15 CHAPTER SECTION 15.1
1. 5. 6. 9. 11. 12.
f = (6x - y) i + (1 - x) j
3.
f = exy [ (xy + 1) i + x2 j]
f = 2y 2 sin(x2 + 1) + 4x2 y 2 cos(x2 + 1) i + 4xy sin(x2 + 1) j f= 2y 2x i+ 2 j 2 +y x + y2
x2
f = (z 2 + 2xy) i + (x2 + 2yz) j + (y 2 + 2zx) k f = e-z (2xy i + x2 j - xy 2 k) f= xyz xyz + yz ln(x + y + z) i + + xz ln(x + y + z) j x+y+z x+y+z + xyz + xy ln(x + y + z) k x+y+z
15. 17. 19. 20. 23.
f = 2y cos(2xy) +
1 2 i + 2x cos(2xy) j + k x z at (2, 3), f = -i + 18j f= x2 + y2 2 4 i+ j 5 5 j, f (1, 1) = 1 - 4 2 i+ 1 j 2
f = (4x - 3y) i + (8y - 3x) j; f= f= 2x 2y i+ 2 j; 2 +y x + y2 x2 xy + y2
x2
at (2, 1), i+
tan-1 (y/x) -
x2
f = -e-x sin (z + 2y) i + 2e-x cos (z + 2y) j + e-x cos (z + 2y) k; at (0, /4, /4), 1 f = - 2 2 (i + 2j + k)
2 2 2
26. 33.
f = - sin(xyz )(yz i + xz j + 2xyz k), f = F(x, y) = 2xy i + 1 + x2 j Now, f = x2 + g (y) = 1 + x2 y
f
2 1 , , -1 = - 4 2
1 i+j- k 4 2
f = 2xy x g (y) = 1
f (x, y) = x2 y + g(y) for some function g. g(y) = y + C, C a constant.
Thus, f (x, y) = x2 y + y (take C = 0) is a function whose gradient is F. 34. f = (2xy + x)i + (x2 + y)j = fx = 2xy + x, Take 39. (a) f (x, y) = x2 y + f = 2x i + 2y j = 0 y2 x2 + 2 2 = x = y = 0; 1 f = 0 at (0, 0). fy = x2 + y
f has an absolute minimum at (0, 0) 40. (a) (c) f= -1 4 - x2 - y 2 (x i + y j) = 0 at (0, 0)
f has a maximum at (0, 0)
SECTION 15.2 1 (3i + 4j), 5
3.
f (ey - yex ) i + (xey - ex ) j, fu (1, 0) = f (1, 0) u =
f (1, 0) = i + (1 - e)j,
u=
1 (7 - 4e) 5 u= 1 (i - 3j), 2
4.
1 (-2yi + 2xj), f (1, 0) = 2j, (x - y)2 fu (1, 0) = f (1, 0) u = - 3 f= 2x 2y i+ 2 j, x2 + y 2 x + y2 2 fu (0, 1) = f u = 65 f=
7.
f (0, 1) = 2 j,
1 u = (8 i + j), 65
9.
f = (y + z)i + (x + z)j + (y + x)k, fu (1, -1, 1) = f (1, -1, 1) u = 2 6 3
f (1, -1, 1) = 2j,
u=
1 6
6 (i + 2j + k),
10.
f = (z 2 + 2xy)i + (x2 + 2yz)j + (y 2 + 2zx)k, 10 fu (1, 0, 1) = f (1, 0, 1) u = 10 f= x2 x y i+ 2 j, 2 +y x + y2 x y k, u= 1 x2 + y2
f (1, 0, 1) = i + j + 2k,
1 u = (3i - k) 10
15.
(-xi - yj) ,
fu (x, y) =
f u=-
1 x2 + y2
18.
z z i - j + ln x y 1 u = (i + j - k); 3 f= f = ey
2
f (1, 1, 2) = 2i - 2j f (1, 1, 2) u = 0 r (t) = i - 2 sin (t - 1) j - 2et-1 k, f (1, 2, -2) u = - 7 5 5
fu (1, 1, 2) =
19.
-z 2
(i + 2xyj - 2xzk),
f (1, 2, -2) = i + 4j + 4k, u=
1 5
5 (i - 2k), fu (1, 2, -2) =
at (1, 2, -2) t = 1,
r (1) = i - 2k,
20.
f = 2xi + zj + yk,
f (1, -3, 2) = 2i + 2j - 3k 2
Direction: r (-1) = -2i+3j-3k,
1 u = (-2i+3j-3k), 22
fu (1, -3, 2) =
f (1, -3, 2) u =
1 22 2
21.
f = (2x + 2yz) i + 2xz - z 2 j + (2xy - 2yz) k,
f (1, 1, 2) = 6 i - 2 k
23.
