Solução_Calculo_Stewart_6e

F.
TX.10 SECTION 15.7 MAXIMUM AND MINIMUM VALUES ET SECTION 14.7 ¤ 203
29. Since f is a polynomial it is continuous on D,soanabsolutemaximumandminimumexist.Heref x =4, f y = −5 so
there are no critical points inside D. Thus the absolute extrema must both occur on the boundary. Along L 1 : x =0and
f(0,y)=1− 5y for 0 ≤ y ≤ 3, a decreasing function in y, so the maximum value is f (0, 0) = 1 and the minimum value
is f(0, 3) = −14. Along L 2 : y =0and f(x, 0) = 1 + 4x for 0 ≤ x ≤ 2,an
increasing function in x, so the minimum value is f(0, 0) = 1 and the maximum
value is f(2, 0) = 9. Along L 3 : y = − 3 2 x +3and f x, − 3 2 x +3 = 23 2 x − 14
for 0 ≤ x ≤ 2, an increasing function in x, so the minimum value is
f(0, 3) = −14 and the maximum value is f(2, 0) = 9. Thus the absolute
maximum of f on D is f(2, 0) = 9 and the absolute minimum is f(0, 3) = −14.
31. f x (x, y) =2x +2xy, f y (x, y) =2y + x 2 , and setting f x = f y =0
gives (0, 0) as the only critical point in D,withf(0, 0) = 4.
On L 1: y = −1, f(x, −1) = 5,aconstant.
On L 2 : x =1, f(1,y)=y 2 + y +5, a quadratic in y which attains its
maximum at (1, 1), f(1, 1) = 7 and its minimum at
1, − 1 2 , f 1, −
1
2 =
19
. 4
On L 3: f(x, 1) = 2x 2 +5which attains its maximum at (−1, 1) and (1, 1)
with f(±1, 1) = 7 and its minimum at (0, 1), f(0, 1) = 5.
On L 4: f(−1,y)=y 2 + y +5with maximum at (−1, 1), f(−1, 1) = 7 and minimum at −1, − 1 2
, f −1, −
1
2 =
19
. 4
Thus the absolute maximum is attained at both (±1, 1) with f(±1, 1) = 7 and the absolute minimum on D is attained at
(0, 0) with f(0, 0) = 4.
33. f(x, y) =x 4 + y 4 − 4xy +2is a polynomial and hence continuous on D,so
it has an absolute maximum and minimum on D.InExercise7,wefoundthe
critical points of f;only(1, 1) with f(1, 1) = 0 is inside D. OnL 1 : y =0,
f(x, 0) = x 4 +2, 0 ≤ x ≤ 3, a polynomial in x which attains its maximum
at x =3, f(3, 0) = 83, and its minimum at x =0, f(0, 0) = 2.
On L 2 : x =3, f(3,y)=y 4 − 12y +83, 0 ≤ y ≤ 2, a polynomial in y
which attains its minimum at y = 3√ 3, f 3, 3√ 3 =83− 9 3√ 3 ≈ 70.0,anditsmaximumaty =0, f(3, 0) = 83.
On L 3 : y =2, f(x, 2) = x 4 − 8x +18, 0 ≤ x ≤ 3, a polynomial in x which attains its minimum at x = 3√ 2,
f 3 √ 2, 2 =18− 6 3√ 2 ≈ 10.4, and its maximum at x =3,f(3, 2) = 75. OnL 4 : x =0, f(0,y)=y 4 +2, 0 ≤ y ≤ 2,a
polynomial in y which attains its maximum at y =2, f(0, 2) = 18, and its minimum at y =0, f(0, 0) = 2. Thus the absolute
maximum of f on D is f(3, 0) = 83 and the absolute minimum is f(1, 1) = 0.