Page 290 - MATLAB an introduction with applications

Optimization ——— 275

>> x=A\b
x =
32.0000
–33.5000
9.5000
>> x=inv(A)*b
x =
32.0000
–33.5000
9.5000
a = 32
0
a = –33.5
1
a = 9.5
2
>> x=[1 2 3];y=[8 3 17]; a2a1a0=polyfit(x,y,2)
a2a1a0 =
9.50000000000001 –33.50000000000002 32.00000000000003
and the function is
F(X) = 32 – 33.5X + 9.5X 2

For finding minimum of the function, we find the first derivative of the function
dF () =− 33.5 19X = 0; for a minima or maxima
X
+
dX
Thus, X = 1.763
2
dF ()
X
To check for minima or maxima, we find second derivative of the function = 19 > 0; thus it’s a
dX 2
minimum.
The minimum value of F(X) at X is
F(X = 1.763) = 2.462

Example E5.4: Fit a polynomial by quadratic approximation and determine the values of X at which F(X) is
minimum.
X F(X)
1 –7
2 5
3 14

Solution:

X F (X)
1 –7
2 5
3 14