1 The vector v = 2 i + j - 3 k is a direction vector for the given line; u = (2 i + j - 3 k) 14 18 is a corresponding unit vector; fu (1, 1, 2) = f (1, 1, 2) u = 14 1 f = (i + j) f (0, 1) = 2 i + 2 j, f = 2 2, f = 2y 2 e2x i + 2ye2x j, f 2 1 f increases most rapidly in the direction u = (i + j); the rate of change is 2 2. 2 1 f decreases most rapidly in the direction v = - (i + j); the rate of change is -2 2. 2 f = (2xzey + z 2 )i + x2 zey j + (x2 ey + 2xz)k Fastest increase in direction u = f (1, ln 2, 2) = 12i + 4j + 6k
26.
1 (6i + 2j + 3k), rate of change f (1, ln 2, 2) = 14 7 1 Fastest decrease in direction v = - (6i + 2j + 3k), rate of change -14 7 (a) The projection of the path onto the xy-plane is the curve C : r(t) = x(t)i + y(t)j which begins at (1, 1) and at each point has its tangent vector in the direction of - f. Since f = 2xi + 6yj, we have the initial-value problems x (t) = -2x(t), x(0) = 1 and y (t) = -6y(t), y(0) = 1.
33.
From Theorem 7.6.1 we find that x(t) = e-2t and y(t) = e-6t .
Eliminating the parameter t, we find that C is the curve y = x3 from (1, 1) to (0, 0). (b) Here x (t) = -2x(t), so that x(t) = e-2t and y(t) = -2e-6t . x(0) = 1 and y (t) = -6y(t), y(0) = -2
Eliminating the parameter t, we find that the projection of the path onto the xy-plane is the curve y = -2x3 from (1, -2) to (0, 0). 3
34.
1 2 x - y2 ; f = xi - 2yj, so the projection r(t) = x(t)i + y(t)j of the path onto the 2 xy-plane must satisfy x (t) = x(t), y(t) = -2y(t) 1 1 (a) With initial point (-1, 1, - 2 ),, we get x(t) = -et , y(t) = e-2t , or y = 2 x from (-1, 1), in the direction of decreasing x. z : f (x, y) = (b) With initial point (1, 0, 1 ), we get x(t) = et , y(t) = 0, or the x-axis 2 from (1, 0), in the direction of increasing x. f 2 + h, (2 + h)2 - f (2, 4) 3(2 + h)2 + (2 + h)2 - 16 = lim h0 h0 h h lim = lim 4
h0
39. (a)
4h + h2 = lim 4(4 + h) = 16 h0 h h+8 4
2
f (b)
h0
lim
h+8 3 , 4 + h - f (2, 4) 4 = lim h0 h = lim
+ (4 + h) - 16 h
3 2 16 h
h0
+ 3h + 12 + 4 + h - 16 h +4 =4 17
= lim 17 (i + 4j), f (2, 4) = 12i + j;
h0
3 16 h
(c) (d)
u=
1 17
fu (2, 4) =
f (2, 4) u =
16 17
The limits computed in (a) and (b) are not directional derivatives. In (a) and (b) we have, in essence, computed is r0 a unit vector. -GM m -GM m (xi + yj + zk) = r 2 + z 2 )3/2 r 3 +y f (2, 4) r0 taking r0 = i + 4j in (a) and r0 = 1 i + j in (b). In neither case 4
40.
f=
(x2
SECTION 15.3 1. f (b) = f (1, 3) = -2; f (a) = f (0, 1) = 0; f = 3x2 - y i - x j; b -a = i+2j f (b) - f (a) = -2 and f (b - a) = 3x2 - y - 2x
The line segment joining a and b is parametrized by x = t, Thus, we need to solve the equation 3t2 - (1 + 2t) - 2t = -2, which is the same as 3t2 - 4t + 1 = 0, 0 t 1 satisfies the equation. y = 1 + 2t, 0t1
The solutions are: t = 1 , t = 1. Thus, c = ( 1 , 5 ) 3 3 3
Note that the endpoint b also satisfies the equation. 4
2.
