Applications of Derivatives 157
Solution:
Step 1: Draw a diagram. (See Figure 8.2-2.)
y = – x + 2
y
y
x x
P(x,y)
0
1
2
Figure 8.2-2
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 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.