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

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Unformatted text preview: +2x) = 0, fy = −2ye−(x +y +2x) = 0; critical point (−1, 0); D = 4e2 &gt; 0 and fxx = −2e &lt; 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 &gt; 0 then D = 3 &gt; 0 and fxx = 2b3 /a3 &gt; 0 at a2 /b, b2 /a , relative minimum; if ab &lt; 0 then D = 3 &gt; 0 and fxx = 2b3 /a3 &lt; 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 &gt; 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 &gt; 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 ) &gt; 0 (along the x-axis) and points where f (x, y ) &lt; 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 &gt; 0 and fxx = −6 &lt; 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 &gt; 0 and fxx = −16 &lt; 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 &lt; y &lt; 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 &gt; 0, y &gt; 0, z &gt; 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 &gt; 0 and Pxx &lt; 0 at (16,16), relative maximum. z = 16 when x = y = 16, the product is maximum for the nu...
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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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