5 Steps to a 5 AP Calculus BC 2019

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.)

8. 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.