2.6 Optimizing Volume and Surface Area Without technology
Practice (page 110-111)
For all rectangular prisms with given/same surface area – cube has max. volume (s3)
For all rectangular prisms with given/same volume – cube has minimum surface area (6s2).
1. Rectangular Prism of SA=96cm2
2. . Rectangular Prism of V=729 cu. In.
with greatest volume:
a) What are it’s dimensions?
SA = 6s2
96 = 6s2
16 = s2
with least S.A.:
a) Dimensions are
V = s3
729 = s3
s = 16
s=4
The length of each side is 4 cm.
The rectangular prism is a cube with edge length 4
cm.
s=
3
729
=9
The length of each side is 9 in.
The rectangular prism is a cube with edge length 9
in.
b) What will be the volume?
b) Surface area is: SA = 6s2
3
V=s
V = (4) 3
V = 64
The volume would be 64 cm3.
= 6(9)2
= 486
Its surface area would be 486 cm2.
3. Two prisms in a) and b) have Volume = 240 cubic cm. Which has less S.A.?
a) SAA = 2(lw + wh + lh)
b) SAB = 2(lw + wh + lh)
= 2((10)(6)+(6)(4)+(10)(4))
= 2((12)(10)+(10)(2)+(12)(2))
= 248
= 328
The surface area would be 248 cm2.
The surface area would be 328 cm2.
4 cm
The rectangular prism from part a had the least surface
area.
2 cm
6 cm
10 cm
10 cm
12 cm
.
4. For each surface area what dimensions give maximum volume?
a) 600 sq. in.
c) 2400 sq. in.
SA = 6s²
600 = 6s²
100 = s²
SA = 6s²
2400 = 6s²
400 = s²
s = 100
s = 10
The rectangular prism is a cube with edge
length 10 in.
s = 400
s = 20
The rectangular prism is a cube with edge
length 20 in.
5. For each volume what dimensions give minimum surface area?
a) 1 m³
c) 8 cubic feet
V = s³
V = s3
1 = s³
s=
3
8 = s³
3
1
s=
=1
8
=2
The rectangular prism is a cube with edge
The rectangular prism is a cube with edge
length 1 m.
length 2 ft.
6. 6s² = 844 cm²
a) s² = 844 cm² / 6 = 144cm, s=12 cm so max. volume is 12³ = 1728 cm³
b) The rectangular prism would have the shape of a cube with side length 12 cm.
7. a) 1 gal  277.42 cu. in.
5 gal = 5  277.42
 1387.1 cu. in.
b) A cube will produce the smallest surface area.
V = s3
1387.1 = s3
s=
3
1387.1
 11.152
The container would be a cube with edge length of about 11.15
in.
SA = 6s2
= 6(11.152)2
= 746.203
Its surface area will be about 746.20 sq. in.
8. a) A cube will use the least amount of material.
3
V=s
274 625 = s3
s=
3
274625
= 65
The container would be a cube with edge
length 65 cm.
b) SA = 6s2
= 6(65)2
= 25 350
The box will need 25 350 cm2 of sheet metal.
c) 1 ft  30.48 cm
1 sq. ft. = (30.48) 2
 929.03 cm2
20 sq. ft. = 20  929.03
 18 580.5
The box needs 25 350 cm2 of sheet metal so there will not be enough sheet metal.
10. The shape should be as close to a cube as possible.
V = s3
144 = s3
s = 3 144
= 5.24
The values must be whole numbers, the closest to a cube has dimensions
6 by 6 by 4.
2.6 Optimizing Volume and Surface Area
part 2 constraints and technology
Practice (page 110-111)
Download the emulator from http://teacherweb.com/ON/Westminster/SRShivers/photo3.stm
For all rectangular prisms with given/same surface area – cube has max. volume (s3)
For all rectangular prisms with given/same volume – cube has minimum surface area (6s2).
