Ru 2 cos u cos v i 2 cos u sin v j 2 sin uk rv 2

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: = 0.11, dw = (2xy − 2y + y 2 )dx + (x2 − 2x + 2yx)dy = −(−0.1) = 0.1 23. dV = 1 1 2 2 xhdx + x2 dh = 2(−0.1) + (0.2) = −0.06667 m3 ; ∆V = −0.07267 m3 3 3 3 3 24. f = (2x + 3y − 6)i + (3x + 6y + 3)j = 0 if 2x + 3y = 6, x + 2y = −1, x = 15, y = −8, D = 3 > 0, fxx = 2 > 0, so f has a relative minimum at (15, −8). 25. f = (2xy − 6x)i + (x2 − 12y )j = 0 if 2xy − 6x = 0, x2 − 12y = 0; if x = 0 then y = 0, and if x = 0 then y = 3, x = ±6, thus the gradient vanishes at (0, 0), (−6, 3), (6, 3); fxx = 0 at all three points, fyy = −12 < 0, D = −4x2 , so (±6, 3) are saddle points, and near the origin we write f (x, y ) = (y − 3)x2 − 6y 2 ; since y − 3 < 0 when |y | < 3, f has a maximum by inspection. 26. f = (3x2 − 3y )i − (3x − y )j = 0 if y = x2 , 3x = y , so x = y = 0 or x = 3, y = 9; at x = y = 0, D = −9, saddle point; at x = 3, y = 9, D = 9, fxx = 18 > 0, relative minimum 571 Chapter 15 27. (a) (b) 5 K P=3 5 4 P=2 4 3 3 P=1 2 2 1 L 1 2 3 4 1 5 0 0 1 2 3 4 5 28. (a) ∂P/∂L = cαLα−1 K β , ∂P/∂K = cβLα K β −1 (b) the rates of change of output with respect to labor and capital equipment, respectively (c) K (∂P/∂K ) + L(∂P/∂L) = cβLα K β + cαLα K β = (α + β )P = P 29. (a) L + K = 200,000, P = 1000L0.6 (200,000 − L)0.4 , L6 (200,000 − L)0.4 − 400 = 0 when L = 120,000, L4 (200,000 − L)0.6 P = 102,033,960.1, which is a maximum because P = 0 at L = 0, 200,000, P > 0 in between, and dP/dL = 0 has only the one solution. dP/dL = 600 (b) Since L + K = 200,000 and L = 120,000, K = 80,000 √ √ 30. (a) y 2 = 8 − 4x2 , find extrema of f (x) = x2 (8 − 4x2 ) = −4x4 + 8x2 defined for − 2 ≤ x ≤ 2. Then f (x) = −16x3 + 16x = 0 when x = 0, ±1, f (x) = −48x2 + 16, so f has a relative √ maximum at x = ±1, y = ±2 and a relative minimum at x = 0, y = ±2 2. At the endpoints √ x = ± 2, y = 0 we obtain the minimum f = 0 again. λy (b) f (x, y ) = x2 y 2 , g (x, y ) = 4x2 + y 2 − 8 = 0, f = 2xy 2 i + 2x2 y j = λ g = 8λxi + 2√ j, so √ solve 2xy 2 = λ8x, 2x2 y = λ2y . If x = 0 then y = ±2 2, and if y = 0 then x = ± 2. In either case f has a relative and absolute minimum. Assume x, y = 0, then y 2 = 4λ, x2 = λ, use g = 0 to obtain x2 = 1, x = ±1, y = ±2, and f = 4 is a relative and absolute maximum at (±1, ±2). 31. Let a corner of the box be at (x, y, z ), so that (x/a)2 + (y/b)2 + (z/c)2 = 1. Maximize V = xyz subject to g (x, y, z ) = (x/a)2 + (y/b)2 + (z/c)2 = 1, solve V = λ g , or yz i + xz j + xy k = (2λx/a2 )i + (2λy/b2 )j + (2λz/c2 )k, a2 yz = 2λx, b2 xz = 2λy, c2 xy = 2λz . For the maximum volume, x, y, z = 0; divide the first and second equations to obtain a2 y 2 = b2 x2 ; the first √ and third to obtain a2 z 2 = c2 x2 , and finally b2 z 2 = c2 y 2 . From g = 0 get 3(x/a)2 = 1, x = ±a/ 3, √ √ 2a 2b 2c and then y = ±b/ 3, z = ±c/ 3. The dimensions of the box are √ × √ × √ , and the 3 3 3 √ maximum volume is 8abc/(3 3). 32. (a) Let f (x, y ) = 3x2 − 5xy + tan xy = 0. Then dy 6x − 5y + y sec2 xy dy df = 6x − 5y + y sec2 xy + (−5x + x sec2 xy ) = 0, so = . dx dx dx 5x − x sec2 xy (b) Let g (x, y ) = x ln y + sin(x − y ) = π, ln y + cos(x − y ) dy = dx −x/y + cos(x − y ) dg = ln y + cos(x − y ) + dx x − cos(x − y ) y dy = 0, dx Chapter 15 Supplementary Exercises 33. F (x, y ) = 0, Fx + Fy 572 dy dy dy d2 y = 0, Fxx + Fxy + Fyx + Fy 2 = 0, thus dx dx dx dx Fx d2 y Fxx + 2Fxy (dy/dx) Fxx Fy − 2Fx Fxy dy =− , 2 =− =− 2 dx Fy dx Fy Fy 34. Denote the currents I1 , I2 , I3 by x, y, z respectively. Then minimize F (x, y, z ) = x2 R1 + y 2 R2 + z 2 R3 subject to g (x, y, z ) = x + y + z − I = 0, so solve F = λ g , 2xR1 i +2yR2 j +2zR3 k = λ(i + j + k), 1 1 1 : : . λ = 2xR1 = 2yR2 = 2zR3 , so the minimum value of F occurs when I1 : I2 : I3 = R1 R2 R3 √ 35. Solve (t − 1)2 /4 + 16e−2t + (2 − t)2 = 1 for t to get t = 1.8332, 2.83984; the particle strikes the surface at the points P1 (0.8332, 0.63959, 0.64603), P2 (1.83984, 0.23374, 0.31482). The velocity √ dy dz dx i + j + k = i − 4e−t j − 1/(2 t)k, and a normal to the surface is vectors are given by v = dt dt dt n = (x2 /4 + y 2 + z 2 ) = x/2i + 2y j + 2z k. At the points Pi these are v1 = i − 0.639589j − 0.369286k, v2 = i − 0.23374j + 0.29670k; n1 = 0.41661i + 1.27918j + 1.29207k and n2 = 0.91992i + 0.46748j + 0.62963k so cos−1 [(vi · ni )/( vi ni )] = 112.3◦ , 61.1◦ ; the acute angles are 67.7◦ , 61.1◦ . 1 ey cos(xey )dy = 36. (a) F (x) = 0 (b) Use a CAS to get x = F π e+1 1 = sin 0 sin(ex) − sin x x π so the maximum value of F (x) is e+1 πy e e+1 dy ≈ 0.909026. 37. Let x, y, z be the lengths of the sides opposite angles α, β, γ , located at A,√ C respectively. Then B, x2 = y 2 + z 2 − 2yz cos α and x2 = 100 + 400 − 2(10)(20)/2 = 300, x = 10 3 and 2x so dy dz dz dy dα dx = 2y + 2z − 2 y cos α + z cos α − yz (sin α) dt dt dt dt dt dt √...
View Full Document

Ask a homework question - tutors are online