MATHEMATICS AND ORIGAMI

Mathematics and Origami

7.6 MAXIMA AND MINIMUMS

The value of x that will make z’ = 0 is the minimum for AB we are trying to find. It is

obvious that the maximum is ∞ (what happens when O is folded over D).

O

1

2

Y

1 X

Y

O

1

2

X

Y

2 3 O X

F P

D

P

O

F

Y

AED X

B

2 2

2 2

2

x (a x)

ax

y

− −

= Then we seek the minimum of 2

2 2

2

2 ax a

ax

x

−

+

making

2

1

2

2

1

2

























−

= +

x

a

x

z x and deriving, we have:









































−

 −











−

+

























−

= +

−

2

(^22)
1
2
2
1
2
2
12
2
2
1
2 2
1
'
x
a
a
x x x
a
x
x
a
x
z x
a
Y
A
y
O
X
x B D
C
In a rectangle of width a, to fold O over the edge CD in such a way that, be-
ing B within OD, the length of fold AB will be minimum. If
y=OA ; x=OB
we must find an x value such that it will make minimum the expres-
sion x^2 +y^2
∆AOB and ∆COD are similar:
CD
a
x
y
= ; in ∆CBD: CD= x^2 −()a−x^2