§4-7 Optimization
Homework：4,10,17,24,29,33,35,51,57.
微積分方法 : 求最大值與最小值，可以用下列兩種方法來解。
(i) f is continuous on [ a, b ] .
比較
x
a
critical points
B
y
(ii) f has exactly one critical point c in the interval I .
If
f′
⇒ f ( c ) is an absolute minimum.
If：
f′
⇒ f ( c ) is an absolute maximum.
Example 1：A cylindrical can is to be made to hold 1 L of oil. Find the
dimensions that will minimize the cost of the metal to
manufacture the can?
Solution：
已知： π r 2 h = 1000
Minimize A = 2π r 2 + 2π rh
2000
, 0<r<∞
= 2π r 2 +
r
(
3
dA
2000 4 π r − 500
= 4π r − 2 =
dr
r
r2
)
1
⎛ 500 ⎞ 3
⇒r =⎜
⎟
⎝ π ⎠
1⎤
⎡
⎛ 500 ⎞ 3 ⎥
⎢
A ⎜
is a absolute minimun.
⎢⎝ n ⎟⎠ ⎥
⎣⎢
⎦⎥
( )
500
π
1
3
⇒
⇒h=
1000
2
⎛ 500 ⎞ 3
π⎜
⎟
⎝ π ⎠
= 23
500
π
= 2r.
Example 2：A box with a square base and open top must have a volume of
32,000 cm3. Find the dimensions of the box that minimize the
amount of material used.
Solution：
已知： x 2 h = 32000
Minimize
128000
x
= : f ( x)
x 2 + 4 xh = x 2 +
(
3
128000 2 x − 64000
=
f ′ ( x) = 2x −
x2
x2
)
, 0< x<∞
f ′( x )
⇒ x = 40
h = 20 .
Example 3：A man launches his boat from point A on a bank of a straight river,
5 km wide, and wants to reach point B, 5 km downstream on the
opposite bank, as quick as possible. If he can row 6 km/h and run
8 km/h, where should he land to reach B as soon as possible?
Solution：
Minimize
25 + x 2 5 − x
+
, 0≤ x≤5
6
8
x
1
T ′( x) =
−
6 25 + x 2 8
T ( x) =
x
T ′( x) = 0 ⇒
6 25 + x 2
15
⇒x=
>5
7
x
0
5
y
5 5
+
6 8
5
2
6
=
1
⇒ 16 x 2 = 9 x 2 + 225
8
He should row directly to B.
Example 4：A cone-shaped drinking cup is made from radius R by cutting
out a sector and joining the edges CA and CB. Find the maximum
capacity of such a cup.
A
C
B
θ
R
R
h
C
r
(
)
1
1
Proof： V ( h ) = π r 2 h = π R 2 − h 2 h
3
3
π
=
3
R2 h −
π
3
h3 , 0<h<R
dV π 2
R − 3h 2 = 0
=
dh 3
(
h=
1
3
)
R
dV
dh
1
R
3
⎛ 1 ⎞ 2π 3
⇒V ⎜
R
R⎟ =
⎝ 3 ⎠ 9 3
is the maximum.
θ
想想看: 為達最大容量，角度
要取多少?
Example 5：The upper right-hand corner of a piece of paper, 12 inch by 8 inch,
as in the figure, is folded over to the bottom edge. How would
you fold it, so as to minimize the length of the fold ? In other
words, how would you choose x to minimize y?
12
D
y
x
θ
θ B
8
x α
α
O
C
8-x
A
Solution： ΔOCD ∼ ΔBAO
8
⇒
y 2 − x2
(
⇒ y 2 − x2
⇒ y2 =
=
4 x−4
x
) ( x − 4) = 4x
2
想想看: why 4＜ x ≤ 8?
x3
4 x2
+ x2 =
=: f ( x ) , 4＜ x ≤ 8 .
x−4
x−4
⇒ f ′( x) =
2 x2 ( x − 6)
( x − 4 )2
=0 ⇒
x=6
f′
Example 6：An observer stands at a point P, one unit away from a track. Two
runners stand at the point S in the figure and run along the track.
One runner runs three times as fast as other. Find the maximum
value of the observer’s angle of sight between runners?
Solution：
α
runner B runs three times as fast as runner A
tan θ = tan (θ + α − α ) =
=
1 + tan (θ + α ) tan α
3x − x
2x
=
=: f ( x ) , 0 ≤ x < ∞
2
1 + 3x
1 + 3x 2
f ′( x) =
(
2 1 − 3x 2
(
1 + 3x 2
)
) =0⇒ x=
2
f′
1
3
tan (θ + α ) − tan α
1
3
( x ≥ 0)
2
⇒ tan θ =
1
3
⎛ 1 ⎞
1 + 3⎜
⎟
⎝ 3⎠
2
=
1
3
.
⇒θ =
π
6
is the max angle of sight.
Example 7：Where should the point p be chosen on the line segment AB so as
to maximize the angle θ ？
β
α
Solution： tan θ = tan (π − (α + β ) ) = − tan (α + β )
5
2
+
⎛ tan α + tan β ⎞
x 3− x
= −⎜
⎟= 5 2
⎝ 1 − tan α tan β ⎠
⋅
−1
x 3− x
3(5 − x )
= 2
=: f ( x ) , 0 ≤ x ≤ 5
x − 3x + 10
f ′( x) =
(
3 x 2 − 10 x + 5
(x
2
− 3x + 10
x = 5−2 5
)
)
2
( 5 + 2 5 ＞3 不合)
f′
⇒ x = 5−2 5
will maximize the angle θ
5−2 5