SOLUTION: You have exactly 36 units of red dye and 28 units of blue dye. Your boss also wants you to make as many gallons of unique color "c" and unique color "d" as possible. Each gallon of

Let X and Y be the numbers of gallons of liquid C and D, respectively.


Then the objective function is  F(X,Y) = X + Y.


The constraint inequalities are

1*x + 6*Y <= 36     (1)  (red dye constraint)
2*X + 1*Y <= 28     (2)  (blue dye constraint)

X >= 0,  Y >= 0.


The feasibility region is shown in the figure below.

It is the quadrilateral in QI, adjacent to x- and y- axes and constrained by the red and the green lines.



  


Plots 1X + 6Y = 36 (red)  and  2X + Y = 28 (green)


The feasibility quadrilateral has the corner  

    P1 = (0,6)      (red line y-intercept);

    P2 = (12,4)     (intersection point of the red and green lines);  and

    P3 = (14,0)     (green line x-intercept).


The objective function has the values

     at P1:  F(X,Y) = 0 + 6 = 6;

     at P2:  F(X,Y) = 12 + 4 = 16;   and

     at P3:  F(X,y) = 14 + 0 = 14.


According to the Linear programming method, it means that the point P2 gives the solution to the given linear minimax problem:


    The most amount of gallons you can create of both color C and color D is 16.

    It happens when 12 gallons of the liquid C and 4 gallons of the liquid D is produced.