f = 4zi - 2yj + (4x + 2z)k, b - a = i + 2j + k,
f (a) = f (0, 1, 1) = 0,
f (b) = f (1, 3, 2) = 3
so we want (x, y, z) such that
f (b - a) = 4z - 4y + 4x + 2z = 6z - 4y + 4x = f (b) - f (a) = 3 Parameterizing the line segment from a to b by x(t) = t, y(t) = 1 + 2t, z(t) = 1 + t, we get t = 1 , or c = ( 1 , 2, 3 ) 2 2 2 7. f = 2xyi + x2 j; f (r(t)) r (t) = 2i + e2t j (et i - e-t j) = et 10. f= 1 (4xi + 3y 2 j) + y3 1 1 8e4t + 1 f (r(t)) r (t) = 4t (4e2t i + 3t2/3 j) (2e2t i + t-2/3 j) = 4t 2e + t 3 2e + 1 2x2 f (r(t)) = (tt - ln t) i + tt ln t + 1 t j 1 i + [1 + ln t] j t 1 t j 1 + ln t + [ln t]2 t + 1 t
11.
f = (ey - ye-x ) i + (xey + e-x ) j;
f (r(t)) r (t) = (tt - ln t) i + tt ln t + 13. f = yi + (x - z)j - yk;
= tt
f (r(t)) r (t) = t2 i + t - t3 j - t2 k i + 2tj + 3t2 k = 3t2 - 5t4 15. f = 2xi + 2yj + k; f (r(t)) r (t) = (2a cos t i + 2b sin t j + k) (-a sin t i + b cos t j + bk) = 2 b2 - a2 sin t cos t + b
17.
u dx u dy du = + = (2x - 3y)(- sin t) + (4y - 3x)(cos t) dt x dt y dt = 2 cos t sin t + 3 sin2 t - 3 cos2 t = sin 2t - 3 cos 2t u dx u dy du = + dt x dt y dt = (ex sin y + ey cos x) = et/2
1 2 1 2
19.
+ (ex cos y + ey sin x) (2)
1 2
sin 2t + 2 cos 2t + e2t
cos 1 t + 2 sin 1 t 2 2
20.
u dx u dy du = + = (4x - y)(-2 sin 2t) + (2y - x) cos t dt x dt y dt = 2 sin 2t(sin t - 4 cos 2t) + cos t(2 sin t - cos 2t) du u dx u dy u dz = + + dt x dt y dt z dt = (y + z)(2t) + (x + z)(1 - 2t) + (y + x)(2t - 2) 5
23.
= (1 - t)(2t) + (2t2 - 2t + 1)(1 - 2t) + t(2t - 2) = 1 - 4t + 6t2 - 4t3 u x u y u = + = (2x - y)(cos t) + (-x)(t cos s) s x s y s = 2s cos2 t - t sin s cos t - st cos s cos t u x u y u = + = (2x - y)(-s sin t) + (-x)(sin s) t x t y t = -2s2 cos t sin t + st sin s sin t - s cos t sin s 32. u x u y u z u = + + s x s y s z s = z 2 y sec xy tan xy(2t) + z 2 x sec xy tan xy + 2z sec xy(2st) = sec[2st(s - t2 )] 2s4 t3 (s - t2 ) tan[2st(s - t2 )] + 2s3 t2 tan[2st(s - t2 )] + 4s3 t2 u = z 2 y sec xy tan xy(2s) + z 2 x sec xy tan xy(-2t) + 2z sec xy(s2 ) t = sec[2st(s - t2 )] 2s5 t2 (s - t2 ) tan[2st(s - t2 )] - 4s5 t4 tan[2st(s - t2 )] + 2s4 t 33. u u x u y u z = + + s x s y s z s = (2x - y)(cos t) + (-x)(- cos (t - s)) + 2z(t cos s) = 2s cos2 t - sin (t - s) cos t + s cos t cos (t - s) + 2t2 sin s cos s u u x u y u z = + + t x t y t z t = (2x - y)(-s sin t) + (-x)(cos (t - s)) + 2z(sin s) = -2s2 cos t sin t + s sin (t - s) sin t - s cos t cos (t - s) + 2t sin2 s d [f (r(t) ) ] = dt r (t) r (t)
29.
35.
f (r(t) )
r (t) where u(t) = r (t) r (t)
= fu(t) (r(t)) r (t)
SECTION 15.4 1. Set f (x, y) = x2 + xy + y 2 . Then, f (-1, -1) = -3i - 3j.
f = (2x + y)i + (x + 2y)j, normal vector i + j; tangent vector i - j tangent line x + y + 2 = 0; normal line x - y = 0 6
2.
