# C 39 c x 4 cos t y 4 sin t 0 t 2 2 0 1 sin

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Unformatted text preview: mbers 16,16,16. 34. Minimize S = x2 + y 2 + z 2 subject to x + y + z = 27, x > 0, y > 0, z > 0. z = 27 − x − y so S = x2 + y 2 + (27 − x − y )2 , Sx = 4x + 2y − 54 = 0, Sy = 2x + 4y − 54 = 0; critical point (9,9); 2 Sxx Syy − Sxy = 12 > 0 and Sxx = 4 > 0 at (9,9), relative minimum. z = 9 when x = y = 9, the sum of the squares is minimum for the numbers 9,9,9. 35. Maximize w = xy 2 z 2 subject to x + y + z = 5, x > 0, y > 0, z > 0. x = 5 − y − z so w = (5 − y − z )y 2 z 2 = 5y 2 z 2 − y 3 z 2 − y 2 z 3 , wy = 10yz 2 − 3y 2 z 2 − 2yz 3 = yz 2 (10 − 3y − 2z ) = 0, wz = 10y 2 z − 2y 3 z − 3y 2 z 2 = y 2 z (10 − 2y − 3z ) = 0, 10 − 3y − 2z = 0 and 10 − 2y − 3z = 0; critical 2 point when y = z = 2; wyy wzz − wyz = 320 > 0 and wyy = −24 < 0 when y = z = 2, relative maximum. x = 1 when y = z = 2, xy 2 z 2 is maximum at (1,2,2). 36. Minimize w = D2 = x2 + y 2 + z 2 subject to x2 − yz = 5. x2 = 5 + yz so w = 5 + yz + y 2 + z 2 , 2 wy = z + 2y = 0, wz = y + 2z = 0; critical point when y = z = 0; wyy wzz − wyz = 3 > 0 and √ 2 wyy = 2 > 0 when y = z = 0, relative minimum. x = 5, x = ± 5 when y = z = 0. The points √ ± 5, 0, 0 are closest to the origin. 37. The diagonal of the box must equal the diameter of the sphere, thus we maximize V = xyz or, for convenience, w = V 2 = x2 y 2 z 2 subject to x2 +y 2 +z 2 = 4a2 , x > 0, y > 0, z > 0; z 2 = 4a2 −x2 −y 2 hence w = 4a2 x2 y 2 − x4 y 2 − x2 y 4 , wx = 2xy 2 (4a2 − 2x2 − y 2 ) = 0, wy = 2x2 y 4a2 − x2 − 2y 2 = 0, √ √ 4a2 − 2x2 − y 2 = 0 and 4a2 − x2 − 2y 2 = 0; critical point 2a/ 3, 2a/ 3 ; √ √ 4096 8 128 4 2 a > 0 and wxx = − a < 0 at 2a/ 3, 2a/ 3 , relative maximum. wxx wyy − wxy = 27 9 √ √ z =√a/ 3 √ 2 when x√ y = 2a/ 3, the dimensions of the box of maximum volume are = 2a/ 3, 2a/ 3, 2a/ 3. 38. Maximize V = xyz subject to x + y + z = 1, x > 0, y > 0, z > 0. z = 1 − x − y so V = xy − x2 y − xy 2 , Vx = y (1 − 2x − y ) = 0, Vy = x(1 − x − 2y ) = 0, 1 − 2x − y = 0 and 1 − x − 2y = 0; critical point 2 (1/3, 1/3); Vxx Vyy − Vxy = 1/3 > 0 and Vxx = −2/3 < 0 at (1/3, 1/3), relative maximum. The maximum volume is V = (1/3)(1/3)(1/3) = 1/27. 563 Chapter 15 39. Let x, y , and z be, respectively, the length, width, and height of the box. Minimize C = 10(2xy ) + 5(2xz + 2yz ) = 10(2xy + xz + yz ) subject to xyz = 16. z = 16/(xy ) so C = 20(xy + 8/y + 8/x), Cx = 20(y − 8/x2 ) = 0, Cy = 20(x − 8/y 2 ) = 0; 2 critical point (2,2); Cxx Cyy − Cxy = 1200 > 0 and Cxx = 40 > 0 at (2,2), relative minimum. z = 4 when x = y = 2. The cost of materials is minimum if the length and width are 2 ft and the height is 4 ft. 40. Maximize the proﬁt P = 500(y − x)(x − 40) + [45, 000 + 500(x − 2y )](y − 60) = 500(−x2 − 2y 2 + 2xy − 20x + 170y − 5400). Px = 1000(−x + y − 10) = 0, Py = 1000(−2y + x + 85) = 0; critical point (65,75); 2 Pxx Pyy − Pxy = 1,000,000 > 0 and Pxx = −1000 < 0 at (65,75), relative maximum. The proﬁt will be maximum when x = 65 and y = 75. 41. (a) x = 0 : f (0, y ) = −3y 2 , minimum −3, maximum 0; ∂f (1, y ) = −6y + 2 = 0 at y = 1/3, minimum 3, x = 1, f (1, y ) = 4 − 3y 2 + 2y, ∂y maximum 13/3; y = 0, f (x, 0) = 4x2 , minimum 0, maximum 4; ∂f (x, 1) = 8x + 2 = 0 for 0 < x < 1, minimum −3, maximum 3 y = 1, f (x, 1) = 4x2 + 2x − 3, ∂x d (b) f (x, x) = 3x2 , minimum 0, maximum 3; f (x, 1−x) = −x2 +8x−3, f (x, 1−x) = −2x+8 = 0 dx for 0 < x < 1, maximum 4, minimum −3 (c) fx (x, y ) = 8x + 2y = 0, fy (x, y ) = −3y + 2x = 0, solution is (0, 0), which is not an interior point of the square, so check the sides: minimum −3, maximum 14/3. 42. Maximize A = ab sin α subject to 2a + 2b = , a > 0, b > 0, 0 < α < π . b = ( − 2a)/2 so A = (1/2)( a − 2a2 ) sin α, Aa = (1/2)( − 4a) sin α, Aα = (a/2)( − 2a) cos α; sin α = 0 so from Aa = 0 we get a = /4 and then from Aα = 0 we get cos α = 0, α = π/2. Aaa Aαα − A2 = 2 /8 > 0 aα and Aaa = −2 < 0 when a = /4 and α = π/2, the area is maximum. 43. Minimize S = xy + 2xz + 2yz subject to xyz = V , x > 0, y > 0, z > 0 where x, y , and z are, respectively, the length, width, and height of the box. z√ V /(xy ) so S = xy + 2V /y + 2V /x, =√ 2 Sx = y − 2V /x2 = 0, Sy = x − 2V /y 2 = 0; critical point ( 3 2V , 3 2V ); Sxx Syy − Sxy = 3 > 0 and Sxx = 2 > 0 at this point so there is a relative minimum there. The length and width are each √ √ 3 2V , the height is z = 3 2V /2. 44. The altitude of the trapezoid is x sin φ and the lengths of the lower and upper bases are, respectively, 27 − 2x and 27 − 2x + 2x cos φ so we want to maximize A = (1/2)(x sin φ)[(27 − 2x) + (27 − 2x + 2x cos φ)] = 27x sin φ − 2x2 sin φ + x2 sin φ cos φ. Ax = sin φ(27 − 4x + 2x cos φ), Aφ = x(27 cos φ − 2x cos φ − x sin2 φ + x cos2 φ) = x(27 cos φ − 2x cos φ + 2x cos2 φ − x). sin φ = 0 so from Ax =...
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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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