Let x be the amount of candy A, in kilograms;

let y be the amount of candy B.


The revenue function (the subject to maximize) is  R(x,y) = 285x + 260y.    (1)


The constraint inequalities are

     +  <= 360     (2)    (dark chocolate)

     <= 430              (3)    (caramel)

     <= 210              (4)    (walnuts)

    x >= 0,  y >= 0.


Constraints can be rewritten equivalently

    3x + 4y <= 2160            (2')

     x      <=  860            (3')

     y      <=  630            (4')


The feasibility domain is shown in the plot below.



    


    Figure.  Plots  3x + 4y = 2160 (red);  x = 860 (blue);  y = 630 (green)



The feasibility domain is the triangle in QI, adjacent to x- and y-axis below the red line.


It is clear from the plot, that the real working constraint is (2').

The constraints (3') and (4') do not work.


In other words, the only working constraint is (2) for dark chocolate.

The other constraints (3) and (4) do not work.


     +------------------------------------------------------------------------------+
     |    It means, that the solution and the strategy in this problem is to use    |
     |    as much dark chocolate as possible (as constraint (2') or (2) allow it).  |
     +------------------------------------------------------------------------------+


Now, to get the solution in x and y variables, you have a linear function of the revenue (1)
over the triangle, which is the feasibility domain.


As the Linear Programming method says (and it is obvious from the common sense of view),
the solution is one of the two vertices of the triangle, 

    (x,y) = (,) = (0,540)  or  (x,y) = (,) = (720,0).


So, you calculate the revenue at these two points

    R(0,540) = 285*0   + 260*540 = 140400

    R(720,0) = 285*720 + 260*0   = 205200.


Comparing it, you see that the optimum solution is  to produce and to sell 720 kilograms of candy A  and  0 kg of candy B,
which gives the maximum revenue of P205200.    ANSWER