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

CHAPTER 14 REVIEW D 243
xy 2 z 3 = 2 so x and z must have the same sign, that is, x = ~ z. Thus g(x, y, z) = 2 implies ~z G z 2 )z 3 = 2 or
z = ±3 1 / 4 and the possible points are (±3- 1 / 4 , 3- 1 / 4 y'2, ±3 1 1 4 ), (±3- 1 / 4 , --,-3...:. 1 ; 4 ._/2, ±3 1 1 4 ) . However at each of these
points f takes on the same value, 2 J3. But (2, 1, 1) also satisfies g(x, y , z) = 2 and / (2, 1, 1) = 6 > 2 v'3. Thus. f has an
absolute minimum value of 2 J3 and no absolute maximum subject to the constraint xy 2 z 3 = 2. ·
2 3 2 . I' ., 2 . . . f( ) 2 2 2 .I
xz
xz
Alternate solution: g ( x , y, z ) = xy z = tmp tes y- = -
3 , so mmtmtze x, z = x + - 3 + z . T 1en
2
fx = 2x -
2
3 , f~ = - --;. + 2z, fxx = 2 + 3
4
3 , f ~~ = 24
5 + 2 and fr~ = 2
X Z X Z X Z' X Z X Z
6
4
• Now f x = 0 implies
2x 3 z 3 -
2 = 0 or z = 1/x. Substituting into / 11 = 0 implies -6x 3 + 2x- 1 = 0 or x .= ~·so the two criti 0, so each point
is a minimum. Finally, y 2 = _2_ 3
, so the four points closest to the origin are (± 4~ , .:fl, ± :y'3), (±+ , - ~ , ± V'3).
= · v3" V3 V3
65.
The area of the triangle is ~ ca sinO and the area of the rectangle is be. Thus,
c
b
the area of the whole object is f(a, b, c)= ~ca sinO+ be. The perimeter of
the object is g( a, b, c) = 2a + 2b + e = P. To simplifY sin 0 in terms of a, b,
· and e notice that a 2 sin 2 o + ( ~ e) 2 = a 2 => sino= ..!. v4a 2 - c2.
c
Thus f(a, b, c)= 4 J4a 2 -
- 2a
. '
c 2 +be. (Instead of using 0, we could just have
used the Py1hagorean Theorem.) As a result, by Lagrange's method, we must find a, b, c, and A by solving "il f = A "il g which
gives the following equations: ca(4a 2 - c 2 ) - 1 1 2 = 2A (1), c = 2A (2), *(
4a 2 - c 2 ) 1 1 2 - i c 2 (4a 2 - c 2 ) - 1 1 2 + b = A
(3),and2a+2b +c = P (4). From(2),A= t candso(l)producesca(4a 2 - c 2 ) ....: 1 1 2 =c => (4a 2 - c 2 ) 1 1 2 = a =>
2
4a 2 - e 2 = a 2 => · c = v'3 a (5). Similarly, since {4a 2 - c 2 ) 112 = a and A= tc, (3) gives ~ - !:__ + b = ~ . so from
4 4a 2
(5), ~ - 3 a + b = .;;a => -~ - v;a = -b => b =% (1 + vf3) (6). Substituting (5) and (6) into (4) we get:
4
2a + a(l + v'3) + v'3a=P => 3a + 2v'3a=P => a = p = 2 J3 - 3 P and thus
3+ 2 J3 3
b = ·(2 J3- 3~ (1 + J3) p = 3 - 6 J3 p and c = (2- V3)P.
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