Solution
= Solution
Put $y_i=1-x_i$. Feasibility gives $y_i\geq0$ and
$$
y_1+y_2+y_3\geq\frac12,
$$
while the objective is
$$
x_1+2x_2+4x_3=7-(y_1+2y_2+4y_3).
$$
To maximize it, the required total deficit $1/2$ must be assigned entirely to the cheapest variable $y_1$. Thus
$$
y_1=\frac12,\qquad y_2=y_3=0,
$$
and the unique optimizer is
$$
\boxed{x^*=\left(\frac12,1,1\right)},
\qquad
\boxed{x_1^*+2x_2^*+4x_3^*=\frac{13}{2}}.
$$
This is the vertex $A$ from part (b).