Engineering Optimization: Theory and Practice, Fourth Edition

12.2 Calculus of Variations 677

By substituting Eq. (E 9 ) in (E 7 ), the variational problem can be restated as

Findy(x)which maximizes

H= 2 h

∫L

0

t (x) dx (E 10 )

subject to the constraint

g(x, t, t′)= 2 ρ

h

k

∫L

0

1

dt/dx

[∫L

x

t (x) dx

]

dx+m= 0 (E 11 )

This problem can be solved by using the Lagrange multiplier method. The functional
Ito be extremized is given by

I=

∫L

0

(H +λg) dx= 2 h

∫L

0

[

t(x)+

λρ

k

1

dt/dx

∫L

x

t (x) dx

]

dx (E 12 )

whereλis the Lagrange multiplier.
By comparing Eq. (E 12 ) with Eq. (12.1) we find that

F(x, t, t′) = 2 ht+

2 hλρ

k

1

t′

∫ L

x

t (x) dx (E 13 )

The Euler–Lagrange equation, Eq. (12.10), gives

h−

λhρ

k

[

2 t′′

(t′)^3

∫ L

x

t (x) dx+

t(x)

(t′)^2

−

∫x

0

dx

t′

]

= 0 (E 14 )

This integrodifferential equation has to be solved to find the solutiont (x). In this case
we can verify that

t (x)= 1 −x

(

λρ

k

) 1 / 2

(E 15 )

satisfies Eq. (E 14 ). The thickness profile of the fin can be obtained from Eq. (E 9 ) as

y(x)= −

h

k

1

t′

∫L

x

t (x) dx=

h

k

(

k

λρ

) 1 / 2 ∫L

x

[

1 −

(

λρ

k

) 1 / 2

x

]

dx

=

h

(kλρ)^1 /^2

[

L−

(

λρ

k

) 1 / 2

L^2

2

−x+

(

λρ

k

) 1 / 2

x^2

2

]

=c 1 +c 2 x+c 3 x^2 (E 16 )

where

c 1 =

h

(kλρ)^1 /^2

[

L−

(

λρ

k

) 1 / 2

L^2

2

]

(E 17 )

c 2 = −

h

(kλρ)^1 /^2

(E 18 )

c 3 =

h

2 (kλρ)^1 /^2

(

λρ

k

) 1 / 2

=

h

2 k

(E 19 )