moriheru said:
Just as a matter of interest, the Lagrange multiplier solution is as follows. Let
$$L = \sum_{i=1}^7 (7-i) x_i + \lambda(1-\sum_{i=1}^7 x_i).$$
For the maximization problem with bound constraints ##0 \leq x_i \leq 1## for all ##i##, the (Karush-Kuhn-Tucker) optimality conditions are:
(1) If ##x_i = 0## then ##\partial L / \partial x_i \leq 0##.
(2) If ##x_i = 1## then ##\partial L / \partial x_i \geq 0## .
(3) If ##0 < x_i < 1## then ##\partial L / \partial x_i = 0##.
Let us guess the solution ##x_1 = 1## all other ##x_i = 0##.
$$\begin{array}{l}
\partial L/ \partial x_1 = 6 - \lambda \geq 0\\
\partial L / \partial x_2 = 5 - \lambda \leq 0 \\
\hspace{1cm} \vdots \\
\partial L / \partial x_6 = 1 - \lambda \leq 0 \\
\partial L / \partial x_7 = - \lambda \leq 0
\end{array}
$$
These conditions are satisfied by choosing any ##\lambda## between 5 and 6. Since the problem is a so-called "convex programming problem" the conditions are also sufficient to guarantee a global constrained maximum. In other words, the intuitive solution ##x = (1,0,0,0,0,0,0)## is provably optimal.
Much simpler, however, is to note that the constraints ##x_i \leq 1## are redundant because if ##\sum_i x_i = 1## and each ##x_i \geq 0## then each ##x_i## is automatically ##\leq 1##. Therefore, the original problem is equivalent to
$$\begin{array}{rcl}
\max Z &=& \sum_{i=1}^7 (7-i) x_i \\
\text{s.t.}&& \sum_{i=1}^7 x_i = 1 \\
&&x_i \geq 0, \; i = 1,2, \ldots, 7
\end{array}
$$
That makes the problem almost trivial: the simplex method takes just one step: use the equality constraint to write
$$x_1 = 1 - \sum_{i=2}^7 x_i ,$$
giving
$$Z = 6 - 1x_2 - 2 x_3 - 3 x_4 - 4 x_5 - 5 x_6 - 6 x_7.$$
We obtain ##Z = 6## if we set all ##x_2 = \cdots = x_7## equal to zero, but get ##Z < 6## if any of the variables ##x_2, \cdots, x_7## is ##> 0##. Therefore, the optimal solution is obtained as ##x = (1,0,,0,0,0,0).##