EXAMPLE 3 Maximizing Volume A supplier of bolts wants to create open boxes for the bolts by cutting a square from each corner of a 12-in. by 12-in. piece of metal and then folding up the sides. What size square should be cut from each corner to produce a box of maximum volume? 0001 APPLY IT SOLUTION Let x represent the length of a side of the square that is cut from each corner, as shown in Figure 8(a). The width of the box is 12 2x, with the length also 12 As shown in Figure 8(b), the depth of the box will be x inches. The volume of the box is given by the product of the length, width, and height. In this example, the volume, V(x), depends on x: 2x. V(x) = x(12 - 2x)(12 - 2x) = 144x – 48x2 + 4x³. %3D %3D Clearly, 0 < x, and since neither the length nor the width can be negative, 0 < 12 – 2x, so x < 6. Thus, the domain of V is the interval 0, 6]. ninmob sdi o1duovitoged Jer ods 0 21 12-2х — 0000 x = depth 002 12 2x 12 2x (a) (b) FIGURE 8 766 CHAPTER 14 Applications of the Derivative Extrema Candidates The derivative is V'(x) = 144 - 96x + 12x². Set this derivative equal to 0. %3D х V(x) 12x2 - 96x + 144 = 0 12(x² - 8x + 12) = 0 12(x - 2)(x - 6) =D 0 %3D 128 Maximum 6. 0. x - 2 = 0 x – 6 = 0 %3D or %3D %3D YOUR TURN 3 Repeat Example 3 using an 8-m by 8-m piece of metal. Find V(x) for x equal to 0, 2, and 6 to find the depth that will maximize the volume. The table indicates that the box will have maximum volume when x = 2 and that the maximum volume will be 128 in. %3D 3 TRY YOUR TURN 3

College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter4: Polynomial And Rational Functions
Section4.1: Quadratic Functions
Problem 64E: Maximizing parking lot area A rectangular parking lot is being constructed for your college football...
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EXAMPLE 3
Maximizing Volume
A supplier of bolts wants to create open boxes for the bolts by cutting a square from each
corner of a 12-in. by 12-in. piece of metal and then folding up the sides. What size square
should be cut from each corner to produce a box of maximum volume?
0001
APPLY IT SOLUTION Let x represent the length of a side of the square that is cut from each corner,
as shown in Figure 8(a). The width of the box is 12 2x, with the length also 12
As shown in Figure 8(b), the depth of the box will be x inches. The volume of the box is
given by the product of the length, width, and height. In this example, the volume, V(x),
depends on x:
2x.
V(x) = x(12 - 2x)(12 - 2x) = 144x – 48x2 + 4x³.
%3D
%3D
Clearly, 0 < x, and since neither the length nor the width can be negative, 0 < 12 – 2x, so
x < 6. Thus, the domain of V is the interval 0, 6].
ninmob sdi o1duovitoged
Jer ods
0 21
12-2х —
0000
x = depth
002
12 2x
12 2x
(a)
(b)
FIGURE 8
Transcribed Image Text:EXAMPLE 3 Maximizing Volume A supplier of bolts wants to create open boxes for the bolts by cutting a square from each corner of a 12-in. by 12-in. piece of metal and then folding up the sides. What size square should be cut from each corner to produce a box of maximum volume? 0001 APPLY IT SOLUTION Let x represent the length of a side of the square that is cut from each corner, as shown in Figure 8(a). The width of the box is 12 2x, with the length also 12 As shown in Figure 8(b), the depth of the box will be x inches. The volume of the box is given by the product of the length, width, and height. In this example, the volume, V(x), depends on x: 2x. V(x) = x(12 - 2x)(12 - 2x) = 144x – 48x2 + 4x³. %3D %3D Clearly, 0 < x, and since neither the length nor the width can be negative, 0 < 12 – 2x, so x < 6. Thus, the domain of V is the interval 0, 6]. ninmob sdi o1duovitoged Jer ods 0 21 12-2х — 0000 x = depth 002 12 2x 12 2x (a) (b) FIGURE 8
766 CHAPTER 14 Applications of the Derivative
Extrema Candidates
The derivative is V'(x) = 144 - 96x + 12x². Set this derivative equal to 0.
%3D
х
V(x)
12x2 - 96x + 144 = 0
12(x² - 8x + 12) = 0
12(x - 2)(x - 6) =D 0
%3D
128
Maximum
6.
0.
x - 2 = 0
x – 6 = 0
%3D
or
%3D
%3D
YOUR TURN 3 Repeat
Example 3 using an 8-m by 8-m
piece of metal.
Find V(x) for x equal to 0, 2, and 6 to find the depth that will maximize the volume. The
table indicates that the box will have maximum volume when x = 2 and that the maximum
volume will be 128 in.
%3D
3
TRY YOUR TURN 3
Transcribed Image Text:766 CHAPTER 14 Applications of the Derivative Extrema Candidates The derivative is V'(x) = 144 - 96x + 12x². Set this derivative equal to 0. %3D х V(x) 12x2 - 96x + 144 = 0 12(x² - 8x + 12) = 0 12(x - 2)(x - 6) =D 0 %3D 128 Maximum 6. 0. x - 2 = 0 x – 6 = 0 %3D or %3D %3D YOUR TURN 3 Repeat Example 3 using an 8-m by 8-m piece of metal. Find V(x) for x equal to 0, 2, and 6 to find the depth that will maximize the volume. The table indicates that the box will have maximum volume when x = 2 and that the maximum volume will be 128 in. %3D 3 TRY YOUR TURN 3
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