Set f (x, y) = (y - x)2 - 2x,
f = -2(y - x + 1)i + 2(y - x)j,
f (2, 4) = -6i + 4j
normal vector -3i + 2j; tangent vector 2i + 3j tangent line 3x - 2y + 2 = 0; normal line 2x + 3y - 16 = 0 5. Set f (x, y) = xy 2 - 2x2 + y + 5x. Then, f (4, 2) = -7i + 17j.
f = (y 2 - 4x + 5) i + (2xy + 1) j, normal vector 7i - 17j; tangent vector 17i + 7j
tangent line 7x - 17y + 6 = 0; normal line 17x + 7y - 82 = 0 7. Set f (x, y) = 2x3 - x2 y 2 - 3x + y.
2 2
Then, f (1, -2) = -5i + 5j.
f = (6x - 2xy - 3) i + (-2x2 y + 1) j, normal vector i - j; tangent vector i + j tangent line x - y - 3 = 0; normal line x + y + 1 = 0 11. Set f (x, y, z) = x3 + y 3 - 3xyz. Then,
f = (3x2 - 3yz) i + (3y 2 - 3xz) j - 3xyk, tangent plane at 1, 2, Normal: 12. x = 1 + 4t,
3 2
f 1, 2, 3 = -6i + 2
3 2
15 2 j
- 6k;
: -6(x - 1) + y = 2 - 5t,
15 2 (y 3 2
- 2) - 6 z - + 4t
= 0, which reduces to 4x - 5y + 4z = 0.
z=
Set f (x, y, z) = xy 2 + 2z 2 . Tangent plane: Normal:
f = y 2 i + 2xyj + 4zk,
f (1, 2, 2) = 4i + 4j + 8k
x + y + 2z - 7 = 0 1 1 1 i+ j+ k 2 4 2
14.
x = 1 + t, y = 2 + t, z = 2 + 2t 1 1 1 Set f (x, y, z) = x + y + z. f = i + j + k, 2 x 2 y 2 z Tangent plane: 2x + y + 2z - 8 = 0 x = 1 + 2t, y = 4 + t, z = 1 + 2t
f (1, 4, 1) =
Normal: 16.
Set f (x, y, z) = x2 + xy + y 2 - 6x + 2 - z. Tangent plane: Normal: x = 4, z = -10 y = -2, z=t
f = (2x + y - 6)i + (x + 2y)j - k,
f (4, -2, -10) = -k
20.
g = (4 - 2x + y)i + (2 + x - 2y)j 8 10 , y= g = 0 = 4 - 2x + y = 0, 2 + x - 2y = 0 = x = 3 3 The tangent plane is horizontal at ( 10 , 8 , 28 ) 3 3 3 Set z = g(x, y) = 2x2 + 2xy - y 2 - 5x + 3y - 2. Then,
z = g(x, y) = 4x + 2y - x2 + xy - y 2 .
23.
g = (4x + 2y - 5) i + (2x - 2y + 3) j. g =0 = 4x + 2y - 5 = 0 = 2x - 2y + 3
1 11 3, 6 , 1 - 12 .
=
x = 1, 3
y=
11 6 .
The tangent plane is horizontal at
7
25. 33.
y - y0 z - z0 x - x0 = = (f /x)(x0 , y0 , z0 ) (f /y)(x0 , y0 , z0 ) (f /z)(x0 , y0 , z0 ) Set f (x, y, z) = x2 y 2 + 2x + z 3 . Then, f (2, 1, 2) = 6i + 8j + 12k.
f = (2xy 2 + 2) i + 2x2 yj + 3z 2 k,
The plane tangent to f (x, y, z) = 16 at (2, 1, 2) has equation 6(x - 2) + 8(y - 1) + 12(z - 2) = 0, or 3x + 4y + 6z = 22. Next, set g(x, y, z) = 3x2 + y 2 - 2z. Then, g(2, 1, 2) = 12i + 2j - 2k.
g = 6xi + 2yj - 2k,
The plane tangent to g(x, y, z) = 9 at (2, 1, 2) is 12(x - 2) + 2(y - 1) - 2(z - 2) = 0, or 6x + y - z = 11. 34. Sphere: f (x, y, z) = z 2 + y 2 + z 2 - 8x - 8y - 6z + 24, f (2, 1, 1) = -4i - 6j - 4k Ellipsoid: g(x, y, z) = x2 + 3y 2 + 2z 2 , g = 2xi + 6yj + 4zk f = (2x - 8)i + (2y - 8)j + (2z - 6)k
g(2, 1, 1) = 4i + 6j + 4k Since their normal vectors are parallel, the surfaces are tangent. 35. The gradient to the sphere at (1, 1, 2) is 2xi + (2y - 4) j + (2z - 2)k = 2i - 2j + 2k. The gradient to the paraboloid at (1, 1, 2) is 6xi + 4yj - 2k = 6i + 4j - 2k. Since (2i - 2j + 2k) (6i + 4j - 2k) = 0, the surfaces intersect at right angles.
