166 STEP 4. Review the Knowledge You Need to Score High
Step 2:
dA
dx
=
3
2
( 2 x− 9 )−^1 /^2 (2)(9−x)
+(−1)(3)(2x−9)^1 /^2.
=
3 ( 9 −√x)− 3 ( 2 x− 9 )
2 x− 9
=
√^54 −^9 x
2 x− 9
Step 3: Set
dA
dx
= 0 ⇒ 54 − 9 x=0;x= 6.
dA
dx
is undefined atx=
9
2
. The
critical numbers are
9
2
and 6.
Step 4: First Derivative Test:
undef undef + 0 –
9/2 6
incr decr
A′
A
Thus, atx=6, the areaAis a
relative maximum.
A(6)=
(
1
2
)
(3)(
√
2(6)−9)(9−6)
= 9
√
3
Step 5: Check endpoints. The domain of
Ais [9/2, 9].A(9/2)=0; and
A(9)=0. Therefore, the
maximum area of an isosceles
triangle with the perimeter of
18 cm is 9
√
3cm^2. (Note that at
x=6, the triangle is an equilateral
triangle.)
- Step 1: Letxbe the number and
1
x
be its
reciprocal.
Step 2: s=x+
1
x
with 0<x<2.
Step 3:
ds
dx
= 1 +(−1)x−^2 = 1 −
1
x^2
Step 4: Set
ds
dx
= 0 ⇒ 1 −
1
x^2
= 0
⇒x=±1, since the domain is
(0, 2), thusx=1.
ds
dx
is defined for allxin (0, 2).
Critical number isx=1.
Step 5: Second Derivative Test:
d^2 s
dx^2
=
2
x^3
and
d^2 s
dx^2
∣∣
∣∣
x= 1
=2.
Thus, atx=1,sis a relative
minimum. Since it is the only
relative extremum, atx=1, it is
the absolute minimum.
9.(See Figure 8.6-5.)
x
x
xx
x
x
x
x x
x x
x x
x
x
15 – 2x
8 – 2x
Figure 8.6-5
Step 1: Volume:V=x(8− 2 x)(15− 2 x)
with 0≤x≤4.
Step 2: Differentiate: Rewrite as
V= 4 x^3 − 46 x^2 + 120 x
dV
dx
= 12 x^2 − 92 x+120.
Step 3: SetV= 0 ⇒ 12 x^2 − 92 x+ 120 = 0
⇒ 3 x^2 − 23 x+ 30 =0. Using the
quadratic formula, you havex= 6
orx=
5
3
and
dV
dx
is defined for all
real numbers.