5 Steps to a 5 AP Calculus BC 2019

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

19. (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.