practice22answers - 2 Practice Exam 2 The multichoice...

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Unformatted text preview: 2 Practice Exam 2 The multichoice answers are all A. Problem (#11). Let f (x, y ) = x2 y + xy 2 + 3xy . Find the critical points of f and tell what type each one is. Solution. A point (x0 , y0 ) is a critical point of f just in case either ∇f (x0 , y0 ) is undefined or ∇f (x0 , y0 ) = 0. We compute, ￿ ￿ ∇f = 2xy + y 2 + 3y, x2 + 2xy + 3x This is defined for all (x, y ), so we only need to worry about the points where the gradient is 0. For these, we solve, ￿ 2xy + y 2 + 3y = 0 2xy + x2 + 3x = 0 Equivalently, we can solve the system: ￿ y (2x + y + 3) x(x + 2y + 3) =0 =0 From the first equation, we see that y = 0 or 2x + y + 3 = 0. We then derive y = 0 =⇒ x(x + 2 · 0 + 3) = 0 =⇒ (x = 0 or x = −3) (2x + y + 3 = 0 and x = 0) =⇒ y = −3 ￿ 2x + y + 3 = 0 (2x + y + 3 = 0 and x ￿= 0) =⇒ =⇒ x = y = −1 x + 2y + 3 = 0 Thus, the critical points of f are (0, 0), (−3, 0), (0, −3), (−1, −1). To classify them, we use the second derivative test. First, note that ￿ ￿ fxx fxy Df (x, y ) = det fyx fyy ￿ ￿ 2y 2x + 2y + 3 = det 2x + 2 y + 3 2x = 4xy − (2x + 2y + 3)2 • Df (0, 0) = −9 < 0, so (0, 0) is a saddle point 4 • Df (−3, 0) = −9 < 0, so (−3, 0) is a saddle point • Df (0, −3) = −9 < 0, so (0, −3) is a saddle point • Df (−1, −1) = 3 > 0 and fxx (−1, −1) = −2 < 0, so (−1, −1) is a local maximum point. Problem. Suppose that (1, 1) is the only critical point of the function f (x, y ) = 2x − 2xy + y 2 + 1. Find the absolute maximum value of the function on the rectangle R = [0, 2] × [0, 3]. Solution. To find the absolute maximum of f over R, we find the critical points of f (which is done for us, here) and find the local maximum points of f on the boundary; then we compare the values of f at these points. The boundary of R has (unfortunately) 4 parts. • L1 = {(x, y ) : y = 0, 0 ≤ x ≤ 2} • L2 = {(x, y ) : y = 3, 0 ≤ x ≤ 2} • L3 = {(x, y ) : x = 0, 0 ≤ y ≤ 3} • L4 = {(x, y ) : x = 2, 0 ≤ y ≤ 3} We will find the local maximum points of f on each of L1 through L4 , and we will do this by parametrizing each segment. . L1 : Maximize f (x, y ) subject to y = 0, 0 ≤ x ≤ 2. Take x(t) = t, y (t) = 0. Then the stated problem is the same as finding the maximum of g1 (t) = f (x(t), y (t)) for t in [0, 2]. Now, g1 (t) = 2t + 1, and the maximum occurs when t = 2. Hence, maximum point of f (x, y ) restricted to L1 is (2, 0). . L2 : Maximize f (x, y ) subject to y = 3, 0 ≤ x ≤ 2. Take x(t) = t, y (t) = 3. Again, the stated problem is the same as finding the maximum of g3 (t) = f (x(t), y (t)) for t in [0, 2]. Now, g2 (t) = 2t − 6t + 32 + 1 = −4t + 10, and the maximum occurs when t = 0. Hence, maximum point of f (x, y ) restricted to L2 is (0, 3). . L3 : Maximize f (x, y ) subject to x = 0, 0 ≤ y ≤ 3. 5 Take x(t) = 0, y (t) = t. Then the stated problem is the same as finding the maximum of g3 (t) = f (x(t), y (t)) for t in [0, 3]. Now, g3 (t) = t2 + 1, and the maximum occurs when t = 3. Hence, maximum point of f (x, y ) restricted to L3 is (0, 3). . L4 : Maximize f (x, y ) subject to x = 2, 0 ≤ y ≤ 3. Take x(t) = 0, y (t) = t. Then the stated problem is the same as finding the maximum of g3 (t) = f (x(t), y (t)) for t in [0, 3]. Now, g3 (t) = 3 − 6t + ￿ t2 + 1 = t2 − 6t + 1. Note that g4 (t) = 2t − 6, and the maximum occurs when t = 0. Hence, maximum point of f (x, y ) restricted to L3 is (2, 0). Finally, we compare the values of f at the points we’ve found: • f (1, 1) = 2 • f (2, 0) = 5 • f (0, 3) = 10 So, the maximum value of f over R is 10, and this occurs at (0, 3). Problem. Find the maximum and minimum values of f (x, y, z ) = 2x − z subject to x2 + 10y 2 + z 2 = 5. Solution. Setting g (x, y, z ) = x2 + 10y 2 + z 2 , we see that we need to solve the system, ￿ ∇f = λ∇g x2 + 10y 2 + z 2 = 5 This expands to 2 0 −1 2 x + 10y 2 + z 2 = 2λ x = 20λy = 2λz =5 To solve this, first note that λ ￿= 0, for if λ = 0, we’d get 2 = 0 in the first equation. Since λ ￿= 0 and 20λy = 0, we now know that y = 0.m Also, λ ￿= 0 means that we can divide by λ, so from the first and third equations we get 1 1 x = λ and z = −λ . Substituting all of this into the last equation, we get 2 ￿ ￿2 ￿ ￿2 1 2 −1 + 10 · 0 =5 λ 2λ 5 so 4λ2 = 5, and so λ = ± 1 . If λ = 1 , we get x = 2, y = 0, z = −1, and if 2 2 λ = − 1 , we get x = −2, y = 0, z = 1. Thus, 2 6 • f (2, 0, −1) = 5 is the maximum of f on the surface x2 + 10y 2 + z 2 = 5 • f (−2, 0, 1) = −5 is the minimum of f on the surface x2 +10y 2 + z 2 = 5 7 ...
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This note was uploaded on 11/16/2011 for the course MULTIVARIA 251 taught by Professor Staff during the Fall '11 term at Rutgers.

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