5 Steps to a 5 AP Calculus AB 2019 - William Ma

MA 3972-MA-Book April 11, 2018 14:46

Applications of Derivatives 185

Solution:

Step 1: Draw a diagram. (See Figure 9.2-2.)

y = – x + 2

y

y

x x

P(x, y)

0

1

2

Figure 9.2-2

Step 2: LetP(x,y) be the vertex of the rectangle on the graph ofy=−

1

2

x+2.

Step 3: Thus, the area of the rectangle is:

A=xyorA=x

(

−

1

2

x+ 2

)

=−

1

2

x^2 + 2 x.

The domain ofAis [0, 4].

Step 4: Differentiate:

dA

dx

=−x+ 2.

Step 5:
dA
dx
is defined for all real numbers.

Set

dA

dx

= 0 ⇒−x+ 2 =0;x=2.

A(x) has one critical numberx=2.

Step 6: Apply the Second Derivative Test:

d^2 A

dx^2

=− 1 ⇒A(x) has a relative maximum point atx=2;A(2)=2.

Sincex=2 is the only relative maximum, it is the absolute maximum. (Note that

at the endpoints:A(0)=0 andA(4)= 0 .)

Step 7: Atx=2,y=−

1

2

(2)+ 2 =1.

Therefore, the length of the rectangle is 2, and its width is 1.