SECTION 15.5 2. f = (2 - 2x) i + (2 + 2y) j = 0 only at (1, -1). The difference f (1 + h, -1 + k) - f (1, -1) = [2(1 + h) + 2(-1 + k) - (1 + h)2 + (-1 + k)2 + 5] - 5 = -h2 + k 2 does not keep a constant sign for small h and k; (1, -1) is a saddle point. 3. f = (2x + y + 3) i + (x + 2y) j = 0 only at (-2, 1). The difference f (-2 + h, 1 + k) - f (-2, 1) = [(-2 + h)2 + (-2 + h)(1 + k) + (1 + k)2 + 3(-2 + h) + 1] - (-2) = h2 + hk + k 2 is positive for all small h and k. To see this, note that 8
h2 + hk + k 2 h2 + k 2 - |h| |k| > 0; there is a local minimum of -2 at (-2, 1). 5. f = (2x + y - 6) i + (x + 2y) j = 0 only at (4, -2). fxx = 2, fxy = 1, fyy = 2.
At (4, -2), D = 3 > 0 and A = 2 > 0 so we have a local min; the value is -10. 7. f = (3x2 - 6y)i + 3y 2 - 6x j = 0 at (2, 2) and (0, 0). fxx = 6x, fxy = -6, fyy = 6y, D = 36xy - 36.
At (2, 2), D = 108 > 0 and A = 12 > 0 so we have a local min; the value is -8. At (0, 0), D = -36 < 0 so we have a saddle point. 9. f = (3x2 - 6y + 6)i + (2y - 6x + 3) j = 0 at 5, fxx = 6x, fxy = -6, fyy = 2, D = 12x - 36.
27 2
and 1, 3 . 2
At 5, 27 , D = 24 > 0 and A = 30 > 0 so we have a local min; the value is - 117 . 2 4 At 1, 3 , D = -24 < 0 so we have a saddle point. 2 10. f = (2x - 2y - 3)i + (-2x + 4y + 5)j = 0 at ( 1 , -1) 2 2f 2f 2f = -2, = 2, = 4; D = 2 4 - (-2)2 > 0, A = 2 = local minimum; x2 yx y 2 the value is - 13 . 4 f = (2xy + 1 + y 2 )i + x2 + 2xy + 1 j = 0 at fxx = 2y, fxy = 2x + 2y, fyy = 2x,
13.
(1, -1) and (-1, 1).
D = 4xy - 4(x + y)2 .
At both (1, -1) and (-1, 1) we have saddle points since D = -4 < 0. 14. x y x2 + y 2 x2 + y 2 1 1 + 2 i+ - 2 - j= i- j 2y y x y x x xy 2 no stationary points, no local extreme values. f= f= ln xy + 1 - 3 x i+ x-3 j = 0 at (3, 1/3) y 2f 3-x = 2 y y2 2f = 0 and D = -9 < 0 = saddle point. y 2 is never 0 ;
20.
2f 3 1 = + 2, 2 x x x At (3, 1/3), 21.
1 2f = , yx y
2f 2 = , x2 3
2f = 3, yx
f = 4x3 - 4x i + 2y j = 0 at fxx = 12x2 - 4, fxy = 0,
(0, 0), (1, 0), and (-1, 0). D = 8 - 24x2 . 9
fyy = 2,
point (0, 0) (1, 0) (-1, 0) f (1, 0) = -3. 25. (a)
A -4 8 8
B 0 0 0
C 2 2 2
D -8 16 16
result saddle loc. min. loc. min.
f = (2x + ky) i + (2y + kx) j and fxy = k, fyy = 2,
f (0, 0) = 0 independent of the value of k.
(b) fxx = 2,
D = 4 - k 2 . Thus, D < 0 for |k| > 2 and (0, 0) is a saddle point
(c) D = 4 - k 2 > 0 for |k| < 2. Since A = fxx = 2 > 0, (0, 0) is a local minimum. (d) The test is inconclusive when D = 4 - k 2 = 0 i.e., for k = 2. 26. (a) (b) (c) (d) 27. f = (2x + ky)i + (kx + 8y)j = 0 at (0, 0). 2f 2f f = k, = 2, = 8; we want 16 - k 2 < 0, or |k| > 4 2 x yx y 2 We want 16 - k 2 > 0, or |k| < 4
2
k = 4 x2 + y 2 + z 2 .
