Unformatted text preview: = 0.11,
dw = (2xy − 2y + y 2 )dx + (x2 − 2x + 2yx)dy = −(−0.1) = 0.1
23. dV = 1
1
2
2
xhdx + x2 dh = 2(−0.1) + (0.2) = −0.06667 m3 ; ∆V = −0.07267 m3
3
3
3
3 24. f = (2x + 3y − 6)i + (3x + 6y + 3)j = 0 if 2x + 3y = 6, x + 2y = −1, x = 15, y = −8, D = 3 > 0,
fxx = 2 > 0, so f has a relative minimum at (15, −8). 25. f = (2xy − 6x)i + (x2 − 12y )j = 0 if 2xy − 6x = 0, x2 − 12y = 0; if x = 0 then y = 0, and
if x = 0 then y = 3, x = ±6, thus the gradient vanishes at (0, 0), (−6, 3), (6, 3); fxx = 0 at all
three points, fyy = −12 < 0, D = −4x2 , so (±6, 3) are saddle points, and near the origin we write
f (x, y ) = (y − 3)x2 − 6y 2 ; since y − 3 < 0 when y  < 3, f has a maximum by inspection. 26. f = (3x2 − 3y )i − (3x − y )j = 0 if y = x2 , 3x = y , so x = y = 0 or x = 3, y = 9; at
x = y = 0, D = −9, saddle point; at x = 3, y = 9, D = 9, fxx = 18 > 0, relative minimum 571 Chapter 15 27. (a) (b) 5 K
P=3 5 4 P=2 4
3 3 P=1
2 2 1
L
1 2 3 4 1 5 0
0 1 2 3 4 5 28. (a) ∂P/∂L = cαLα−1 K β , ∂P/∂K = cβLα K β −1
(b) the rates of change of output with respect to labor and capital equipment, respectively
(c) K (∂P/∂K ) + L(∂P/∂L) = cβLα K β + cαLα K β = (α + β )P = P
29. (a) L + K = 200,000, P = 1000L0.6 (200,000 − L)0.4 ,
L6
(200,000 − L)0.4
− 400
= 0 when L = 120,000,
L4
(200,000 − L)0.6
P = 102,033,960.1, which is a maximum because P = 0 at L = 0, 200,000, P > 0 in between,
and dP/dL = 0 has only the one solution. dP/dL = 600 (b) Since L + K = 200,000 and L = 120,000, K = 80,000
√
√
30. (a) y 2 = 8 − 4x2 , ﬁnd extrema of f (x) = x2 (8 − 4x2 ) = −4x4 + 8x2 deﬁned for − 2 ≤ x ≤ 2.
Then f (x) = −16x3 + 16x = 0 when x = 0, ±1, f (x) = −48x2 + 16, so f has a relative
√
maximum at x = ±1, y = ±2 and a relative minimum at x = 0, y = ±2 2. At the endpoints
√
x = ± 2, y = 0 we obtain the minimum f = 0 again.
λy
(b) f (x, y ) = x2 y 2 , g (x, y ) = 4x2 + y 2 − 8 = 0, f = 2xy 2 i + 2x2 y j = λ g = 8λxi + 2√ j, so
√
solve 2xy 2 = λ8x, 2x2 y = λ2y . If x = 0 then y = ±2 2, and if y = 0 then x = ± 2. In
either case f has a relative and absolute minimum. Assume x, y = 0, then y 2 = 4λ, x2 = λ,
use g = 0 to obtain x2 = 1, x = ±1, y = ±2, and f = 4 is a relative and absolute maximum
at (±1, ±2).
