Part II: The Answers
328
Answers
501–600
506.
3 5 in. × 6 5 in.
You want to maximize the printed area. Let x and y be the length and width of the
poster. Because the total area must be 90 square inches, you have A = xy = 90.
The poster has a 1-inch margin at the bottom and sides and a 3-inch margin at the top,
so the printed area of the poster is
A x y
x
y
( , ) = −
−
2
4
(
)(
)
Using y x
= 90 , the printed area is
A x
x
x
x
x
x
x
( ) = ( − )
−
= −
−
+
= −
−
2 90 4
90 4
180 8
98 4
180
( )
Find the derivative of the function for the printed area:
′
A x
x
x
x
( ) = − +
= −
+
4 180
4
180
2
2
2
Set the derivative equal to zero and solve for x: –4x
2
+ 180 = 0 so that x
2
= 45. Keeping
the positive solution, you have x =
=
=
45
9 5
3 5
( )( )
.
Using the first derivative test, you can verify that x = 3 5 gives you a maximum. From
the equation y x
= 90 , you get the y value:
y =
=
=
=
90
3 5
30
5
30 5
5
6 5
Therefore, the dimensions are 3 5 inches × 6 5 inches (approximately 6.7 inches ×
13.4 inches).
507.
− 3
2
and 3
2
You want to locate the maximum slope of f
 
(x) = 2 + 20x
3
– 4x
5
. The slope of the tangent
line is given by the derivative
′
f x
x
x
( ) =
−
60
20
2
4
At a point p, the slope is
s p
p
p
( ) =
−
60
20
2
4