Let P (x, y, z) be a point in the plane. We want to find the minimum of f (x, y, z) =
However, it is sufficient to minimize the square of the distance: F (x, y, z) = x2 + y 2 + z 2 . It is clear that F has a minimum value, but no maximum value. Since P lies in the plane, 2x - y + 2z = 16 which implies y = 2x + 2z - 16 = 2(x + z - 8). Thus, we want to find the minimum value of F (x, z) = x2 + 4(x + z - 8)2 + z 2 Now, F = [2x + 8(x + z - 8)] wi + [8(x + z - 8) k The gradient is 0 when 8(x + z - 8) + 2z = 0 16 32 , from which it follows that y = - . The only solution to this pair of equations is: x = z = 9 9 The point in the plane that is closest to the origin is P 32 , - 16 , 32 . 9 9 9 The distance from the origin to the plane is: F (P ) = Check using (12.6.7): d(P, 0) = |2 0 - 0 + 2 0 - 16| 22 + (-1)2 + 22
16 3 .
2x + 8(x + z - 8) = 0 and
=
16 . 3
SECTION 15.6 1. f = (4x - 4) + i (2y - 2) j = 0 at (1, 1) in D; 10
f (1, 1) = -1 Next we consider the boundary of D. We parametrize each side of the triangle: C1 : r1 (t) = t i, t [ 0, 2 ], t [ 0, 4 ], t [ 0, 2 ],
y 4
C2 : r2 (t) = 2 i + t j, C3 : r3 (t) = t i + 2t j,
1
Now, f1 (t) = f (r1 (t)) = 2(t - 1)2 , t [ 0, 2 ]; critical number: t = 1, t [ 0, 4 ]; t [ 0, 2 ]; critical number: t = 1, critical number: t = 2 . 3
2
x
f2 (t) = f (r2 (t)) = (t - 1)2 + 1, f3 (t) = f (r3 (t)) = 6t2 - 8t + 2,
Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: f1 (0) = f3 (0) = f (0, 0) = 2; f2 (1) = f (2, 1) = 2; f1 (1) = f (1, 0) = 0; f1 (2) = f2 (0) = f (2, 0) = 2;
2 f3 (2/3) = f (2/3, 4/3) = - 3 .
f2 (4) = f3 (2) = f (2, 4) = 10;
f takes on its absolute maximum of 10 at (2, 4) and its absolute minimum of -1 at (1, 1). 5. f = (2x + 3y) i + (2y + 3x) j = 0 at (0, 0) in D; Next we consider the boundary of D. C : r(t) = 2 cos t i + 2 sin t j, f (0, 0) = 2
We parametrize the circle by:
t [ 0, 2 ]
The values of f on the boundary are given by the function F (t) = f (r(t)) = 6 + 12 sin t cos t, F (t) = 12 cos2 t - 12 sin2 t : F (t) = 0 = t [ 0, 2 ] = t = 1 , 4
3 5 7 4 , 4 , 4
cos t = sin t
Evaluating F at the endpoints and critical numbers, we have: F (0) = F (2) = f (2, 0) = 6; F 1 = F 5 = f 2, 2 = f ( - 2, - 2 = 12; 4 4 F 3 = f - 2, 2 = F 7 = f ( 2, - 2 = 0 4 4 f takes on its absolute maximum of 12 at 2, 2 and at - 2, - 2 ; f takes on its absolute 2, - 2 . minimum of 0 at - 2, 2 and at
8.
f = (y + 1) i + (x - 1) j = 0 at (1, -1) which is not in the interior of D. Next we consider the boundary of D. We parametrize the boundary by: 11
C1 : r1 (t) = t j + t2 j, C2 : r2 (t) = t i + 4 j, and evaluate f :
t [ -2, 2 ], t [ -2, 2 ],
f1 (t) = f (r1 (t)) = t3 - t2 + t + 3, f2 (t) = f (r2 (t)) = 5t - 1,
t [ -2, 2 ]; no critical numbers, no critical numbers.
t [ -2, 2 ];
Evaluating these functions at the endpoints of their domains, we find that: f1 (-2) = f2 (-2) = f (-2, 4) = -11; f1 (2) = f2 (2) = f (2, 4) = 9.
f takes on its absolute maximum of 9 at (2, 4) and its absolute minimum of -11 at (-2, 4). 11. f = (4 - 4x) cos y i + (4x - 2x2 ) sin y j = 0 at (1, 0) in D: f (1, 0) = 2 Next we consider the boundary of D. We parametrize each side of the rectangle: C1 : r1 (t) = t j, t - 1 , 1 4 4 t [ 0, 2 ]
1 t - 4 , 1 4
C2 : r2 (t) = t i - 1 j, 4 C3 : r3 (t) = 2 i + t j, C4 : r4 (t) = t i + 1 j, 4 Now, f1 (t) = f (r1 (t)) = 0;
t [ 0, 2 ]
2 (4t - 2t2 ), 2 f3 (t) = f (r3 (t)) = 0; 2 (4t - 2t2 ), f4 (t) = f (r4 (t)) = 2 f2 (t) = f (r2 (t)) =
t [ 0, 2 ];
critical number: t = 1;
t [ 0, 2 ];
critical number: t = 1;
1 f2 (1) = f4 (1) = f 1, - 4 = f 1, 1 = 4
2.
f at the vertices of the rectangle has the value 0;
f takes on its absolute maximum of 2 at (1, 0) and its absolute minimum of 0 along the lines x = 0 and x = 2.
