Page 318 - MATLAB an introduction with applications

Optimization ——— 303

2
2 2
Example E5.21: Minimize f (x , x ) = 90 (x –x ) + (1 – x ) using the starting point
1
1
2
1
2
x = –1.1, x = 1.0
1
2
Solution:
The minimum = 0 at x = 1, y = 1
EDU>> Run_EN_5_21
x =
1.0000 1.0000
fval =
7.2476e–010
Run_EN_5_21
y=@(x)90*(x(2)–x(1)^2)^2+(1–x(1))^2;
x,fval]=fminsearch(y,[–1.1,1.0])

2
–1
Example E5.22: Minimize f = 0.60 – 0.70/(1. + x ) – 0.5*tan (1/x)
Solution:
The minimum value of the function is f = – 0.8286 at x = 0.1001

EDU>> Run_EN_5_22
f =
@(x)0.60–0.70/(1. + x.^2)–0.5*atan(1/x)
x =
0.1001
f =z
–0.8286
Run_EN_5_22
f = @(x)0.60 – 0.70/(1. + x.^2)– 0.5*atan(1/x)
x = fminbnd(f, .1,1.0)
f = 0.60 – 0.70/(1. + x.^2) – 0.5*atan(1/x)

Example E5.23: Use the MATLAB fminbnd function to find the maximum of
2
y = (3*(sin(x)) – (x /12)
Solution:
The graph of this function is given below. By inspection, the maximum value of the function is close to 2.8
and occurs at approximately x = 1.5.