# C is a potential function thus x y 2 x y 2 12 b

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Unformatted text preview: 0 we get cos φ = (4x − 27)/(2x), x = 0 so from Aφ = 0 we get (27 − 2x + 2x cos φ) cos φ − x = 0 which, for cos φ = (4x − 27)/(2x), yields 4x − 27 − x = 0, x = 9. If x = 9 then cos φ = 1/2, φ = π/3. The critical point occurs when x = 9 and φ = π/3; √ Axx Aφφ − A2 = 729/2 > 0 and Axx = −3 3/2 < 0 there, the area is maximum when x = 9 and xφ φ = π/3. Exercise Set 15.8 45. (a) 564 ∂g = ∂m n n i=1 n x2 i i=1 = 0 if xi yi i=1 n xi m+ i=1 n xi − x2 + b i i=1 n ∂g = ∂b n 2 (mxi + b − yi ) xi = 2 m xi yi , b= i=1 i=1 n n n 2 (mxi + b − yi ) = 2 m i=1 n xi + bn − i=1 = 0 if yi i=1 n xi i=1 m + nb = yi i=1 (b) The function z = g (m, b), as a function of m and b, has only one critical point, found in part (a), and tends to +∞ as either |m| or |b| tends to inﬁnity. Thus the only critical point must be a minimum. n n n 2 (xi − x) = ¯ 46. (a) i=1 x ¯ x2 − 2¯xi + x2 = i i=1 i=1 n x2 − i = i=1 n x2 i = i=1 2 n 1 − n 2 n xi + i=1 1 n xi + nx2 ¯ i=1 2 n xi i=1 2 n n xi 2 n x2 i > 0 so n i=1 − i=1 n xi >0 i=1 n x2 , gbb = 2n, gmb = 2 i (b) gmm = 2 n x2 − 2¯ x i i=1 xi , i=1 2 D = gmm gbb − gmb = 4 n n 2 n x2 − i xi i=1 > 0 and gmm > 0 i=1 (c) g (m, b) is of the second-degree in m and b so the graph of z = g (m, b) is a quadric surface. (d) The only quadric surface of this form having a relative minimum is a paraboloid that opens upward where the relative minimum is also the absolute minimum. 3 3 xi = 3, 47. n = 3, i=1 4 5 i=1 14 36 x+ 35 5 4 x2 = 30, i i=1 xi yi = 23, n = 4; m = 0.5, b = 0.8, y = 0.5x + 0.8. i=1 5 5 xi yi = 39.8, n = 5; m = −0.55, b = 4.67, y = 4.67 − 0.55x x2 = 55, i yi = 15.1, i=1 xi yi = −2, y = − i=1 4 5 xi = 15, 19 3 x+ 4 12 4 x2 = 21, i i=1 yi = 8.2, i=1 x2 = 11, y = i i=1 4 yi = 4, 4 i=1 50. i=1 i=1 xi = 10, 3 xi yi = 13, 4 xi = 7, i=1 49. yi = 7, i=1 4 48. n = 4, 3 i=1 i=1 565 Chapter 15 51. (a) y = 57 8843 + t ≈ 63.1643 + 0.285t 140 200 (b) 80 0(1930) (c) y = 60 60 2909 ≈ 83.1143 35 52. (a) y ≈ 119.84 − 1.13x 90 (b) 35 50 60 (c) about 52 units 53. (a) P = 2798 171 + T ≈ 133.2381 + 0.4886T 21 350 (b) 190 0 130 (c) T ≈ − 120 139,900 =≈ −272.7096◦ C 513 54. (a) for example, z = y (b) For example, on 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 let z = y if 0 < x < 1, 0 < y < 1 1/2 if x = 0, 1 or y = 0, 1 55. f (x0 , y0 ) ≥ f (x, y ) for all (x, y ) inside a circle centered at (x0 , y0 ) by virtue of