F r 1r3 f r 3r4 divrr3 31r3 r3r4

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: +2x) = 0, fy = −2ye−(x +y +2x) = 0; critical point (−1, 0); D = 4e2 > 0 and fxx = −2e < 0 at (−1, 0), relative maximum. 20. fx = y − a3 /x2 = 0, fy = x − b3 /y 2 = 0; critical point a2 /b, b2 /a ; if ab > 0 then D = 3 > 0 and fxx = 2b3 /a3 > 0 at a2 /b, b2 /a , relative minimum; if ab < 0 then D = 3 > 0 and fxx = 2b3 /a3 < 0 at a2 /b, b2 /a , relative maximum. 21. 2 1 0 -1 -2 -2 -1 0 1 2 f = (4x − 4y )i − (4x − 4y 3 )j = 0 when x = y, x = y 3 , so x = y = 0 or x = y = ±1. At (0, 0), D = −16, a saddle point; at (1, 1) and (−1, −1), D = 32 > 0, fxx = 4, a relative minimum. 22. 5 140 4 100 -100 60 3 -60 20 2 -20 1 0 0 0 -1 -2 40 20 20 -3 0 60100 -20 -4 140 -40 -5 -5 -4 -3 -2 -1 0 1 2 3 4 5 f = (2y 2 − 2xy + 4y )i + (4xy − x2 + 4x)j = 0 when 2y 2 − 2xy + 4y = 0, 4xy − x2 + 4x = 0, with solutions (0, 0), (0, −2), (4, 0), (4/3, −2/3). At (0, 0), D = −16, a saddle point. At (0, −2), D = −16, a saddle point. At (4, 0), D = −16, a saddle point. At (4/3, −2/3), D = 16/3, fxx = 4/3 > 0, a relative minimum. 23. (a) critical point (0,0); D = 0 (b) f (0, 0) = 0, x4 + y 4 ≥ 0 so f (x, y ) ≥ f (0, 0), relative minimum. 24. (a) critical point (0,0); D = 0 (b) f (0, 0) = 0, inside any circle centered at (0, 0) there are points where f (x, y ) > 0 (along the x-axis) and points where f (x, y ) < 0 (along the y -axis) so (0, 0) is a saddle point. 561 Chapter 15 25. (a) fx = 3ey − 3x2 = 3 ey − x2 = 0, fy = 3xey − 3e3y = 3ey x − e2y = 0, ey = x2 and e2y = x, x4 = x, x x3 − 1 = 0 so x = 0, 1; critical point (1, 0); D = 27 > 0 and fxx = −6 < 0 at (1, 0), relative maximum. (b) lim f (x, 0) = lim x→−∞ x→−∞ 3x − x3 − 1 = +∞ so no absolute maximum. 26. fx = 8xey − 8x3 = 8x(ey − x2 ) = 0, fy = 4x2 ey − 4e4y = 4ey (x2 − e3y ) = 0, x2 = ey and x2 = e3y , e3y = ey , e2y = 1, so y = 0 and x = ±1; critical points (1,0) and (−1, 0). D = 128 > 0 and fxx = −16 < 0 at both points so a relative maximum occurs at each one. 27. fx = y − 1 = 0, fy = x − 3 = 0; critical point (3,1). Along y = 0 : u(x) = −x; no critical points, along x = 0 : v (y ) = −3y ; no critical points, 4 along y = − x + 4 : 5 4 27 w(x) = − x2 + x − 12; critical point (27/8, 13/10). 5 5 (x, y ) (3, 1) f (x, y ) −3 (0, 0) 0 (5, 0) −5 (0, 4) −12 (27/8, 13/10) −231/80 Absolute maximum value is 0, absolute minimum value is −12. 28. fx = y − 2 = 0, fy = x = 0; critical point (0,2), but (0,2) is not in the interior of R. Along y = 0 : u(x) = −2x; no critical points, along x = 0 : v (y ) = 0; no critical points, along y = 4 − x : w(x) = 2x − x2 ; critical point (1, 3). (x, y ) (0, 0) f (x, y ) 0 (0, 4) 0 (4, 0) −8 (1, 3) 1 Absolute maximum value is 1, absolute minimum value is −8. 29. fx = 2x − 2 = 0, fy = −6y + 6 = 0; critical point (1,1). Along y = 0 : along y = 2 : along x = 0 : along x = 2 : u1 (x) = x2 − 2x; critical point (1, 0), u2 (x) = x2 − 2x; critical point (1, 2) v1 (y ) = −3y 2 + 6y ; critical point (0, 1), v2 (y ) = −3y 2 + 6y ; critical point (2, 1) (x, y ) (1, 1) f (x, y ) 2 (1, 0) −1 (1, 2) −1 (0, 1) 3 (2, 1) 3 (0, 0) 0 (0, 2) 0 (2, 0) 0 (2, 2) 0 Absolute maximum value is 3, absolute minimum value is −1. 30. fx = ey − 2x = 0, fy = xey − ey = ey (x − 1) = 0; critical point (1, ln 2). Along y = 0 : along y = 1 : along x = 0 : along x = 2 : u1 (x) = x − x2 − 1; critical point (1/2, 0), u2 (x) = ex − x2 − e; critical point (e/2, 1), v1 (y ) = −ey ; no critical points, v2 (y ) = ey − 4; no critical points (for 0 < y < 1). (x, y ) (0, 0) f (x, y ) −1 (0, 1) −e (2, 1) e−4 (2, 0) −3 (1, ln 2) (1/2, 0) −1 −3/4 Absolute maximum value is −3/4, absolute minimum value is −3. (e/2, 1) e(e − 4)/4 ≈ −0.87 Exercise Set 15.8 562 31. fx = 2x − 1 = 0, fy = 4y = 0; critical point (1/2, 0). √ Along x2 + y 2 = 4 : y 2 = 4 − x2 , u(x) = 8 − x − x2 for −2 ≤ x ≤ 2; critical points (−1/2, ± 15/2). (x, y ) (1/2, 0) f (x, y ) −1/4 √ −1/2, 15/2 33/4 √ −1/2, − 15/2 33/4 (−2, 0) (2, 0) 6 2 Absolute maximum value is 33/4, absolute minimum value is −1/4. 32. fx = y 2 = 0, fy = 2xy = 0; no critical points in the interior of R. Along y = 0 : u(x) = 0; no critical points, along x = 0 : v (y ) = 0; no critical points along x2 + y 2 = 1 : w(x) = x − x3 for 0 ≤ x ≤ 1; critical point (x, y ) (0, 0) (0, 1) (1, 0) f (x, y ) 0 0 0 Absolute maximum value is √ 1/ 3, 2/3 . √ 1/ 3, 2/3 √ 2 3/9 2√ 3, absolute minimum value is 0. 9 33. Maximize P = xyz subject to x + y + z = 48, x > 0, y > 0, z > 0. z = 48 − x − y so P = xy (48 − x − y ) = 48xy − x2 y − xy 2 , Px = 48y − 2xy − y 2 = 0, Py = 48x − x2 − 2xy = 0. But 2 x = 0 and y = 0 so 48 − 2x − y = 0 and 48 − x − 2y = 0; critical point (16,16). Pxx Pyy − Pxy > 0 and Pxx < 0 at (16,16), relative maximum. z = 16 when x = y = 16, the product is maximum for the nu...
View Full Document

This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

Ask a homework question - tutors are online