Section 4.5, continued:
Problem: Which rectangle with area 1 has smallest
perimeter?
Goal: Minimize
P
= 2
x
+ 2
y
subject to
xy
= 1
Write
y
in terms of
x
:
..?.
.
y
= 1/
x
P
(
x
,
y
) =
P
(
x
) = 2
x
+ 2/
x
(for
x
> 0)
Critical points:
P
′
= 2 – 2/
x
2
which is undefined up at
x
= 0 (which we’ve excluded from
the domain) and vanishes at …
..?.
.
x
2
= 1
⇒
x
= 1 (ignore extraneous root –1)
⇒
y
= 1/1 = 1
What kind of critical point?
P
′′
= 4/
x
3
= 4 > 0
⇒
local minimum (the only one)
0.5
1.0
1.5
2.0

2
0
2
4
6
8
10
So the best rectangle is a square.
Can we use the First Derivative test here?
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.
Yes.
2 – 2/
x
2
is negative for
x
< 1 (with
x
> 0) and positive
for
x
> 1, so the function
P
(
x
) has a local minimum at
x
= 1.
Here’s a geometrical way to see the link between the
problem “Maximize
A
(
x
,
y
) =
xy
subject to 2
x
+2
y
=
p
” and
“Minimize
P
(
x
,
y
) = 2
x
+2
y
subject to
xy
=
a
”.
The lines
below are the loci of 2
x
+2
y
=
p
with
p
= 2, 4, and 6, and the
curves are the loci of
xy
=
a
with
a
= 1/4, 1, and 9/4:
0.5
1.0
1.5
2.0
2.5
3.0
0.5
1.0
1.5
2.0
2.5
3.0
Maximizing
xy
with 2
x
+2
y
fixed is asking “Among all the
given curves, what’s the highest one that crosses a
particular line?”, and the answer is, it’s the one that’s
tangent to the line.
Minimizing 2
x
+2
y
with
xy
fixed is asking “Among all the
given lines, what’s the lowest one that crosses a particular
curve?”, and the answer is, it’s the one that’s tangent to the
curve.
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 Fall '11
 Staff
 Calculus, Critical Point, Optimization

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