geometry
geometry
Maximizing Volume
MAP4C: Foundations of College Mathematics
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A square-based prism must have a surface area of 96 cm2 .
What are the dimensions of the prism that produce the
maximum volume, and what is the volume?
Optimizing Volume and Surface Area
Part 1: Square-Based Prisms
J. Garvin
J. Garvin — Optimizing Volume and Surface Area
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geometry
geometry
Optimizing Volume and Surface Area
Optimizing Volume and Surface Area
The surfaces of a square-based prism are two congruent
squares (A = s 2 ) and four congruent rectangles (A = sh).
The volume of the square-based prism is V = s 2 h.
Substitute the previous expression for h into this equation.
SA − 2s 2
V = s2
4s
The total surface area, then, is SA = 2s 2 + 4sh.
This can be rearranged to isolate h:
SA − 2s 2 = 4sh
Substitute the surface area of 96 cm2 into the equation.
96 − 2s 2
V =s
4
96s − 2s 3
V =
4
V = 24s − 12 s 3
SA − 2s 2
=h
4s
J. Garvin — Optimizing Volume and Surface Area
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J. Garvin — Optimizing Volume and Surface Area
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geometry
geometry
Optimizing Volume and Surface Area
Optimizing Volume and Surface Area
Use technology to determine the maximum volume, and the
side length of the base.
Dimensions of a Square-Based Prism Given a Specific
Surface Area
The maximum volume for a square-based prism occurs when
the prism is a cube. The side length is s, and the maximum
volumecan be determined
by finding the maximum value of
SA − 2s 2
V =s
for some surface area, SA.
4
This implies another theorem about the maximum volume.
The maximum volume of 64 cm3 occurs when the side length
of the base is 4 cm (the value above is slightly off due to
hardware limitations).
Minimum Surface Area of a Square-Based Prism Given a
Specific Volume
J. Garvin — Optimizing Volume and Surface Area
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J. Garvin — Optimizing Volume and Surface Area
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The minimum surface area of a square-based prism
√ occurs
when the prism is a cube with side length s = 3 V for some
volume, V . The surface area is SA = 6s 2 .
geometry
geometry
Optimizing Volume and Surface Area
Optimizing Volume and Surface Area
Example
What if the prism with a surface area of 96 cm2 had no top?
How are the dimensions affected?
A square-based prism has a volume of 50
minimum surface area.
in3 .
Determine the
The total surface area, then, is SA = s 2 + 4sh.
The minimum surface√area occurs when the prism is a cube
.
with side length s = 3 50 = 3.684 in.
.
.
The surface area is SA = 6 × 3.6842 = 81.43 in2 .
J. Garvin — Optimizing Volume and Surface Area
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This can be rearranged to isolate h:
SA − s 2 = 4sh
SA − s 2
=h
4s
J. Garvin — Optimizing Volume and Surface Area
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geometry
Optimizing Volume and Surface Area
The volume of the square-based prism is V =
geometry
Optimizing Volume and Surface Area
s 2 h.
Use technology to determine the maximum volume, and the
side length of the base.
Substitute the previous expression for h into this equation.
SA − s 2
V = s2
4s
Substitute the surface area of 96 cm2 into the equation.
96 − s 2
V =s
4
96s − s 3
V =
4
V = 24s − 14 s 3
The maximum volume of approximately 90.51 cm3 occurs
when the side length of the base is approximately 5.657 cm.
J. Garvin — Optimizing Volume and Surface Area
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J. Garvin — Optimizing Volume and Surface Area
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geometry
geometry
Optimizing Volume and Surface Area
Optimizing Volume and Surface Area
The height of the prism can be determined by working
backward using the formula for volume.
Dimensions of a Square-Based Prism With No Top Given
a Specific Surface Area
V = s 2h
V
h= 2
s
. 90.51
h=
5.6572
.
h = 2.828
The height of the prism is approximately 2.828 cm.
The side length of the base, 5.657 cm, is twice this height.
J. Garvin — Optimizing Volume and Surface Area
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The maximum volume for a square-based prism with no top
occurs when the side length of the base is twice its height.
The side length is s, and the maximum volume can be
determined
the maximum value of
by finding
SA − s 2
V =s
.
4
And its corresponsing theorem. . .
Minimum Surface Area of a Square-Based Prism With
No Top Given a Specific Volume
The minimum surface area of a square-based prism with no
top occurs when the side length of the base is twice its
height.
J. Garvin — Optimizing Volume and Surface Area
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geometry
Questions?
J. Garvin — Optimizing Volume and Surface Area
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