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Unformatted text preview: +∞ ln x ≤ x − 1 on (0, +∞). 41. Let m = slope at x, then m = f (x) = 3x2 − 6x + 5, dm/dx = 6x − 6; critical point for m is x = 1,
minimum value of m is f (1) = 2 42. (a) −64 cos3 x + 27 sin3 x
64 cos x 27 sin x
+
, f (x) = 0 when
=
2
2x
cos
sin x
sin2 x cos2 x
27 sin3 x = 64 cos3 x, tan3 x = 64/27, tan x = 4/3 so the critical point is x = x0 where
tan x0 = 4/3 and 0 < x0 < π/2. To test x0 ﬁrst rewrite f (x) as
27 cos x(tan3 x − 64/27)
27 cos3 x(tan3 x − 64/27)
=
;
f (x) =
sin2 x cos2 x
sin2 x
if x < x0 then tan x < 4/3 and f (x) < 0, if x > x0 then tan x > 4/3 and f (x) > 0 so f (x0 ) is
the minimum value. f has no maximum because lim+ f (x) = +∞.
f (x) = − x→0 (b) If tan x0 = 4/3 then (see ﬁgure)
sin x0 = 4/5 and cos x0 = 3/5
so f (x0 ) = 64/ sin x0 + 27/ cos x0
= 64/(4/5) + 27/(3/5)
= 80 + 45 = 125 5 4 x0
3 43. √
2x(x3 − 24x2 + 192x − 640)
; real root of x3 − 24x2 + 192x − 640 at x = 4(2 + 3 2). Since
3
(x − 8)
lim+ f (x) = lim f (x) = +∞ and there is only one relative extremum, it must be a minimum. f (x) = x→+∞ x→8 44. (a) K
ln(a/b)
dC
dC
=
ae−at − be−bt so
= 0 at t =
. This is the only stationary point and
dt
a−b
dt
a−b
C (0) = 0, lim C (t) = 0, C (t) > 0 for 0 < t < +∞, so it is an absolute maximum.
x→+∞ (b) 0.7 0 10
0 Exercise Set 6.2 182 45. The slope of the line is −1, and the slope of the tangent to y = −x2 is −2x so −2x = −1, x = 1/2.
The line lies above the curve so the vertical distance is given by F (x) = 2 − x + x2 ; F (−1) = 4,
F (1/2) = 7/4, F (3/2) = 11/4. The point (1/2, −1/4) is closest, the point (−1, −1) farthest. 46. The slope of the line is 4/3; and the slope of the tangent to y = x3 is 3x2 so 3x2 = 4/3, x2 = 4/9,
x = ±2/3. The line lies below the curve so the vertical distance is given by F (x) = x3 − 4x/3 + 1;
F (−1) = 4/3, F (−2/3) = 43/27, F (2/3) = 11/27, F (1) = 2/3. The closest point is (2/3, 8/27), the
farthest is (−2/3, −8/27). 47. The absolute extrema of y (t) can occur at the endpoints t = 0, 12 or when dy/dt = 2 sin t = 0, i.e.
t = 0, 12, kπ , k = 1, 2, 3; the absolute maximum is y = 4 at t = π, 3π ; the absolute minimum is y = 0
at t = 0, 2π . 48. (a) The absolute extrema of y (t) can occur at the endpoints t = 0, 2π or when
dy/dt = 2 cos 2t − 4 sin t cos t = 2 cos 2t − 2 sin 2t = 0, t = 0, 2π, π/8, 5π/8, 9π/8, 13π/8;
the absolute maximum is y = 3.4142 at t = π/8, 9π/8; the absolute minimum is y = 0.5859
at t = 5π/8, 13π/8. (b) The absolute extrema of x(t) occur at the endpoints t = 0, 2π or when 49. 2 sin t + 1
dx
=−
= 0,
dt
(2 + sin t)2
t = 7π/6, 11π/6. The absolute maximum is x = 0.5774 at t = 11π/6 and the absolute minimum
is x = −0.5774 at t = 7π/6. f (x) = 2ax + b; critical point is x = −
f (x) = 2a > 0 so f
f b
2a =a − f
50. −
− b
2a ≥ 0, b
2a − b
2a b
2a is the minimum value of f , but 2 +b − b
2a +c= −b2 + 4ac
thus f (x) ≥ 0 if and only if
4a −b2 + 4ac
≥ 0, −b2 + 4ac ≥ 0, b2 − 4ac ≤ 0
4a Use the proof given in the text, replacing “maximum” by “minimum” and “largest” by “smallest” and
reversing the order of all inequality symbols. EXERCISE SET 6.2
1. Let x = one number, y = the other number, and P = xy where x + y = 10. Thus y = 10 − x so
P = x(10 − x) = 10x − x2 for x in [0, 10]. dP/dx = 10 − 2x, dP/dx = 0 when x = 5. If x = 0, 5, 10
then P = 0, 25, 0 so P is maximum when x = 5 and, from y = 10 − x, when y = 5. 2. Let x and y be nonnegative numbers and z the sum of their squares, then z = x2 + y 2 . But x + y = 1,
y = 1 − x so z = x2 + (1 − x)2 = 2x2 − 2x + 1 for 0 ≤ x ≤ 1. dz/dx = 4x − 2, dz/dx = 0 when x = 1/2.
If x = 0, 1/2, 1 then z = 1, 1/2, 1 so
(a) z is as large as possible when one number is 0 and the other is 1.
(b) 3. z is as small as possible when both numbers are 1/2. If y = x + 1/x for 1/2 ≤ x ≤ 3/2 then dy/dx = 1 − 1/x2 = (x2 − 1)/x2 , dy/dx = 0 when x = 1. If
x = 1/2, 1, 3/2 then y = 5/2, 2, 13/6 so
(a) y is as small as possible when x = 1. (b) y is as large as possible when x = 1/2. 183 4. Chapter 6 A = xy where x + 2y = 1000 so y = 500 − x/2 and A = 500x − x2 /2
for x in [0, 1000]; dA/dx = 500 − x, dA/dx = 0 when x = 500. If
x = 0 or 1000 then A = 0, if x = 500 then A = 125, 000 so the area
is maximum when x = 500 ft and y = 500 − 500/2 = 250 ft. Stream y
x 5. 6. Let x and y be the dimensions shown in the ﬁgure and A the area,
then A = xy subject to the cost condition 3(2x) + 2(2y ) = 6000, or
y = 1500 − 3x/2. Thus A = x(1500 − 3x/2) = 1500x − 3x2 /2 for x in
[0, 1000]. dA/dx = 1500 − 3x, dA/dx = 0 when x = 500. If x = 0 or
1000 then A = 0, if x = 500 then A = 375, 000 so the area is greatest
when x = 500 ft and (from y = 1500 − 3x/2) when y = 750 ft. Heavyduty y Standard
x Let x and y be the dimensions shown in the ﬁgure and A the area of
the rectangle, then A = xy and, by similar triangles,
x/6 = (8...
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