1. MAXIMIZING AREA AND VOLUME
Unit 5 Lesson 6A Maximizing Area and Volume
MAXIMIZING AREA AND VOLUME

2. Unit 5 Lesson 6A Maximizing Area and Volume
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?
A = lw
shed
P = 2w + l
A = w(60 – 2w) = 60w – 2w 2
60 = 2w + l
𝒅𝑨 𝒅𝒘 =𝟔𝟎−𝟒𝒘 w w
l = 60 – 2w
0 = 60 – 4 w l
4w = 60
w = 15
When w = 15, l = 60 – 2(15) = 30
The Starks should make their run 30 m by 15m
to get a maximum area of 450m2
𝒅𝑨 𝒅𝒘 =𝟔𝟎−𝟒𝒘 so 𝒅 𝟐 𝑨 𝒅 𝒘 𝟐 =−𝟒
The second derivative is negative;
thus, the area is a maximum.

3. Unit 5 Lesson 6A Maximizing Area and Volume
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
A = lw
A = w(600 – 1.5w) = 600w – 1.5w2
P = 3w+ 2l
𝒅𝑨 𝒅𝒘 =𝟔𝟎𝟎−𝟑𝒘
1200 = 3w+ 2l
0 = 600 – 3 w
2l = 1200 – 3 w
3w = 600
l = 600 – 1.5w
w = 200
When w = 200, l = 600 – 1.5(200) = 300

4. Unit 5 Lesson 6A Maximizing Area and Volume
𝒅𝑨 𝒅𝒘 =𝟔𝟎𝟎−𝟑𝒘
The second derivative is negative;
thus, the area is a maximum.
𝒅 𝟐 𝑨 𝒅 𝒘 𝟐 =−𝟑 l w
The dimensions of the field should be 300m by 200m to get a maximum area of m2

5. Unit 5 Lesson 6A Maximizing Area and Volume
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? y x
volume = l x w x h
V = x2 y
𝑽= 𝒙 𝟐 𝟐𝟗𝟒−𝟐 𝒙 𝟐 𝟒𝒙
surface area = 2x2 + 4xy
𝑽= 𝟐𝟗𝟒 𝒙 𝟐 −𝟐 𝒙 𝟒 𝟒𝒙
294 = 2x2 + 4xy
294 – 2x2 = 4xy
𝑽=𝟕𝟑.𝟓𝒙−𝟎.𝟓 𝒙 𝟑
𝑦= 𝟐𝟗𝟒−𝟐 𝒙 𝟐 𝟒𝒙

6. Unit 5 Lesson 6A Maximizing Area and Volume
𝑽=𝟕𝟑.𝟓𝒙−𝟎.𝟓 𝒙 𝟑
When x = 7 cm
𝒅𝑽 𝒅𝒙 = 𝟕𝟑.𝟓−𝟏.𝟓𝒙𝟐
𝒚= 𝟐𝟗𝟒−𝟐 (𝟕) 𝟐 𝟒(𝟕) =𝟕
73.5 – 1.5x2 = 0
Box is 7cm x 7cm x 7 cm
Max Vol = 343 cm3
1.5x2 = 73.5
x2 = 49
𝒅 𝟐 𝑽 𝒅 𝒙 𝟐 =−𝟑𝒙 = 𝟑 𝟕 =−𝟐𝟏
x = 7
The second derivative is negative;
thus, the volume is a maximum.

7. Unit 5 Lesson 6A Maximizing Area and Volume
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? y x
surface area = x2 + 4xy
volume = l x w x h
192 = x2 + 4xy
𝑽= 𝒙 𝟐 𝟏𝟗𝟐− 𝒙 𝟐 𝟒𝒙
𝒚= 𝟏𝟗𝟐− 𝒙 𝟐 𝟒𝒙
𝑽= 𝟏𝟗𝟐 𝒙 𝟐 − 𝒙 𝟒 𝟒𝒙
𝑽=𝟒𝟖𝒙−𝟎.𝟐𝟓 𝒙 𝟑

8. Unit 5 Lesson 6A Maximizing Area and Volume
𝒚= 𝟏𝟗𝟐− 𝒙 𝟐 𝟒𝒙
𝑑𝑉 𝑑𝑥 = 48 – 0.75𝑥2
𝒚= 𝟏𝟗𝟐− (𝟖) 𝟐 𝟒(𝟖) =𝟒
48 – 0.75x2 = 0
Box is 8 cm x 8 cm x 4 cm
Max Vol = 256 cm3
0.75x2 = 48
x2 = 64
𝑑 2 𝑉 𝑑 𝑥 2 =−1.5𝑥=−1.5 𝟖 =−12
x = 8
The second derivative is negative;
thus, the volume is a maximum.

9. Unit 5 Lesson 6A Maximizing Area and Volume
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 printed area are (x – 4) by (y – 6) x y 3 2
Area of page: x y = 600
𝒚= 𝟔𝟎𝟎 𝒙 =𝟔𝟎𝟎 𝒙 −𝟏
A = (x – 4)(y – 6)
A = (x – 4)( 600 x –1 – 6)
A = 600 – 6x – 2400x –1 + 24
A = 624 – 6x – 2400 x –1

10. Unit 5 Lesson 6A Maximizing Area and Volume
A = 624 – 6x – 2400 x -1
𝒅𝑨 𝒅𝒙 =−𝟔+𝟐𝟒𝟎𝟎 𝒙 −𝟐
When x = 20, 𝒚= 𝟔𝟎𝟎 𝟐𝟎 =𝟑𝟎
The dimensions of the page
are 20 cm by 30 cm.
𝟎=−𝟔+𝟐𝟒𝟎𝟎 𝒙 −𝟐
𝟔= 𝟐𝟒𝟎𝟎 𝒙 𝟐
𝒅 𝟐 𝑨 𝒅 𝒙 𝟐 =−𝟒𝟖𝟎𝟎 𝒙 −𝟑 =−𝟒𝟖𝟎𝟎 (𝟐𝟎) −𝟑 =−𝟎.𝟔
𝟔 𝒙 𝟐 =𝟐𝟒𝟎𝟎
x2 = 400
The second derivative is negative when x = 20;
thus, the area is a maximum.
x = 20

11. Unit 5 Lesson 6A Maximizing Area and Volume
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?
18 – 2x
48 – 2x x
48 cm
18 cm
𝒅𝑽 𝒅𝒙 = 𝟏𝟐𝒙𝟐 −𝟐𝟔𝟒𝒙 + 𝟖𝟔𝟒
V = lwh = x(18 – 2x)(48 – 2x)
12x2 – 264x = 0
V = x (864 – 132x + 4x2)
V = 864x – 132x 2 + 4x 3
x2 – 22x + 72 = 0
(x – 18)(x – 4) = 0
x = 18 or x = 4
The squares should be 4 cm by 4 cm to create a
maximum volume of 4 x 40 x 10 = 1600 cm3

12. The second derivative is negative; thus, the volume is a maximum.
𝒅𝑽 𝒅𝒙 = 𝟏𝟐𝒙𝟐 −𝟐𝟔𝟒𝒙 + 𝟖𝟔𝟒
𝒅 𝟐 𝑽 𝒅 𝒙 𝟐 =𝟐𝟒𝒙 −𝟐𝟔𝟒
𝒅 𝟐 𝑽 𝒅 𝒙 𝟐 =𝟐𝟒 𝟒 −𝟐𝟔𝟒=−𝟏𝟔𝟖
The second derivative is negative;
thus, the volume is a maximum.