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11BM2revsol - econ/ams 11b winter 2010 Midterm 2 Review...

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econ/ams 11b winter 2010 Midterm 2 Review — Solutions 1. Find the critical point(s) of the functions below subject to the given constraint. a. h ( u, v ) = 5 u + 8 v ; 2 u 2 + 3 v 2 = 30. The Lagrangian in this case is F ( u, v, λ ) = 5 u + 8 v - λ (2 u 2 + 3 v 2 - 30) , and the first order equations are F u = 0 = 5 - 4 λu = 0 F v = 0 = 8 - 6 λv = 0 F λ = 0 = 2 u 2 + 3 v 2 = 30 Solving the first two equations for λ (and cross multiplying) gives λ = 5 4 u = 4 3 v = 15 v = 16 u = v = 16 u 15 . Finally, substituting the last expression for v into the third equation (the constraint) gives 2 u 2 + 3 16 u 15 2 = 30 = 450 u 2 + 768 u 2 = 6750 = u = ± r 1125 203 . Thus, there are two critical points: ( u * , v * , λ * ) = ± r 1125 203 , r 1280 203 , r 203 720 ! ≈ ± (2 . 345 , 2 . 511 , 0 . 531) . b. k ( x, y, z ) = xy 2 z 3 ; 5 x + 8 y + 12 z = 300. Comment: To simplify the analysis, we will assume the additional condition xyz 6 = 0. I.e., none of the variables x , y or z can be zero. The Lagrangian in this case is F ( x, y, z, , λ ) = xy 2 z 3 - λ (5 x + 8 y + 12 z - 300) and the first order equations are F x = 0 = y 2 z 3 - 5 λ = 0 F y = 0 = 2 xyz 3 - 8 λ = 0 F z = 0 = 3 xy 2 z 2 - 12 λ = 0 F λ = 0 = 5 x + 8 y + 12 z = 300 1
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Solving the first three equations for λ gives λ = y 2 z 3 5 = xyz 3 4 = xy 2 z 2 4 . Comparing the first two expressions above, multiplying by 20 and dividing by yz 3 , we find that 4 y = 5 x = y = 5 x 4 . Comparing the first and third expressions above, multiplying by 20 and dividing by y 2 z 2 , we find that 4 z = 5 x = z = 5 x 4 . Substituting these expressions for y and z into the constraint gives 5 x + 8 5 x 4 + 12 5 x 4 = 300 = 30 x = 300 = x = 10 . Thus, (excluding the points where x, y or z are 0), there is one critical point: ( x * , y * , z * , λ * ) = (10 , 12 . 5 , 12 . 5 , 61035 . 15625) . Comment: (continued) There are infinitely many other critical points for this La- grangian, where one or two of x , y or z are equal to 0. However, at all of these critical points, the objective function k ( x, y, z ) = xy 2 z 3 is identically 0 (why?), so these critical points are not very interesting. 2. A household’s utility function is given by U ( q A , q B , q C ) = 5 ln q A +12 ln q B +18 ln q C , where q A , q B and q C are the quantities of goods of type A , B and C that the household consumes per year. The average prices per unit of these goods are p A = 10, p B = 15 and p C = 25, respectively, and the household’s disposable income is Y d = $60 , 000. a. Assuming that the household spends all of its disposable income on these three goods, how many units of each type should the household consume to maximize its utility ?
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