Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Name: Solutions Math 16B: Quiz 2 July 3, 2014 1. (4 points) Sketch the level curves of the function f (x, y) = (y − x2 )3 at heights −2, 8 and 20 (on the same axes). For level curve at height c, we have c = f (x, y) = (y − x2 )3 ⇒ c1/3 = y − x2 ⇒ y = x2 + c1/3 . Plugging in the values of c: y = x2 − 21/3 , y = x2 + 2 and y = x2 + 201/3 . 20 15 y 10 c = 20 5 c=8 0 c = −2 −5 −4 −3 −2 −1 0 x 1 2 3 4 Figure 1 2 −y 2 2. (7 points) Find all the relative maxima and minima of f (x, y) = e4y−x mine their nature. and deter- We find the partial derivatives of f and set them equal to zero. We have: ∂f 2 2 = (−2x)e4y−x −y = 0 ∂x ∂f 2 2 = (4 − 2y)e4y−x −y = 0 ∂y ̸ 0, the first equation implies that x = 0; from the second, we get Since e4y−x −y = 4 − 2y = 0 ⇒ y = 2. Thus, the only possible extreme point of this function is at (0, 2). 2 2 2 To determine the nature of (0, 2), we use the second-derivative test. We have: ∂∂xf2 = 2 2 2 2 2 2 2 2 2 ∂2f −2e4y−x −y + 4x2 e4y−x −y , ∂∂yf2 = −2e4y−x −y + (4 − 2y)2 e4y−x −y and ∂x∂y = (4 − 4y−x2 −y 2 4 4 2 8 2y)(−2x)e . At (0, 2), we have D = (−2e )(−2e ) − (0) = 4e > 0. Moreover, ∂f 4 since ∂x2 = −2e < 0, f has a relative maximum at (0, 2). 1 3. (9 points) Find three nonnegative numbers whose sum is 12 and the sum of whose squares is as small as possible. (Finding a relative minimum is fine; 5 bonus points if you can show that the relative minimum is also an absolute minimum.) Let the three numbers be x, y and z. We want to minimize S = x2 + y 2 + z 2 subject to x + y + z = 12 ⇒ z = 12 − x − y. Thus, S = x2 + y 2 + (12 − x − y)2 . To find the critical points, we set the partial derivatives of S equal to zero: ∂S = 2x − 2(12 − x − y) = −24 + 4x + 2y = 0 ∂x ∂S = 2y − 2(12 − x − y) = −24 + 2x + 4y = 0 ∂y Add (−2) times the first equation to the second to get 24 − 6x = 0 ⇒ x = 4. Thus, y = 4 and z = 4. At these values, S = 48. 2 To see whether or not this minimizes S, we use the Hessian test. We have: ∂∂xS2 = 4, ∂2S ∂2S = 4 and ∂x∂y = 2. Thus, D = (4)(4) − (2)2 = 16 − 4 = 12 > 0. Moreover, since ∂y 2 ∂2S ∂x2 > 0, it follows that this choice of x, y and z relatively minimizes S. Bonus part: Since z ≥ 0, the domain of S is the region given by x, y ≥ 0 and x + y ≤ 12. See Figure 2 for the domain. 12 10 8 y x + y = 12 6 4 2 0 0 2 4 6 x 8 10 12 Figure 2 (a) On the x = 0 boundary, we get S = y 2 + (12 − y)2 = 2y 2 − 24y + 144. This is a concave-up parabola so it is minimized at its critical point. Differentiate this w.r.t y and set it equal to zero to get 4y − 24 = 0 ⇒ y = 6. At y = 6, S = 36 + 36 = 72. Thus, S ≥ 72 on this boundary. (b) On the y = 0 boundary, we get S = x2 + (12 − x2 ). Same reasoning as above shows that S never goes below 72 on this boundary. (c) On x + y = 12, use y = 12 − x and plug into S to get S = x2 + (12 − x)2 . Same reasoning as above shows that S never goes below 72 on this boundary. Thus, comparing the value of S at the critical point and at the boundaries, we see that S attains its least value of 48 at the critical point where x = y = z = 4. 2