MATH 31
UNIT 5: LESSON 6A
GEOMETRIC APPLICATIONS: MAXIMIZING AREA AND VOLUME
NAME ANSWERS
Page 1 of 9
EXAMPLE 1: The Starks have 60 metres of fencing with which to make a rectangular dog run. If they
use a side of the shed as one side of the run, what dimensions will give the maximum area?
P = 2w + l
60 = 2w + l
l = 60 – 2w
shed
A = lw
A = w (60 – 2w)
A = 60w – 2w2
𝒅𝑨
= 𝟔𝟎 − 𝟒𝒘
𝒅𝒕
𝟎 = 𝟔𝟎 − 𝟒𝒘
𝒘 = 𝟏𝟓
If w = 15 then l = 60 – 2(15) = 30
The Starks should make their run 30 m by 15 m to get a maximum area of 450 m2
EXAMPLE 2:
A rectangular field is to be enclosed and divided into 2 smaller plots by a fence parallel
to one of the sides.
Find the dimensions of the largest such field if 1200 m of fencing material is to be used.
l
w
P = 3w + 2l
w
w
1200 = 3w + 2l
2l = 1200 – 3w
l = 600 – 1.5 w
A = lw
A = w(600 – 1.5w) = 600w – 1.5w2
𝒅𝑨
= 𝟔𝟎𝟎 − 𝟑𝒘
𝒅𝒕
𝟎 = 𝟔𝟎𝟎 − 𝟑𝒘
𝒘 = 𝟐𝟎𝟎
The dimensions of the field should be 300 m by 200 m to get a maximum area of 60 000 m2
U5 L6A Maximizing Area and Vol ANS
MATH 31
UNIT 5: LESSON 6A
GEOMETRIC APPLICATIONS: MAXIMIZING AREA AND VOLUME
NAME ANSWERS
Page 2 of 9
EXAMPLE 3:
A box has square ends and the sides are congruent rectangles. The total area of the four sides and two
ends is 294 cm2. What are the dimensions of the box if the volume is a maximum and what is the
maximum volume?
Surface area = 2x2 + 4xy
volume = lwh
x
x
294 = 2x2 + 4xy
V  x2 y
y
 294  2 x 2  294 x 2  2 x 4
V x 

4x
4x


V  73.5 x  0.5 x3
294  2 x 2
y
4x
2
dV
 73.5  1.5 x 2
dx
73.5  1.5 x 2  0
x 2  49
x7
294  2(7) 2
If x = 7 then y 
7
4(7)
The box should be 7 cm by 7 cm by 7cm to produce a maximum volume of 343 cm3
U5 L6A Maximizing Area and Vol ANS
MATH 31
UNIT 5: LESSON 6A
GEOMETRIC APPLICATIONS: MAXIMIZING AREA AND VOLUME
NAME ANSWERS
Page 3 of 9
EXAMPLE 4
An open (it has no top) box has square ends and the sides are congruent rectangles.
The total area of the four sides and one end is 192 cm2.
What are the dimensions of the box if the volume is a maximum and what is the maximum volume?
Surface area = x2 + 4xy
volume = lwh
192 = x2 + 4xy
x
V  x2 y
x
y
y
 192  x 2  192 x 2  x 4
V  x2 

4x
 4x 
3
V  48 x  0.25 x
192  x 2
4x
dV
 48  0.75 x 2
dx
48  0.75 x 2  0
x 2  64
x 8
If x = 8 then
y
192  82
4
4(8)
The box should be 8 cm by 8 cm by 4 cm to produce a maximum volume of 256 cm3
PRACTICE QUESTION
1. A storage box has square ends and no top.
The total surface area is 9600 cm2.
What are the dimensions of the box if the volume is a maximum and what is the
maximum volume?
No top here
U5 L6A Maximizing Area and Vol ANS
MATH 31
UNIT 5: LESSON 6A
GEOMETRIC APPLICATIONS: MAXIMIZING AREA AND VOLUME
NAME ANSWERS
Page 4 of 9
EXAMPLE 5
A page contains 600 cm2. The margins at the top and bottom are 3 cm. The margins at each side are to
be 2 cm. What are the dimensions of the paper if the printed area is a maximum?
Dimensions of the printed area are (x – 4) by (y – 6)
3
2
y
Area of whole page
A = xy
600 = xy
x
Area of the print
A = (x – 4) (y – 6)
A = (x – 4) (
– 6)
A = 600 – 6x – 2400x -1 + 24
A = 624 – 6x – 2400x -1
When x = 20,
The dimensions of the page are 20 cm by 30 cm
U5 L6A Maximizing Area and Vol ANS
MATH 31
UNIT 5: LESSON 6A
GEOMETRIC APPLICATIONS: MAXIMIZING AREA AND VOLUME
NAME ANSWERS
Page 5 of 9
EXAMPLE 6
A rectangular open-topped box is to be made from a piece of material 18 cm by 48 cm by cutting a
square from each corner and turning up the sides. What size squares must be removed to maximize
the capacity of the box?
x
18 – 2x
18 cm
48 – 2x
48 cm
V = lwh= x(18 – 2x)(48 –2 x) = x (864 – 132x + 4x2) = 864x – 132x 2 + 4x3
V′ = 12x2 – 264x + 864
12x2 – 264x + 864 = 0
x2 – 22x + 72 = 0
(x – 18)(x – 4) = 0
x = 18 or x = 4
Too high
The squares should be 4 cm by 4 cm to create a maximum volume of 4 x 40 x 10 = 1600cm3
U5 L6A Maximizing Area and Vol ANS
MATH 31
UNIT 5: LESSON 6A
GEOMETRIC APPLICATIONS: MAXIMIZING AREA AND VOLUME
NAME ANSWERS
ASSIGNMENT QUESTIONS
Page 6 of 9
1. A carpenter is building a rectangular room with a fixed perimeter of 54 feet. What
are the dimensions of the largest room that can be built? What is its area?
P = 2l + 2w
54 = 2l + 2w
27 = l + w
l = 27 – w
A = lw
A = w(27 – w ) = 27w – w 2
A′ = 27 – 2w
27 – 2w = 0
w = 13.5 therefore l = 27 – 13.5 = 13.5
The dimensions are 13.5 ft by 13.5 ft. and the maximum area is 182.25 square feet
2. A cattle farmer wants to build a rectangular fenced enclosure divided into five
rectangular pens, as shown in the diagram. A total length of 120 m of fencing
material is available. Find the overall dimensions of the enclosure that will make the
total area a maximum.
A = lw
P = 6w + 2l
A = w (60 – 3w) = 60w – 3w2
A ′ = 60 – 6w
120 = 6w + 2l
60 – 6w = 0
2l = 120 – 6w
w = 10
l = 60 – 3w
l = 60 – 3(10) = 30
The enclosure should be 10 m by 30 m to make the area a maximum
U5 L6A Maximizing Area and Vol ANS
MATH 31
UNIT 5: LESSON 6A
GEOMETRIC APPLICATIONS: MAXIMIZING AREA AND VOLUME
NAME ANSWERS
Page 7 of 9
3. A box has square ends, and the sides are congruent rectangles. The total area of the
four sides and two ends is 216 cm2. What are the dimensions of the box if the
volume is a maximum, and what is the maximum volume?
Surface area = 2x2 + 4xy
volume = lwh
V  x2 y
216 = 2x2 + 4xy
216  2 x 2
y
4x
 216  2 x 2  216 x 2  2 x 4
V x 

