SME
Grade 10 Math · Problem type
Designing a can with the least material
Use this for “A can must hold 500 cm³. Which radius uses the least metal?”
Table
Radius (r)
(cm)
Surface area (S)
(cm²)
S against r, worked out for every r from 1 to 7 cm
The least S ≈ 348.7 cm² comes at r ≈ 4.301 cm
This is an example. Type your own values to start a new problem.
V
Volume
example
r
Radius
example
h
Height
calculated
S
Surface area
calculated
rbest
Radius with the least metal
worked out (not typed)
Sₘᵢₙ
Least surface area
worked out (not typed)
Formulas
V
=
π
×
r
2
×
h
500
=
π
×
4
2
×
9.9472
rbest
=
∛(
V
2 × π
)
4.3013
=
∛(
500
2 × π
)
Sₘᵢₙ
=
2
×
π
×
rbest
2
+
2
×
V
rbest
348.7342
=
2
×
π
×
4.3013
2
+
2
×
500
4.3013
S
=
2
×
π
×
r
2
+
2
×
π
×
r
×
h
350.531
=
2
×
π
×
4
2
+
2
×
π
×
4
×
9.9472
Assumptions
•
The can is a closed cylinder: V = πr²h, and its metal is the surface S = 2πr² + 2πrh.
•
Hold V and try radii: the table and its graph show S dip and then rise again.
•
The least metal comes when the height equals the diameter, h = 2r, so V = 2πr³.
Step-by-step
WE KNOW
V = 500 cm³, r = 4 cm
FIND
height (h), radius with the least metal (rbest), least surface area (Sₘᵢₙ), surface area (S)
Step 1 · Find height (h)
Use V = π × r² × h
Divide by the base’s area.
h
=
V
π ×
r
2
h
=
500
π ×
4
2
h
=
500
16 × π
h
=
500
50.27
h
=
9.9472
cm
Step 2 · Find radius with the least metal (rbest)
Use rbest = ∛(V ÷ (2 × π))
At
the
best
can
h
=
2
r
,
so
V
=
πr²
×
2
r
=
2πr³:
divide
by
2π
and
take
the
cube
root.
rbest
=
∛(
500
2 × π
)
rbest = ∛79.58
rbest = 4.3013 cm
Step 3 · Find least surface area (Sₘᵢₙ)
Use Sₘᵢₙ = 2 × π × rbest² + 2 × V ÷ rbest
The
side
2πrh
is
2
V
r
,
because
h
=
V
πr²
:
put
in
the
best
radius.
Sₘᵢₙ
=
2
×
π
×
4.3013
2
+
2
×
500
4.3013
Sₘᵢₙ
=
2
×
18.5
×
π
+
2
×
500
4.3013
Sₘᵢₙ
=
37
×
π
+
1,000
4.3013
Sₘᵢₙ = 116.2 + 232.5
Sₘᵢₙ = 348.7342 cm²
Step 4 · Find surface area (S)
Use S = 2 × π × r² + 2 × π × r × h
Two
circles
for
the
top
and
bottom,
and
the
side.
The
side
2πrh
is
2
V
r
,
since
h
=
V
πr²
.
S
=
2
×
π
×
r
2
+
2
×
500
r
S
=
2
×
π
×
4
2
+
2
×
500
4
S
=
32π
+
250
S
=
350.531
cm²
Check
Put the numbers back in. Both sides should match.
500
≈
π
×
4
2
×
9.9472
✓
4.3013
≈
∛(
500
2 × π
)
✓
348.7342
≈
2
×
π
×
4.3013
2
+
2
×
500
4.3013
✓
350.531
≈
2
×
π
×
4
2
+
2
×
π
×
4
×
9.9472
✓
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