corresponding points are 2 4 and 2 4 The segment AB On this segment y 4 2 x 2

# Corresponding points are 2 4 and 2 4 the segment ab

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corresponding points are ( 2 , 4 ) and ( 2 , 4 ) . The segment AB : On this segment y = 4, 2 x 2, hence f(x, 4 ) = 2 · x · 4 x 4 = 7 x 4. The maximum value occurs at x = 2 and the minimum value at x 2. The corresponding points on the segment AB are ( 2 , 4 ) and ( 2 , 4 ) Step 3. Conclusions. Since the global extrema occur either at critical points in the interior of the domain or on the boundary of the domain, the candidates for global extrema are the following points: ³ 1 2 , 1 2 ´ ,( 2 , 4 ), ( 2 , 4 ) We compute the values of f = 2 xy x y at these points: f ³ 1 2 , 1 2 ´ = 2 · 1 2 · 1 2 1 2 1 2 1 2 f( 2 , 4 ) = 2 · ( 2 ) · 4 + 2 4 18 2 , 4 ) = 2 · 2 · 4 2 4 = 10 We conclude that the global maximum is 2 , 4 ) = 10 and the global minimum is 2 , 4 ) 18. Find the maximum of f(x,y,z) = xyz subject to the constraint g(x,y,z) = 2 x + y + 4 z = 1.
Chapter Review Exercises 453 55. Use Lagrange multipliers to ﬁnd the minimum and maximum values of f(x,y) = 3 x 2 y on the circle x 2 + y 2 = 4. solution Step 1. Write out the Lagrange Equations. The constraint curve is g(x,y) = x 2 + y 2 4 = 0, hence g = ± 2 x, 2 y ² and f = ± 3 , 2 ² . The Lagrange Condition f = λ g is thus ± 3 , 2 ² = λ ± 2 2 y ² . That is, 3 = λ · 2 x 2 = λ · 2 y Note that λ ³= 0. Step 2. Solve for x and y using the constraint. The Lagrange equations gives 3 = λ · 2 x 2 = λ · 2 y x = 3 2 λ y =− 1 λ (1) We substitute x and y in the equation of the constraint and solve for λ . We get ± 3 2 λ ² 2 + ± 1 λ ² 2 = 4 9 4 λ 2 + 1 λ 2 = 4 1 λ 2 · 13 4 = 4 λ = 13 4 or λ 13 4 Substituting in (1), we obtain the points x = 6 13 ,y 4 13 x 6 13 = 4 13 The critical points are thus P 1 = ± 6 13 , 4 13 ² P 2 = ± 6 13 , 4 13 ² Step 3. Calculate the value at the critical points. We ﬁnd the value of = 3 x 2 y at the critical points: f(P 1 ) = 3 · 6 13 2 · 4 13 = 26 13 2 ) = 3 · 6 13 2 · 4 13 = 26 13 Thus, the maximum value of f on the circle is 26 13 , and the minimum is 26 13 . Find the minimum value of = xy subject to the constraint 5 x y = 4 in two ways: using Lagrange multipliers and setting y = 5 x 4in . 57. Find the minimum and maximum values of = x 2 y on the ellipse 4 x 2 + 9 y 2 = 36. solution We must ﬁnd the minimum and maximum values of = x 2 y subject to the constraint = 4 x 2 + 9 y 2 36 = 0. Step 1. Write out the Lagrange Equations. The gradient vectors are f = ³ 2 xy, x 2 ´ and g = ± 8 18 y ² , hence the Lagrange Condition f = λ g gives ³ 2 xy, x 2 ´ = λ ± 8 18 y ² = ± 8 λx, 18 λy ² We obtain the following Lagrange Equations: 2 xy = 8 λx x 2 = 18
454 CHAPTER 14 DIFFERENTIATION IN SEVERAL VARIABLES (LT CHAPTER 15) Step 2. Solve for λ in terms of x and y .If x = 0, the equation of the constraint implies that y 2. The points ( 0 , 2 ) and ( 0 , 2 ) satisfy the Lagrange Equations for λ = 0. If x ²= 0, the second Lagrange Equation implies that y ²= 0. Therefore the Lagrange Equations give 2 xy = 8 λx λ = y 4 x 2 = 18 λy λ = x 2 18 y Step 3. Solve for x and y using the constraint. We equate the two expressions for