This problem came up when I was trying to solve a bigger, probabilistic problem. So at the end it boils down to this: how can we maximize the function
$f(x_2,x_3,x_4) = \frac{18}{100}x_2 + \frac{42}{100}x_3 + \frac{35}{100}x_4$
subject to:
$x_2 + x_3 + x_4 = C \gt 0$
$x_2 \gt 0$
$x_3 \gt 0$
$x_4 \gt 0$
I need some explanation/reference for this particular problem (I hope to understand this one now), and also some general reference for this type of problems (maybe I will need some more time to understand that part). Many thanks in advance.
According to the constraints and objective function, it seems that the objective function can get the maximum value on the boundary of feasible solutions, and since it is an open set, I think the objective function does not have any maximum in this case.
Now, let replace the last three constraints as given below.
$x_2\geq 0$, $x_3\geq 0$, and $x_4\geq 0$
Since the coefficients in the objective functions are positive and the sum of variables are constant, so we just need to find the maximum coefficient in the objective function and set it as $C$ and put the others $0$.
$$\max\{\frac{18}{100},\ \frac{42}{100},\ \frac{35}{100}\}=\frac{42}{100}$$
So, the maximum value can be obtained setting $x_2=x_4=0$ and $x_3=C$ which is $\frac{42}{100}C$.