31. Let a corner of the box be at (x, y, z ), so that (x/a)2 + (y/b)2 + (z/c)2 = 1. Maximize V = xyz
subject to g (x, y, z ) = (x/a)2 + (y/b)2 + (z/c)2 = 1, solve V = λ g , or
yz i + xz j + xy k = (2λx/a2 )i + (2λy/b2 )j + (2λz/c2 )k, a2 yz = 2λx, b2 xz = 2λy, c2 xy = 2λz . For the
maximum volume, x, y, z = 0; divide the ﬁrst and second equations to obtain a2 y 2 = b2 x2 ; the ﬁrst
√
and third to obtain a2 z 2 = c2 x2 , and ﬁnally b2 z 2 = c2 y 2 . From g = 0 get 3(x/a)2 = 1, x = ±a/ 3,
√
√
2a
2b
2c
and then y = ±b/ 3, z = ±c/ 3. The dimensions of the box are √ × √ × √ , and the
3
3
3
√
maximum volume is 8abc/(3 3).
32. (a) Let f (x, y ) = 3x2 − 5xy + tan xy = 0. Then
dy
6x − 5y + y sec2 xy
dy
df
= 6x − 5y + y sec2 xy + (−5x + x sec2 xy )
= 0, so
=
.
dx
dx
dx
5x − x sec2 xy
(b) Let g (x, y ) = x ln y + sin(x − y ) = π,
ln y + cos(x − y )
dy
=
dx
−x/y + cos(x − y ) dg
= ln y + cos(x − y ) +
dx x
− cos(x − y )
y dy
= 0,
dx Chapter 15 Supplementary Exercises 33. F (x, y ) = 0, Fx + Fy 572 dy
dy
dy
d2 y
= 0, Fxx + Fxy
+ Fyx
+ Fy 2 = 0, thus
dx
dx
dx
dx Fx d2 y
Fxx + 2Fxy (dy/dx)
Fxx Fy − 2Fx Fxy
dy
=− , 2 =−
=−
2
dx
Fy dx
Fy
Fy
34. Denote the currents I1 , I2 , I3 by x, y, z respectively. Then minimize F (x, y, z ) = x2 R1 + y 2 R2 + z 2 R3
subject to g (x, y, z ) = x + y + z − I = 0, so solve F = λ g , 2xR1 i +2yR2 j +2zR3 k = λ(i + j + k),
1
1
1
:
:
.
λ = 2xR1 = 2yR2 = 2zR3 , so the minimum value of F occurs when I1 : I2 : I3 =
R1 R2 R3
√
35. Solve (t − 1)2 /4 + 16e−2t + (2 − t)2 = 1 for t to get t = 1.8332, 2.83984; the particle strikes
the surface at the points P1 (0.8332, 0.63959, 0.64603), P2 (1.83984, 0.23374, 0.31482). The velocity
√
dy
dz
dx
i + j + k = i − 4e−t j − 1/(2 t)k, and a normal to the surface is
vectors are given by v =
dt
dt
dt
n = (x2 /4 + y 2 + z 2 ) = x/2i + 2y j + 2z k. At the points Pi these are
v1 = i − 0.639589j − 0.369286k, v2 = i − 0.23374j + 0.29670k;
n1 = 0.41661i + 1.27918j + 1.29207k and n2 = 0.91992i + 0.46748j + 0.62963k so
cos−1 [(vi · ni )/( vi ni )] = 112.3◦ , 61.1◦ ; the acute angles are 67.7◦ , 61.1◦ .
1 ey cos(xey )dy = 36. (a) F (x) =
0 (b) Use a CAS to get x =
F π
e+1 1 = sin
0 sin(ex) − sin x
x π
so the maximum value of F (x) is
e+1
πy
e
e+1 dy ≈ 0.909026. 37. Let x, y, z be the lengths of the sides opposite angles α, β, γ , located at A,√ C respectively. Then
B,
x2 = y 2 + z 2 − 2yz cos α and x2 = 100 + 400 − 2(10)(20)/2 = 300, x = 10 3 and
2x so dy
dz
dz
dy
dα
dx
= 2y
+ 2z
− 2 y cos α + z cos α − yz (sin α)
dt
dt
dt
dt
dt
dt
√...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.
 Spring '14
 The Land

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