Applications of Derivatives 167
Step 4: Second Derivative Test:
d^2 V
dx^2
= 24 x−92;
d^2 V
dx^2
∣∣
∣∣
x= 6
=52 and
d^2 V
dx^2
∣∣
∣∣
x=^53
=− 52.
Thus, atx=
5
3
is a relative
maximum.
Step 5: Check endpoints.
Atx=0,V=0 and atx=4,
V=0. Therefore, atx=
5
3
,Vis
the absolute maximum.
- (See Figure 8.6-6.)
Figure 8.6-6
Step 1: Distance formula:
Z=
√
(x− 2 )^2 +
(
y−
(
−
1
2
)) 2
=
√
(x− 2 )^2 +
(
−x^2 +
1
2
) 2
=
√
x^2 − 4 x+ 4 +x^4 −x^2 +
1
4
=
√
x^4 − 4 x+
17
4
Step 2: LetS=Z^2 , sinceSandZhave
the same maximums and
minimums.
S=x^4 − 4 x+
17
4
;
dS
dx
= 4 x^3 − 4
Step 3: Set
dS
dx
=0;x=1 and
dS
dx
is
defined for all real numbers.
Step 4: Second Derivative Test:
d^2 S
dx^2
= 12 x^2 and
d^2 S
dx^2
∣∣
∣
∣x= 1 =12.
Thus, atx=1,Zhas a minimum,
and since it is the only relative
extremum, it is the absolute
minimum.
Step 5: Atx=1,
Z=
√
( 1 )^4 −4(1)+
17
4
=
√
5
4
.
Therefore, the shortest distance is
√
5
4
.
- Step 1: Average cost:
C=
C(x)
x
=
3 x^2 + 5 x+ 12
x
= 3 x+ 5 +
12
x
.
Step 2:
dC
dx
= 3 − 12 x−^2 = 3 −
12
x^2
Step 3: Set
dC
dx
= 0 ⇒ 3 −
12
x^2
= 0 ⇒
3 =
12
x^2
⇒x=±2. Sincex>0,x= 2
andC(2)= 17.
dC
dx
is undefined at
x=0, which is not in the domain.
Step 4: Second Derivative Test:
d^2 C
dx^2
=
24
x^3
and
d^2 C
dx^2
∣∣
∣∣
x= 2
= 3
Thus, atx=2, the average cost is
a minimum.
