For Example, 3, show that the optimal value of x is x^∗ = 1.

Data from Example 3

and we will search for an optimal value of c in the closed interval [0, 3]. We choose a tolerance t = 0.2. We apply the Golden Section Search Method until an interval of uncertainty is less than 0.2. The results are summarized in Table 7.5.

Table 7.5

 The length of the final interval is less than 0.2. We can estimate c^∗ = (0.8115 + 0.9787) / 2 = 0.8951, with f (0.8951) = 2.5804. In the problem set, we ask you to show analytically that the optimal value for c is c = 8/ 9 ≈ 0.8889.