Figure 3.2-3 shows the cross section of an irrigation channel. A preliminary analysis has shown that the cross-sectional area of the channel should be 100ft2 to,carry the desired water-flow rate. To minimize the cost of concrete used to line’ the channel,we want to
minimize the length of the channel’s perimeter. Find the values of d, b, ‘and ()that minimize
this length .
• Solution
The perimeter length L can be written in terms of the base b, depth d, and angle () as
follows:
2d
L=b+- sine
The area.of the trapezoidal cross section is
d2
l00=db+ -tan ()
The variables to be selected are b, d, and (). We can reduce the number of variables by solving the latter equation for b to obtain b =.!. (100-~) d tan ()
Substitute this expression into the equation for L. The result is
100 d 2d L= + d tan e sin B
,We must now find the values of d and () to minimize L.

First define the function file for the perimeter length. Let the vector x be [d OJ.
function L = channel(x)
L = 100./x(1) – x(l) ./tan(x(2)) + 2*X”(1~~.~~(X(2));
Then use the fmin search function. Using a guess of d = 20 and 0 = I rad, the
session is
7.5984 1.0472
»x = fmins(‘channel’, [20,1))
x
Thus the minimum perimeter length is obtained with d = 7.5984 ft and 0 = 1.0472 rad, or 0 = 60°. Using a different guess, d = 1,0 = 0.1, produces the same answer. The value of the base b corresponding to these values is b = 8.7738. .I However, using the guess d = 20,0 = 0.1 produces the physically meaningless result
d = -781,0 = 3.1416. The guess d = 1,0 = 1.5 produces the physically meaningless. result d = 3.6058,0 = -3.1416.
The equation for L is a function of the two variables d and 0, and it forms a surface when L is plotted versus d and 0 on a three-dimensional coordinate system. This ‘surface might have multiple peaks, multiple valleys, and “mountain passes” called saddle points that can fool a minimization technique. Different initial guesses fo;the
solution vector can cause the minimization technique to find different valleys and thus report different results. We can use the surface-plotting functions covered in Chapter 5 to’ look for multiple valleys, or we can use a.large number of initial values for d and 0, say, over the physically realistic ranges 0 < d < 30 and 0 < 0 < n/2. If all the physically meaningful answers are identical, then we can be reasonably sure that we have found
the minimum.