9. Tori’s hanging shelf: V=400 cu. In., depth must be 4 in., what are minimum dimensions and surface area.
400 = 4x² (closest to cube if l=w)
x² = 100, x = 10
Dimensions are 4 x 10 x 10
SA = 2 sides + top and bottom
= 2 ( 4 x 10 ) + 2 (4 x 10)
= 160 sq. in.
12. USE TI-83
750 ml = 750 cm³
min SA and dimensions can be found using 2 equations / finding minimum of SA curve
i)
Volume of cylinder
750 = πr²h
h=750/ πr²
let Y1 = height, X = radius
Y1= 750/ πX²
ii) SA = 2 πr² + 2 πrh
let Y2 be S.A.
Y2 = 2πX² + 2 πXY1
window
Xmin=0
Xmax=10
Xscl=1
Ymin=-10
Ymax=750
Yscl=50
to get Y1 use keys: (vars)(>)(1)(1)
Graph the equations, set window, use 2nd trace to find minimum of Y2 curve minY2=min SA,
and minimum X = min radius.
From Y2 curve: Min SA = 456.97 sq. cm At radius = 4.93cm
And from Y1 curve at h = 9.85 cm.
13. similar to 12 except volume 1800 cu. Ft
h=1800/ πr²
let Y1 = height, X = radius
Y1= 1800/(πX²)
window
Xmin=0
Xmax=20
Xscl=1
Ymin=-10
Ymax=2000
Yscl=100
SA = 2 πr² + 2 πrh
let Y2 be S.A.
Y2 = πX² + 2 πXY1
From Y2 curve: Min SA = 819.1 sq.ft At radius = 6.6 ft And from Y1 curve at h = 13.2 ft
14.
height is approx. equal to diameter for optimal cylinders – maximize volume minimizing
surface area
15.
Rect. Prism vs Cylinder
R. prism
512 = s³
s = 8 cm
SA = 6(8)² = 384 cm²
Cylinder
512 = πr²h
use TI83
Y1= 512/ (πX²)
Y2 = 2πX² + 2 πXY1
Minimum Y2 = 354.3 cm²
16.
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17.
a) In an equilateral triangle, A = 60°, h = height;
tan A = h/(s/2)
tan 60
=
2h/s
tan 60 = 3
=
2h/s
3s
h
=
2
1
Area of triangle = 2 by base by height
1
1
3s
3s2
= by s by h = × s by
=
2
2
2
4
s
h
s
s
b) From the table, the side length of the base are 10.0 cm, and the height is 5.8 cm.
c) From the table, the surface area is 259.8 cm3.
d) As the side length increased, the height decreased by roughly the same amount.
18.
a)
V = πr²h = 355 cm 3
X 12 cans = 12πr²h = 4260 cm3
window
Xmin=0
Xmax=10
Xscl=1
Ymin=-10
Ymax=750
Yscl=50
h = 4260 / 12πr² = 355 / πr²
Y1 = h, X=r
Y1 = 355/ πX²
Y2 = 2πX² + 2 πXY1
to get Y1 use keys: (vars)(>)(1)(1)
Graph the equations, set window, use 2nd trace to find minimum of Y2 curve minY2=min SA,
and minimum X = min radius.
From Y2 curve: Min SA of each can is 277.5 sq. cm
At radius = 3.84cm and from Y1 curve at h = 7.67 cm.
b) The diameter of the cans is 7.68 cm
In terms of cans, the closest to a square for 12 cans would be an arrangement of 3 layers of 2 rows
of 2 cans each.
The dimensions of the box would be:
l = 2  7.68 = 15.36 cm
w = 2  7.68 = 15.36 cm
h = 3  7.67 = 23.01 cm
SA = 2(lw + wh + lh)
= 2((15.4)(15.4)+(15.4)(23)+(15.4)(23))
= approx. 1891.1
The minimum amount of cardboard for the box is about 1891.1 cm2.
19. skip