4x
4x


V  54 x  0.5 x3
2
dV
 54  1.5 x 2
dx
54  1.5 x 2  0
If x = 6 then y 
216  2(6) 2
6
4(6)
x 2  36
x6
The box should be 6 cm by 6 cm by 6cm to produce a maximum volume of 216 cm3
4. A storage box has square ends and no top.
The total surface area is 9600 cm2.
What are the dimensions of the box if the volume is a maximum and what is the
maximum volume?
No top here
S = 2x2 + 3xy
V = x2 y
 9600  2 x 2 
V  x2 

3x


9600 = 2x2 + 3xy
9600  2 x 2
y
3x
V
V '  3200  2 x 2
0  3200  2 x
The dimensions of the box
should be 40 cm by 53 ⅓ cm
and the maximum volume will
be
2 x 2  3200
x 2  1600
x  40
40 x 40 x 53 ⅓ = 85 333 ⅓ cm3
U5 L6A Maximizing Area and Vol ANS
2
9600 x  2 x3
2
 3200 x  x3
3
3
9600  2 x 2
3x
9600  2(40) 2
1
y
 53
3  40 
3
y
MATH 31
UNIT 5: LESSON 6A
GEOMETRIC APPLICATIONS: MAXIMIZING AREA AND VOLUME
NAME ANSWERS
Page 8 of 9
5. A rectangular page is to contain 150 cm2 of printing. The margins at the top and
bottom of the page are 3 cm. The margins at each side are 2 cm. What should the
dimensions of the page be if the minimum amount of paper is used?
Dimensions of total page are x + 4
Area of print: xy = 150
y
150
x
 150

A  ( x  4) 
 6
 x

1
A  150  6 x  600 x  24
A  6 x  600 x 1  174
dA
 6  600 x 2
dx
600
6 2  0
x
600
6 2
x
2
x  100
x  10
y
150
 15
10
The dimensions of the page should be 10 + 4 = 14 cm
by15 + 6 = 21 cm in order to use the minimum amount
of paper.
U5 L6A Maximizing Area and Vol ANS
by y + 6
MATH 31
UNIT 5: LESSON 6A
GEOMETRIC APPLICATIONS: MAXIMIZING AREA AND VOLUME
NAME ANSWERS
Page 9 of 9
6. From a thin piece of cardboard 30 cm by 30 cm, square corners are cut out so the
sides can be folded up to make a box. What dimensions will yield a box of
maximum volume? What is the maximum volume?
𝑽 = 𝒙(𝟑𝟎 − 𝟐𝒙)(𝟑𝟎 − 𝟐𝒙)
𝟐
𝟑
V = x(30 – 2x)(30 – 2x) = 900 x – 𝑽 = 𝟗𝟎𝟎𝒙 − 𝟏𝟐𝟎𝒙 + 𝟒𝒙
V’ = 900 – 240x + 12x2
0 = 12(x2 – 20x + 75)
0 = (x – 5)(x – 15)
x = 5 or x = 15
30 – 2(15) = 0 so x = 15 is invalid
The dimensions of the box are 5 cm, by 20 cm by 20 cm.
The maximum volume of the box is 2000 cm3 if the squares are 5 cm by 5 cm.
U5 L6A Maximizing Area and Vol ANS