Show that among all rectangles of a given perimeter square has maximum area
Answers
Let
P
be the fixed perimeter of a rectangle with length
x
and height
y
, so that
P
=
2
x
+
2
y
. The area is
A
=
x
y
. We can write this as a function of
x
by solving
P
=
2
x
+
2
y
for
y
and substituting:
P
=
2
x
+
2
y
⇒
2
y
=
P
−
2
x
⇒
y
=
P
2
−
x
⇒
A
=
f
(
x
)
=
x
(
P
2
−
x
)
=
(
P
2
)
x
−
x
2
(for
0
<
x
<
P
2
).
Don't be bothered by the fact that
P
is in this last equation. It's just a constant.
The derivative is
d
A
d
x
=
f
'
(
x
)
=
P
2
−
2
x
. This is zero when
x
=
P
4
and, in fact, changes sign from positive to negative as
x
increases through
x
=
P
4
. By the First Derivative Test, this implies that
x
=
P
4
gives a local maximum value of the area
A
=
f
(
x
)
. In fact, since
f
is quadratic, it's actually a global maximum.
When
x
=
P
4
, then
y
=
P
2
−
P
4
=
P
4
as well. This implies that the dimensions of the rectangle are all equal and it's actually a square. In other words, for all rectangles of a given perimeter
P
, the square of side length
P
4
has the greatest area