SME
Grade 10 Math · Problem type
The most area for a fixed perimeter
Use this for “40 m of fence makes a rectangular pen. Which sides give the most area?”
Table
Length (x)
(m)
Area (A)
(m²)
A against x, worked out for every x from 2 to 14 m
The most A ≈ 100 m² comes at x ≈ 10 m
This is an example. Type your own values to start a new problem.
P
Perimeter
example
x
Length
example
y
Width
calculated
A
Area
calculated
s
Side of the best pen
worked out (not typed)
Aₘₐₓ
Most area
worked out (not typed)
Formulas
s
=
P
4
10
=
40
4
Aₘₐₓ
=
s
2
100
=
10
2
P
=
2
×
x
+
2
×
y
40 = 2 × 12 + 2 × 8
A
=
x
×
y
96 = 12 × 8
Assumptions
•
The fence is the perimeter: P = 2x + 2y, so x + y is half of P.
•
The pen’s area is A = xy; a longer side means a shorter one.
•
For a fixed perimeter the square, x = y = P ÷ 4, holds the most area.
Step-by-step
WE KNOW
P = 40 m, x = 12 m
FIND
side of the best pen (s), most area (Aₘₐₓ), width (y), area (A)
Step 1 · Find side of the best pen (s)
Use s = P ÷ 4
The best pen is a square: the fence makes four equal sides.
s
=
40
4
s
=
10
m
Step 2 · Find most area (Aₘₐₓ)
Use Aₘₐₓ = s²
The
square’
s
area
is
its
side
times
itself;
any
other
sides
give
less.
Aₘₐₓ
=
10
2
Aₘₐₓ = 100 m²
Step 3 · Find width (y)
Use P = 2 × x + 2 × y
Half the fence is one length and one width.
y
=
P
2
−
x
y
=
40
2
−
12
y
=
20
−
12
y
=
8
m
Step 4 · Find area (A)
Use A = x × y
The area of a rectangle is length × width.
A
=
12
×
8
A
=
96
m²
Check
Put the numbers back in. Both sides should match.
10
=
40
4
✓
100
=
10
2
✓
40 = 2 × 12 + 2 × 8 ✓
96 = 12 × 8 ✓
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