Question 1151913
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            First,  I will write / (copy) all constraints from the post by @Theo.


            Then I will show that the problem can be solved analytically in a very simple manner and without using any tools from the outside.



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x = liters of A;  y = liters of B.

The objective function is r = 1200x + 2000y.

r stands for revenue.

This is what you want to maximize.

The constraint functions are:

3x + 1.8y >= 5400.                         (1)

2x + 3y <= 9600.                           (2)

x >= 1000                                  (3)

x >= 400                                   (4)

x + y <= 4000                              (5)

x + y >= 2700 (equals .675 * 4000)         (6)

y = 2x                                     (7)

x >= 0,  y >= 0                            (8)



I start from constraint (1).  I substitute there y = 2x  from (7) to get

    3x + 1.8y >= 5400  --->  3x + 1.8*(2x) >= 5400  --->  6.6x >= 5400  --->  x >= {{{5400/6.6}}} = 818.18       (9)



Similarly, I treat constraint (2).

    2x + 3y <= 9600  --->  2x + 3*(2x) <= 9600  --->  8x <= 9600  --->  x <= 1200      (10)



Similarly, I treat constraint (5).

    x + y <= 4000  --->  x + 2x <= 4000  --->  3x <= 4000  --->  x <= {{{4000/3}}} = 1333.33       (11)



Similarly, I treat constraint (6).

    x + y >= 2700  --->  x + 2x >= 2700  --->  3x >= 2700  --->  x >= {{{2700/3}}} = 900       (12)


Thus from (3), (9) - (12)  I have finally  1000 <= x <= 1200.      (13)


With the constraint (7), the revenue function is  r = 1200x + 2000y = 1200x + 2000*(2x) = 1200x + 4000x = 5200x.


So, I want to find maximum value of the revenue function  r = 5200x  at the interval  1000 <= x <= 1200.


It is absolutely clear that the maximum is at the endpoint x= 1200  and the maximum value is then r = 5200*1200 = 6240000.


<U>ANSWER</U>.  The optimum solution is  x= 1200 liters brand A;  y= 2x = 2400 liters of brand B  and the maximum revenue is 6240000 PhP.
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The solution is completed, &nbsp;<U>with all imposed constraints</U> (!)



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The lesson to learn from my solution is <U>THIS</U> :


<pre>
    The constraint (7) reduces the problem from 2D (two variable on the plane) to 1D (one variable ONLY (!) )
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