Basic Mathematics for Economists - Rosser - Chapter 5 ppsx

... price p ∗ s = p d (55 + 4q)(1 + t) = 3 75 − 2.5q 55 + 55 t + 4q + 4qt = 3 75 −2.5q 6.5q + 4qt = 320 − 55 t q(6 .5 + 4t) = 320 − 55 t q = 320 − 55 t 6 .5 + 4t This reduced form equation for equilibrium q ... + 85. 5 = 120 x = 120 −109 .5 x = 10 .5 Therefore, the solutions are x = 10 .5, y = 2,z = 28 .5. Example 5. 11 Solve for x, y and z in the following set of simultaneous e...

Ngày tải lên: 06/07/2014, 07:20

... rule d 2 TR dq 2 = u dv dq + v du dq = ( 650 − 0.25q) 0 .5 (−0.6 25) + ( 650 − 0.625q)(−0.1 25) ( 650 − 0.25q) −0 .5 = ( 650 − 0.25q)(−0.6 25) + ( 650 − 0.625q)(−0.1 25) ( 650 − 0.25q) 0 .5 (2) Substituting the value q = ... =−0.3 75( 650 −0.25q) 0 .5 and also dv dq = 1 Therefore, using the product rule dTR dq = u dv dq + v du dq = ( 650 − 0.25q) 1 .5 + q(−0.3 75) ( 650 − 0.25q) 0 .5 =...

Ngày tải lên: 06/07/2014, 07:20

... Rosser TestYourself,Exercise4.10 Fortheproductionfunctionsbelow,assumefractionsofaunitofKandLcanbe used,and (a)deriveafunctionfortheisoquantrepresentingthespecifiedoutputlevelinthe formK=f(L) (b)findthelevelofKrequiredtoachievethegivenoutputlevelifL=100,and (c)saywhattypeofreturnstoscalearepresent. 1.Q=9K 0 .5 L 0 .5 ,Q=36 2.Q=0.3K 0.4 L 0.6 ,Q=24 3.Q=25K 0.6 L 0.6 ,Q=800 4.Q=42K 0.6 L 0. 7...

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... 24 .5 -6 82 35. 5 -2 97 10 3 -3 7 14 -5 98 25 -6 75 36 -2 68 11 3 .5 -7 3 14 .5 -6 12 25. 5 -6 67 36 .5 -2 38 12 4 -1 08 15 -6 25 26 -6 58 37 -2 07 13 4 .5 -1 42 15. 5 -6 37 26 .5 -6 48 37 .5 -1 75 14 5 -1 75 16 -6 48 ... 11 -4 93 22 -7 02 33 -4 27 5 0 .5 158 11 .5 -5 13 22 .5 -7 00 33 .5 -4...

Ngày tải lên: 06/07/2014, 07:20

... 0.613913 0 .55 83 95 0 .50 8349 0.463193 0.422411 0.6 755 64 0.613913 0 .55 83 95 0 .50 8349 0.463193 0.422411 11 0.64 958 1 0 .58 4679 0 .52 6788 0.4 750 93 0.428883 0.38 753 3 0.64 958 1 0 .58 4679 0 .52 6788 0.4 750 93 0.428883 ... 0.667263 0 .54 5061 0.363699 0.017497 0.006363 1. 75 0.812 057 0. 659 438 0 .53 550 1 0. 353 130 0.0 155 50 0.0 054 91 1.80 0.807284 0. 651 708 0 .52 6114 0....

Ngày tải lên: 06/07/2014, 07:20

... −0.5q 0.3q 2 − 0.5q − 48 = 0 Using the formula for the solution of quadratic equations q = −(−0 .5) ±  (−0 .5) 2 − 4 ×0.3 × (−48) 2 × 0.3 = 0 .5 ± √ 0. 25 + 57 .6 0.6 = 0 .5 ± √ 57 . 85 0.6 Disregarding the negative ... demand function q = 60 − 2p 0 .5 2p 0 .5 = 60 − q p 0 .5 = 30 − 0.5q p = (30 − 0.5q) 2 = 900 − 30q + 0.25q 2 © 1993, 2003 Mike Rosser Solution Slope = dy dx = 3...

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... Rosser Thereforethesecond-orderconditionforaminimumvalueofACissatisfiedwhenqis5. TheactualvalueofACatitsminimumpointisfoundbysubstitutingthisvalueforqinto theoriginalACfunction.Thus AC=25q −1 +0.1q 2 = 25 5 +0.1× 25= 5+2 .5= 7 .5 TestYourself,Exercise9.3 Findwhetheranystationarypointsexistforthefollowingfunctionsforpositive valuesofq,andsaywhetherornotthestationarypointsareattheminimumval...

Ngày tải lên: 06/07/2014, 07:20