Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

224 0 CHAPTER 14 PARTIAL DERIVATIVES
f(x, x) > 0 for 0 < x < 1r while f(x, x) < 0 for 1r < x < 21r. Thus every disk with center (1r, 1r) contains points where f is
positive as well as points where f is negat~v e, so the graph crosses its tangent plane (z = 0) there and (1r, 1r) is a saddle point.
D(~, ~) = * > 0 and fxx (~,·?f) < 0 so/( ~,~ ) = .tf! is a local maximum while D( 5 3 ", 6 ;) = ~ > 0 and
f xx ( 5 ;, 5 ;) > 0, so f ( 6 3 " , 6 ;) = -¥ is a local minimum.
4x(x 2 - 2y) = 0, sox= 0 or x 2 = 2y. lfx = 0 then substitution into f 11
= 0 gives 4y 3 = -2 :::::}
• 1
y = - 1i2, so
( 0, - ~) is a critical point. Substituting x 2 = 2y into fv = 0 gives 4y 3 - 8y + 2 = 0. Using a graph, solutions are
approximately y = - 1.526, 0.259, and 1.267. (Alternatively, we could have used a calculator or a CAS to find these roots.)
We have x 2 = 2y => x = ±.J2Y, soy = -1.526 gives no real-valued solution for x, but
y = 0.259 => x ~ ± 0.720 andy = ,1.267 => x ~ ±1.592. Thus to three decimal places, the critical points are
( 0, -~) ~ (0, - 0.794), (±0.720, 0.259), and (±1.592, 1.267). Now since f xx = 12x 2 - 8y, f ., 11 • = - 8x, f 1111 = 12y 2 ,
and D = (12x 2 - 8y)(12y 2 )- 64x 2 , we have D(O, -0.794) > 0, fxx(O, - 0.794) > 0, D(±0.720, 0.259) < 0,
D(±1.592, 1.267) > 0, and f xx (± 1.592, 1.267) > 0. Therefore /(0, _-0.794) ~ - 1.191 and /(±1.592, 1.267) ~ - 1.310
are local minima, and (± 0. 720, 0.259) are saddle points. There is no highest point on the graph, but the lowest points are
approximately (±1.592, 1.267, -1.310).
20
10
-6
0
y
27. f(x, y) = x 4 +y 3 - 3x 2 +y 2 +x - 2y + 1 => f x (x, y) = 4x 3 - 6x + 1 and f 11 (x, y) = 3y 2 + 2y- 2. From the
graphs, we see thatto three decimal places, f ., = 0 when x ~ -1.301, 0.170, or 1.131, and / 11
= 0 when y ~ -1.215 or
0.549. (Alternatively, we could have used a calculator or a CAS to find these roots. We could also use the quadratic formula to
find the solutions of / 11 = 0.) So, to three decimal places, f has critical points at (- 1.301, - 1.215), (- 1.301, 0.549),
(0.170, -1.215), (0.170, 0.549), (1.131, - 1.215), and (1.131, 0.549). Now since /zz = 12x 2 - 6, fx 11 .= 0, f 11 y = 6y + 2,
and D = (12x 2 - 6)(6y + 2), we have D( -1.301, -1.215) < 0, D( - 1.301, 0.549) > 0, f xx( -1.301, 0.549) > 0,
D(0.170, - 1.215) > 0, f xx (0.170, - 1.215) < 0, D(0.170, 0.549) < 0, D(l.131, -1.215) < 0, D(l.131, 0.549) > 0, and
/ xz (l.131, 0.549) > 0. Therefore, to three decimal places, f( -1.301, 0.549) ~ -3.145 and /(1.131, 0.549) ~ - 0.701 are
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