170 STEP 4. Review the Knowledge You Need to Score High
Step 2: Enter:y 1 =
2500
x
+. 02 +. 004 ∗x
Step 3: Use the [Minimum] function in
the calculator and obtainx= 790 .6.
Step 4: Verify the result with the First
Derivative Test. Entery 2 =
d(2500/x+. 02 +. 004 x,x).
Use the [Zero] function and
obtainx= 790 .6. Thus,
dC
dx
=0,
atx= 790 .6.
Apply the First Derivative Test:
f ′ –0 +
f decr
rel. min
incr
0 790.6
Thus, the minimum average cost
per unit occurs atx= 790 .6. (The
graph of the average cost function
is shown in Figure 8.6-10.)
Figure 8.6-10
- (See Figure 8.6-11.)
(x,y)
y
2
–2
–5 x 5 x
y
Figure 8.6-11
Step 1: AreaA=(2x)(2y); 0≤x≤5 and
0 ≤y≤ 2.
Step 2: 4 x^2 + 25 y^2 =100;
25 y^2 = 100 − 4 x^2.
y^2 =
100 − 4 x^2
25
⇒y=±
√
100 − 4 x^2
25
Sincey≥0,
y=
√
100 − 4 x^2
25
=
√
100 − 4 x^2
5
.
Step 3: A=(2x)
(
2
5
)(√
100 − 4 x^2
)
=
4 x
5
√
100 − 4 x^2
Step 4: Entery 1 =
4 x
5
√
100 − 4 x^2.
Use the [Maximum] function and
obtainx= 3 .536 andy 1 =20.
Step 5: Verify the result with the First
Derivative Test.
Enter
y 2 =d
(
4 x
5
√
100 − 4 x^2 ,x
)
.
Use the [Zero] function and
obtainx= 3 .536.
Note that:
f′
f incr
rel. max
decr
0 3.536
+– 0
The functionfhas only one
relative extremum. Thus, it is the
absolute extremum. Therefore, at
x= 3 .536, the area is 20 and the
area is the absolute maxima.