
Since the conditional optimization of 1, 2, 3 and 4th steps already performed in Table. 3, then left the table to finish the implementation of constrained optimization on the 0th step, and only for g_i = 200. The above calculations are made in the Table. 5. Optimization of 0th step during unconstrained optimization we Zmax = Z0 * (200) = 27. The maximum is reached at two optimal controls x0 = 40 and O = £ 80. Having taken the first option, we consistently £ 0 * = li *x0 * = 20040 = 160 and * i * = 40 (from Table. 2). Furthermore, gi * = 16040 = 120, of Table. 2, we get x2 = 80 (or 40). Again find £ 2 = 12080 = 40 (or% = 12040 = 80) and respectively x3 = 0 (or x 3 = 40). Finally,  3 = 400 = 40 (or 8040 = 40) and x4 = 40. Thus, we get two alternative optimal control f / i * = (40, 40, 80, 0, 40) and t / 2 * == (40, 40, 40, 40, 40). Similarly, selecting the 0th step of £ 0 = 80, we get a third alternative optimal control t / 3 = (80, 40, 40, 0, 40). 