16.
f = (2x + 1) i + (8y - 2) j = 0 at - 1 , 1 in D; 2 4 Next we consider the boundary of D. C : r(t) = 2 cos t i + sin t j,
1 f -1, 1 = -2 2 4
We parametrize the ellise by:
t [ 0, 2 ]
The values of f on the boundary are given by the function F (t) = f (r(t)) = 4 cos2 t + 4 sin2 t + 2 cos t - 2 sin t = 4 + 2 cos t - 2 sin t, F (t) = -2 sin t - 2 cos t : F (t) = 0 = 12 cos t = - sin t = t [ 0, 2 ]
t = 3 , or 7 4 4
Evaluating F at the endpoints and critical numbers, we have: F (0) = F (2) = f (2, 0) = 6; 2, - 2/ = 4 + 2 2 F 3 = f - 2, 2/2 = 4 - 2 2; F 7 = f 4 4 2, - 2/2 ; f takes on its absolute minimum of f takes on its absolute maximum of 4 + 2 2 at - 1 at - 1 , 1 . 2 2 4 19. Using the hint, we want to find the maximum value of f (x, y) = 18xy - x2 y - xy 2 . The gradient of f is: D = 18y - 2xy - y 2 i + 18x - x2 - 2xy j The gradient is 0 when 18y - 2xy - y 2 = 0 and 18x - x2 - 2xy = 0 The solution set of this pair of equations is: (0, 0), (18, 0), (0, 18), (6, 6). It is easy to verify that f is a maximum when x = y = 6. The three numbers that satisfy x+y+z = 18 and maximize the product xyz are: x = 6, y = 6, z = 6. 21. f (x, y) = xy(1 - x - y), 0 x 1, 0 y 1 - x.
[ dom (f ) is the triangle with vertices (0, 0), (1, 0), (0, 1).] f = (y - 2xy - y 2 )i + x - 2xy - x2 j = 0 = x = y = 1. 3
(Note that [ 0, 0 ] is not an interior point of the domain of f .) fxx = -2x, At
1 1 3, 3
fxy = 1 - 2x - 2y,
1 3
fyy = -2x,
D = (1 - 2x - 2y)2 - 4xy.
, D=
> 0 and A < 0 so we have a local max; the value is 1/27.
Since f (x, y) = 0 at each point on the boundary of the domain, the local max of 1/27 is also the absolute max. 24. C = 4xy + 3(2xz + 2yz) = 4xy + 6z(x + y). Since xyz = 12, we need to minimize C(x, y) = 4xy + C = (4y - dimensions 3 18 3 18 12 . 182/3 72 72 )i + (4x - 2 )j = 0 x2 y 72 (x + y), x > 0, y > 0. xy
at (181/3 , 181/3 )
33.
96 = xyz, C = 30xy + 10(2xz + 2yz) = 30xy + 20(x + y) 13 96 . xy
C(x, y) = 30 xy +
64 64 + , x y = x = y = 4.
C = 30(y - 64x-2 )i + 30(x - 64y -2 ) j = 0 Cxx = 128x-3 ,
Cxy = 1,
Cyy = 128y -3.
When x = y = 4, we have D = -3 < 0 and A = 2 > 0 so the cost is minimized by making the dimensions of the crate 4 4 6 meters.
35.
Let x, y and z be the length, width and height of the box. The surface area is given by S = 2xy + 2xz + 2yz, so z = S - 2xy , 2(x + y) where S is a constant, and x, y, z > 0.
Now, the volume V = xyz is given by: V (x, y) = xy and V =y 2(x + y)(-2y) - (S - 2xy)(2) S - 2xy + xy 2(x + y) 4(x + y)2 + x Setting i j S - 2xy 2(x + y)
2(x + y)(-2x) - (S - 2xy)(2) S - 2xy + xy 2(x + y) 4(x + y)2
V V = = 0 and simplifying, we get the pair of equations x y 2S - 4x2 - 8xy = 0 2S - 4y 2 - 8xy = 0
from which it follows that x = y = maximum value at ( S/6,
S/6. From practical considerations, we conclude that V has a S/6
S/6). Substituting these values into the equation for z, we get z =
and so the box of maximum volume is a cube. (S - xy) , x > 0, y > 0, xy < S. 2(x + y)
36.
