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 =...

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... 7? 51 ? 39? It all depends on which order you perform the calculations and the type of calculator you use. There are set rules for the order in which basic arithmetic operations should be performed, whichareexplainedinChapter2.Nowadays,theseareprogrammedintomostcalculators but ... quantification of economic predictions requires the use of mathematics. Although non-mathematical economic...

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... 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...

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... 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....

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... −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