Question 1184212
make a table such as the one below:


<pre>

product name           R80           FS20
number of units         x              y        x >= 0, y >= 0, x <= 12
plastic                 6              2        6x + 2y <= 140
metal                   2              4        2x + 4y <= 200
profit                  5              7        maximize 5x + 7y

</pre>


objective function is profit = 5x + 7y
this is what you want to maximize.


constraint functions:
x >= 0
y >= 0
x <= 12
6x + 2y <= 140
2x + 4y <= 200


using the <a href = "https://www.desmos.com/calculator" target = "_blank">desmos.com calculator</a>, you graph the opposite of the constraint functions and evaluate each corner point of the feasible region with the objective function.


the feasible region is the area of the graph that is not shaded.


here's the graph.


<img src = "http://theo.x10hosting.com/2021/091304.jpg" >


the corner points of the feasible region and the value of the objective function at those corner points are:


(0,50) = 5x + 7y becomes 5*0 + 7*50 = 350
(8,46) = 5x + 7y becomes 5*8 + 46*7 = 362 ***** maximum profit is here.
(12,34) = 5x + 7y becomes 5*12 + 7*34 = 298
(12,0) = 5x + 7y becomes 5*12 + 7*0 = 60


the maximum profit is at (8,46), where it is equal to 362.
(8,46) means the value of x is 8 and the value of y is 46.


the constraint functions must all be true at this point.

x <= 12 is true.
x >= 0, y >= 0 is true.
6x + 2y becomes 6*8 + 2*46 = 140 which is <= 140, so this is true.
2x + 4y becomes 2*8 + 4*46 = 200 which is <= 200, so this is true.


all constraints are true at the maximum profit point of (8,46).
this confirms the solution at (8,46) is valid.