5 Steps to a 5 AP Calculus BC 2019

Applications of Derivatives 171

20. (See Figure 8.6-12.)

y

y

x (x,0)

(0,y) (0.5,4)

x

l

0

Figure 8.6-12

Step 1: Distance formula:

l^2 =x^2 +y^2 ;x> 0 .5 andy>4.

Step 2: The slope of the hypotenuse:

m=

y− 4

0 − 0. 5

=

− 4

x− 0. 5

⇒(y−4)(x− 0 .5)= 2

⇒xy− 0. 5 y− 4 x+ 2 = 2

y(x− 0 .5)= 4 x

y=

4 x

x− 0. 5

.

Step 3: l^2 =x^2 +

(

4 x

x− 0. 5

) 2

;

l=±

√

x^2 +

(

4 x

x− 0. 5

) 2

Sincel>0,l=

√

x^2 +

(

4 x

x− 0. 5

) 2

.

Step 4: Entery 1 =

√

x^2 +

(

4 x

x− 0. 5

) 2

and use the [Minimum] function

of the calculator and obtain

x= 2 .5.

Step 5: Apply the First Derivative Test.

Entery 2 =d(y1(x),x) and use

the [Zero] function and obtain

x= 2 .5.

Note that:

3

f ′ – 0 +

f decr

rel. min

incr

Sincefhas only one relative

extremum, it is the absolute

extremum.

Step 6: Thus, atx= 2 .5, the length of the

hypotenuse is the shortest. At

x= 2 .5,y=

4(2.5)

2. 5 − 0. 5

=5. The

vertices of the triangle are

(0, 0), (2.5, 0), and (0, 5).

8.7 Solutions to Cumulative Review Problems

21. Rewrite:y=[sin(cos( 6 x− 1 ))]^2

Thus,

dy

dx

= 2 [sin(cos( 6 x− 1 ))]

×[cos(cos( 6 x− 1 ))]

×[−sin(6x−1)](6)

=−12 sin(6x−1)

×[sin(cos( 6 x− 1 ))]

×[cos(cos( 6 x− 1 ))].

22. Asx→∞, the numerator

100

x

approaches 0 and the denominator

increases without bound (i.e.,∞).

Thus, the limx→∞

100 /x

− 4 +x+x^2

=0.