S A R where S g R is an open set be a twice continuously di erentiable function let 9 be a continuously di erentiable function Consider the bordered Hessian matrix BHx associated with the constrained maximization problem just described 1 ff has a local maximum at is subject to the constraint then 71iBHvx 2 0 i 1 n 2 Assume that 1 1X 0 239 1 and 1A 1X 0 If 71iBHvx gt 0 239 1 n then at is a local maximum of f subject to the constraint That 71iBHix 2 0 i 1 n in the rst part of this result is referred to as the second order necessary condition for the constrained maximization problem That 71iBHix gt 0 i n in the rst part of this result is referred to as the second order su icient condition for the constrained maximization problem Solution Method for Some Equation Systems Cramer s Rule Proposition 14 Cramer s Rule Consider the system of equations a11x1a1jjquottl1nn b1 M1961aijxjamn bi an10Lnjj 39llnnn bin or in matrix form 1111 a1 aln 901 b1 1121 am am at 172 am anj am xn bn Let A be the determinant associated with the matrix A if A 7E 0 the unique solutions to xj j 17 n7 exist and are given by 1111 quot39 111071 1 1 111041 quot39 aln 1 avj m an quot39 11204 bi 112041 quot39 am am 39 39 39 angel bn anj1 39 39 39 ann DIFFERENTIALS De nition 15 Let f S A R where S g R be a di erentiable function The di erential of f denoted df is de ned as was was h 1 If we let y fx dy dfx and dx h we can rewrite I as dy f x dx dy is the di erential of the dependent variable y and dx is the di erential of the independent variable x Differential Rules You should be able to derive the following rules do 0 dx nxn 1dx 53919gt93PH 2m 3 N a gt1 d 1 9df f2d9 900 8 If f fu7 u ux7 then df f uu xdx l dlogx dx7 dlogux x 3 1 l d 10 deg e Cdx7 dew e u xdx De nition 16 Let 2 The total di ei ential of z is dz fidac fydy More generally ify facl an the total di ei ential ofy is dy fldacl fndacn Example Consider a consumer s constrained utility maximization problem 1 Let F7 be a function Which is de ned using the form of the jth FOO 12 associated With the problem that has interior solutions When the budget constraint is satis ed With equality Find F 171967117 Mu UAMI 7 Am dF1 Facy Apdac Fy1acy Apdy F ac y Apd Fplzacy Apdp 3 dFl 7 Umacydac Uryacydy 7 pm 7 Mp Similarly de ne 172967117 Amy UyWy 7 Am and nd that dF2 F3acy Apydac 750531 Apydy Ffacy Apyd Fp2yacy pydpy 3 dF2 7 Uwxydx Uwacydy 7 pydA 7 Adpy Let G be a function Which is de ned using the form of the last FOO associated With the problem that has interior solutions When the budget constraint is satis ed With equality Find dG N Gx7y7px7py71 1 7177596 717111 dG Gdaammwndl 7 Gxx7y7pmpy71d Gyx7y7px7pyvldy 7Gpx7y7px7py71dpx 7 prx7y7pmpy71dpy 7 dG d 7 pidac 7 pydy 7 xdpi 7 ydpy ECONOMIC APPLICATIONS THE METHOD OF COMPARA TIVE STATICS THE ISLM FRAMEWORK Goods Market Consumption function GYdi Where Yd disposable income Y 7 T T taxes assumed to be exogenous i interest rate Investment function Yd Government expenditure G assumed to be exogenous Speci cation of partial derivatives Cy2 Gigi i yE 12 0ltCy1ylt1 Ci1ilt0 De nition 17 The S curve is the locus of Y points such that the goods market is in equilibrium That is it is the Yi combinations such that Y 7 CYd 239 10622 G 2 Along the IS curve G and T are held xed Money Market Nominal money supply M Price level P M Real money supply Real demand for money LYi Speci cation of partials 8L 8L L E 0 L E 0 Y av gt Z 8239 lt De nition 18 The LM curve is the locus of Y7 points such that the money market is in equilibrium That is it is the Y7i combinations such that M LY7 Along an LM curve M is held constant P Combining Both Markets The ISLM Framework Yi equilibrium income and interest is determined by the intersection of the IS and LM curves Problem a Determine the impact on Yi of an increase in taxes b Determine the impact on Yi of an increase in government spending nanced by an increase in the real money supply c Determine the impact on Yi of an increase in government spending nanced by an increase in taxes Solution First7 totally differentiate both 2 and Totally differentiating the equation for the IS curve7 we nd that dY Cded Cidi ded idi 16 dY Cy IydY 7 1T Ci Iidi dG x 17Cy 71ydY7CiIidi 7CylydTdG 4 Notice that terms involving endogenous exogenous variables are on the left right hand side of By the way7 notice that 4 can be used to determine the slope of the IS curve by setting dT d and rearranging Totally differentiate the relationship de ning the LM curve we nd that 7 PdM 7 MdP M Lde Lidi d P2 5 Notice that terms involving endogenous exogenous variables are on the left right hand side of Now express 4 and 5 in matrix form 1 7 0y 7 1y 7 Ci Hi dY CY H dT dG M 6 LY 239 dz d P PdM 7 MdP P2 I Let D denote the determinant associated with the left hand matrix in Note that In some s1tuations7 1t may be more useful to write 1 1n 6 as D Li1 7 Cy 71y LyCi 12 lt 0 M a In 6 set dG d 0 By Cramer s rule z a i CYi IY 7Cili LleY l IY lt0 3T M Guam D 0 Li D I F V You should verify that quotq 8quot L C I Z Y Y Y lt 0 BT gem g M b In 6 set dG d and dT 0 By Cramer s Rule BYquot 1 1 7Ci1i L 0 12 8G 5 v D gt 039 Tconst dGd Z V By Cramer s Rule 7 7 7 I z quAK gt a il ycyily 1 7170y71y7Ly 7 0 8G Tconst dGd D Y 1 D lt c I ll leave this for you CONSUMER DEMANDS In a two good world a consumer has the utility function U007 21 u1 u217 where u l gt 0u 2 gt 0u 1I lt 0 and Mg lt 0 You know nothing else about the utility function that is you do not know the speci c functional form Nonetheless you may be able to determine the signs on the comparative statics terms 81 81 81 831 831 831 81757 apy7 8 7 prl 81757 BM in the following way First lay out the F00 acy u1x u2y A M 717751 ipyy 12412212 X u lw 7 Mp 0 13yx7y7 Myquot 7 pr 0 Aacy M ipgcaf ipyyquot 0 Totally differentiate the F00 and rearrange u 1 xdac 7 pidA X dpi u39239ydy 7 pydk Vdpy ipfdac ipydy dng ydpy 7 dM Arrange in matrix form 1400 0 72775 dx Vdm 0 u y 7171 dy y Pz Py 0 d 1dpx ydpy 7 dM I leave it to you to determine that 1400 0 px D 0 ipy gt 0 pw 71 8 To determine 82 set dpi dpy 0 in the right hand matrix Use Cramer s Rule 8 1 0 p 7 H y 0 0 7 M gt 0 8M D W D 71775 71 0 V I Will leave you to determine the other partials