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11Breview3sol - econ 11b ucsc ams 11b Review Questions 3...

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econ 11 b ucsc ams 11 b Review Questions 3 Solutions Note: In these problems, you may generally assume that the critical point(s) you find produce the required optimal value(s). At the same time, you should see if you can find an argument to justify this assumption in each example. 1a. Find the minimum value of f ( x, y ) = x 2 + y 2 subject to 3 x + 5 y = 68. Lagrangian : F ( x, y, λ ) = x 2 + y 2 - λ (3 x + 5 y - 68). ‘Structural’ equations : F x = 2 x - 3 λ = 0 F y = 2 y - 5 λ = 0 Solving these equations for λ gives λ = 2 x 3 = 2 y 5 = x = 3 y 5 . Substituting this into the constraint gives 3 3 y 5 + 5 y = 68 = 34 y = 340 = y 0 = 10 and x 0 = 6 . The minimum value of x 2 + y 2 subject to 3 x + 5 y = 68 is therefore obtained at the point (6 , 10), giving 6 2 + 10 2 = 136. 1b. Find the maximum value of g ( x, y, z ) = 20 x 1 / 2 y 1 / 3 z 1 / 6 , subject to the constraint 5 x + 4 y + 7 z = 1680. Lagrangian: F ( x, y, z, λ ) = 20 x 1 / 2 y 1 / 3 z 1 / 6 - λ (5 x + 4 y + 7 z - 1680). ‘Structural’ equations : F x = 10 x - 1 / 2 y 1 / 3 z 1 / 6 - 5 λ = 0 F y = (20 / 3) x 1 / 2 y - 2 / 3 z 1 / 6 - 4 λ = 0 F z = (10 / 3) x 1 / 2 y 1 / 3 z - 5 / 6 - 7 λ = 0 . Solving these equations for λ yields the triple equation ( λ =) 2 y 1 / 3 z 1 / 6 x 1 / 2 = 5 x 1 / 2 z 1 / 6 3 y 2 / 3 = 10 x 1 / 2 y 1 / 3 21 z 5 / 6 . Comparing the first and second expressions, canceling the factor of z 1 / 6 from both and clearing denominators gives 6 y = 5 x = y = 5 x 6 .
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Comparing the first and third expressions, canceling the factor of y 1 / 3 from both and clearing denominators gives 42 z = 10 x = z = 5 x 21 . Substituting the expressions for y and z into the constraint gives 5 x + 4 5 x 6 + 7 5 x 21 = 1680 = 420 x = 70560 = x 0 = 168 , y 0 = 140 , z 0 = 40 . Thus, the maximum value of f ( x, y, z ) = 20 x 1 / 2 y 1 / 3 z 1 / 6 , subject to the constraint 5 x +4 y +7 z = 1680 is given by f (168 , 140 , 40) 2489 . 262 .
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