158
Chapter 3
EXAMPLE 3.6. Find the minimum surface area of an open tank
with volume no less than 1 unit. The radius is, r, and the height is h.
(a) The criterion function is the area of the tank
go(X) = 2 + 2rtrh
(3.33)
(b) The functional constraint is that of the volume
g~(x) = ~zr2h > 1
(3.34)
The constraint is rewritten
gj(x) is ~tr2h- 1 > 0
(3.35)
This is done to comply with the format the computer accepts as
input. This must be studied carefully. The answer for r and h
are shown in Example 3.11. Always try a problem with known
answers to check a new or questionable computer routine.
Another example sets up the equations for a nonlinear optimization problem.
(c) There is a regional constraint in here as the shape of the tank is
round
EXAMPLE 3.7. A steel spherical tank holds 250 gallons and is
fabricated in two hemispheres and welded to two flanges which are bolted
together. The steel for the two halves (neglecting the flanges) is 0.50 dollars
per cubic inch and the weld cost is 1.50 dollars/t around the two flanges.
The allowable stress is 15 Kpsi for a thin wall analysis.
(a) Criterion function
(b)
Cost = material cost + welding cost/2 flanges
= 0.50(4~RZt) + ~-~ (2~R)
go = 2~zR2t + 3~R
t
Functional constraint 1
4~R 3
(231 in3’~
Volume is T >- 250 gallon
\~/
47zR 3
1
>I
gJ= 3 57,750in
3(3.36)
(3.37)