Thomas Calculus 13th [Solutions]

290 Chapter 4 Applications of Derivatives
4.6 APPLIED OPTIMIZATION
1. Let and w represent the length and width of the rectangle, respectively. With an area of 16 in. , we have that
2
1
1
2( 16)
( )( w) 16 w 16 the perimeter is P 2 2w 2 32 and P ( ) 2
32
. Solving
2 2
P ( ) 0
2( 4)( 4)
2
0 4, 4. Since 0 for the length of a rectangle, must be 4 and w 4 the
perimeter is 16 in., a minimum since P ( )
16
3
0.
2. Let x represent the length of the rectangle in meters (0 x 4) . Then the width is 4 x and the area is
A( x) x(4 x) 4 x
2
x . Since A ( x) 4 2 x , the critical point occurs at x 2. Since, A ( x ) 0 for 0 x 2
and A ( x ) 0 for 2 x 4, this critical point corresponds to the maximum area. The rectangle with the largest
area measures 2 m by 4 2 2 m, so it is a square.
Graphical Support:
2
3. (a) The line containing point P also contains the points (0, 1) and (1, 0) the line containing P is y 1 x
a general point on that line is ( x, 1 x).
(b) The area A( x) 2 x(1 x ), where 0 x 1.
2
(c) When A( x) 2x 2 x , then A ( x) 0 2 4x 0 x
1
. Since A (0) 0 and A (1) 0, we conclude
2
A sq units is the largest area. The dimensions are 1 unit by 1 2 unit.
that
1 1
2 2
4. The area of the rectangle is A 2xy 2 x (12 x ),
2
where 0 x 12. Solving A ( x) 0 24 6x
0
x 2 or 2. Now 2 is not in the domain, and
since A (0) 0 and A 12 0, we conclude that
A (2) 32 square units is the maximum area. The
dimensions are 4 units by 8 units.
2
5. The volume of the box is V ( x) x(15 2 x)(8 2 x)
2 3
120x 46x 4 x , where 0 x 4. Solving
2
V ( x) 0 120 92x 12x 4(6 x)(5 3 x) 0
x
5
or 6, but 6 is not in the domain. Since
3
V (0) V (4) 0, V
5 2450
91 in 3 must be the
3 27
maximum volume of the box with dimensions
14 35 5
3 3 3 inches.
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