Solução_Calculo_Stewart_6e

F.
TX.10 SECTION 15.7 MAXIMUM AND MINIMUM VALUES ET SECTION 14.7 ¤ 201
21. f(x, y) =x 2 + y 2 + x −2 y −2
From the graphs, there appear to be local minima of about f(1, ±1) = f(−1, ±1) ≈ 3 (and no local maxima or saddle
points). f x =2x − 2x −3 y −2 , f y =2y − 2x −2 y −3 , f xx =2+6x −4 y −2 , f xy =4x −3 y −3 , f yy =2+6x −2 y −4 .Then
f x =0implies 2x 4 y 2 − 2=0or x 4 y 2 =1or y 2 = x −4 . Note that neither x nor y can be zero. Now f y =0implies
2x 2 y 4 − 2=0,andwithy 2 = x −4 this implies 2x −6 − 2=0or x 6 =1. Thus x = ±1 and if x =1, y = ±1;ifx = −1,
y = ±1. So the critical points are (1, 1), (1, −1),(−1, 1) and (−1, −1). NowD(1, ±1) = D(−1, ±1) = 64 − 16 > 0 and
f xx > 0 always, so f(1, ±1) = f(−1, ±1) = 3 are local minima.
23. f(x, y) =sinx +siny +sin(x + y), 0 ≤ x ≤ 2π, 0 ≤ y ≤ 2π
From the graphs it appears that f has a local maximum at about (1, 1) with value approximately 2.6, a local minimum
at about (5, 5) with value approximately −2.6, and a saddle point at about (3, 3).
f x =cosx +cos(x + y), f y =cosy +cos(x + y), f xx = − sin x − sin(x + y), f yy = − sin y − sin(x + y),
f xy = − sin(x + y). Setting f x =0and f y =0andsubtractinggivescos x − cos y =0or cos x =cosy. Thusx = y
or x =2π − y. Ifx = y, f x =0becomes cos x +cos2x =0or 2cos 2 x +cosx − 1=0,aquadraticincos x. Thus
cos x = −1 or 1 and x = π, π ,or 5π , yielding the critical points (π, π), π
, π
2 3 3 3 3 and 5π , 5π
3 3 . Similarly if
x =2π − y, f x =0becomes (cos x)+1=0and the resulting critical point is (π, π). Now
D(x, y) =sinx sin y +sinx sin(x + y)+siny sin(x + y). SoD(π, π) =0and the Second Derivatives Test doesn’t apply.
However, along the line y = x we have f(x, x) =2sinx +sin2x =2sinx +2sinx cos x =2sinx(1 + cos x), and
f(x, x) > 0 for 0