V = xyz, V =
S = xy + 2xz + 2yz
=
V (x, y) = xy
y 2 (S - x2 - 2xy) x2 (S - y 2 - 2xy) i+ j 2(x + y)2 2(x + y)2 = x= s , y= 3 s ; 3 dimensions for maximum volume: s 3 s 1 3 2 s 3
V =0 39. (a)
Let x and y be the cross-sectional measurements of the box, and let l be its length. Then V = xyl, where 2x + 2y + l 108, x, y > 0
To maximize V we will obviously take 2x+2y+l = 108. Therefore, V (x, y) = xy(108-2x-2y) and V = [y(108 - 2x - 2y) - 2xy] i + [x(108 - 2x - 2y) - 2xy] j 14
Setting
V V = = 0, we get the pair of equations x y V = 108y - 4xy - 2y 2 = 0 x V = 108x - 4xy - 2x2 = 0 y = l = 36.
from which it follows that x = y = 18 Now, at (18, 18), we have A = Vxx = -4y = -72 < 0, C = Vyy = -4x = -72,
B = Vxy = 108 - 4x - 4y = -36, and D = (36)2 - (72)2 < 0.
Thus, V is a maximum when x = y = 18 and l = 36. (b) Let r be the radius of the tube and let l be its length. Then V = r2 l, where 2 r + l 108, r>0 To maximize V we take 2 r + l = 108. Then V (r) = r2 (108 - 2 r) = 108 r2 - 2 2 r3 . Now dV = 216 r - 6 2 r2 dr dV = 0, we get Setting dr 36 = l = 36 216 r - 6 2 r2 = 0 = r = Now, at r = 36/, we have d2 V 36 = - 216 < 0 = 216 - 12 2 dr2 Thus, V is a maximum when r = 36/ and l = 36.
SECTION 15.7 3. f (x, y) = xy, f = yi + xj, f = g = g(x, y) = b2 x2 + a2 y 2 - a2 b2 g = 2b2 xi + 2a2 yj. y = 2b2 x and x = 2a2 y.
Multiplying the first equation by a2 y and the second equation by b2 x, we get a2 y 2 = 2a2 b2 xy = b2 x2 . 1 Thus, ay = bx. From g(x, y) = 0 we conclude that x = 1 a 2 and y = 2 b 2. 2 Since f is continuous and the ellipse is closed and bounded, the minimum exists. It occurs at 1 1 1 1 1 2 a 2, - 2 b 2 and - 2 a 2, 2 b 2 ; the minimum is - 2 ab. pb 8 f (x, y, z) = xyz, g(x, y, z) = x2 + y 2 + z 2 - 1 15
f = yzi + xzj + xyk, f = g = yz = 2x,
g = 2xi + 2yj + 2zk xz = 2y, xy = 2z = x2 = y 2 = z 2 .
1 1 1 From g(x, y, z) = 0 we get 3x2 = 1 = x = 3 , y = 3 , z = 3 1 Minimum of xyz is: - 3 9
11.
Since the sphere is closed and bounded and 2x + 3y + 5z is continuous, the maximum exists. f (x, y, z) = 2x + 3y + 5z, f = 2i + 3j + 5k, f = g = g(x, y, z) = x2 + y 2 + z 2 - 19 g = 2xi + 2yj + 2zk. 2 = 2x, 3 = 2y, 5 = 2z.
Since = 0 here, we solve the equations for x, y and z: x= 1 , y= 3 , 2 z= 5 , 2
and substitute these results in g(x, y, z) = 0 to obtain 1 9 25 + 2 + 2 - 19 = 0, 2 4 4 38 - 19 = 0, 42 = 1 2. 2
The positive value of will produce positive values for x, y, z and thus the maximum for f. We get x = 2, y = 3 2, z = 5 2, and 2x + 3y + 5z = 19 2. 2 2 14. Maximize area A = xy given that the perimeter P = 2x + 2y f (x, y) = xy, f = yi + xj, g(x, y) = 2x + 2y - P g = 2i + 2j; f = g = y = 2, x = 2 = x = y.
The rectangle of maximum area is a square. 17. It suffices to maximize and minimize the square of the distance from (2, 1, 2) to the sphere. Clearly, these extreme values exist. f (x, y, z) = (x - 2)2 + (y - 1)2 +...