Deﬁnition 15.8.1. If r is the radius of the circle, then in particular f (x0 , y0 ) ≥ f (x, y0 ) for all x satisfying |x − x0 | < r so f (x, y0 ) has a relative maximum at x0 . The proof is similar for the function f (x0 , y ). EXERCISE SET 15.9 1. (a) xy = 4 is tangent to the line, so the maximum value of f is 4. (b) xy = 2 intersects the curve and so gives a smaller value of f . (c) Maximize f (x, y ) = xy subject to the constraint g (x, y ) = x + y − 4 = 0, f = λ g , y i + xj = λ(i + j), so solve the equations y = λ, x = λ with solution x = y = λ, but x + y = 4, so x = y = 2, and the maximum value of f is f = xy = 4. Exercise Set 15.9 566 2. (a) x2 + y 2 = 25 is tangent to the line at (3, 4), so the minimum value of f is 25. (b) A smaller value of f yields a circle of a smaller radius, and hence does not intersect the line. (c) Minimize f (x, y ) = x2 + y 2 subject to the constraint g (x, y ) = 3x + 4y − 25 = 0, f = λ g , 2xi + 2y j = 3λi + 4λj, so solve 2x = 3λ, 2y = 4λ; solution is x = 3, y = 4, minimum = 25. 3. y = 8xλ, x = 16yλ; y/(8x) = x/(16y ), x2 = 2y 2 so 4 2y 2 + 8y 2 = 16, y 2 = 1, y = ±1. Test √ √ √ √ √ √ √ √ ± 2, −1 and (± 2, 1). f − 2, −1 = f 2, 1 = 2, f − 2, 1 = f 2, −1 = − 2. √ √ √ √ √ √ Maximum 2 at − 2, −1 and 2, 1 , minimum − 2 at − 2, 1 and 2, −1 . 2 4. 2x = 2xλ, −1 = 2yλ. If x = 0 then λ = 1 and y = −1/2 so x2 + (−1/2)√= 25, √ 2 2 2 x = 99/4, x = ±3 11/2. If x = 0 then 0 + y = 25, y = ±5. Test ±3 11/2, −1/2 and (0, ±5). √ √ f ±3 11/2, −1/2 = 101/4, f (0, −5) = 5, f (0, 5) = −5. Maximum 101/4 at ±3 11/2, −1/2 , minimum −5 at (0,5). 5. 12x2 = 4xλ, 2y = 2yλ. If y = 0 then λ = 1 and 12x2 = 4x, 12x(x − 1/3) = 0, x = 0 or x = 1/3 √ that so from 2x2 + y 2 = 1 we ﬁnd √ y = ±1 when x = 0, y = ± 7/3 when x = 1/3. If y = 0 √ √ then 2x2 + (0)2 = 1, x = ±1/ 2. Test (0, ±1), 1/3, ± 7/3 , and ±1/ 2, 0 . f (0, ±1) = 1, √ √ √ √ √ √ √ f 1/3, ± 7/3 = 25/27, f 1/ 2, 0 = 2, f −1/ 2, 0 = − 2. Maximum 2 at 1/ 2, 0 , √ √ minimum − 2 at −1/ 2, 0 . 6. 1 = 2xλ, −3 = 6yλ; 1/(2x) = −1/(2y ), y = −x so x2 + 3(−x)2 = 16, x = ±2. Test (−2, 2) and (2, −2). f (−2, 2) = −9, f (2, −2) = 7. Maximum 7 at (2, −2), minimum −9 at (−2, 2). 7. 2 = 2xλ, 1 = 2yλ, −2 = 2zλ; 1/x = 1/(2y ) = −1/z thus x = 2y , z = −2y so (2y )2 + y 2 + (−2y )2 = 4, y 2 = 4/9, y = ±2/3. Test (−4/3, −2/3, 4/3) and (4/3, 2/3, −4/3). f (−4/3, −2/3, 4/3) = −6, f (4/3, 2/3, −4/3)...
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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