Anyone know anything about this type of problem?

Question

A manufacturer produces two models of elliptical cross-training exercise machines. The times for assembling, finishing, and packaging model X are 3 hours, 3 hours, and 0.8 hour, respectively. The times for model Y are 4 hours, 2.5 hours, and 0.4 hour. The total times available for assembling, finishing, and packaging are 6300 hours, 4500 hours, and 1080 hours, respectively. The profits per unit are $300 for model X and $375 for model Y. What is the optimal production level for each model? What is the optimal profit?

Answer 1

Recognize that this is a Linear Programming problem. We are 

attempting to model optimal values (production and profit). The
keyword "optimal" tells that we will be using inequalities.

Let....
"x" and "y" refer to each model, quantity of machine throughout,
            variables
"A" refer to Assembly
"F" refer to Fabrication
"P" refer to Packaging
"$MP" refer to maximum profit

From the problem statement, we can generate three inequality
objective functions for the total hours for A, F, and P which we
will simultaneously graph. From each, identify "m" (slope) and
"y-intercept" (0,b) for graphing purposes. They are...

3x+4y≤6300....................EqA, assembly hours (not to exceed)
      4y≤-3x+6300
        y≤(-3/4)x+1575.......EqA rewritten in "slope-intercept" form

3x+2.5y≤4500.................EqF, fabrication hours (not to exceed)
      2.5y≤-3x+4500
           y≤(-6/5)+1800......EqF, rewritten in "slope-intercept" form

0.8x+0.4y≤1080..............EqP, packaging hours (not to exceed)
         0.4y≤-0.8x+1575
              y≤-2x+3937.5....EqP, rewritten in "slope-intercept" form

Maximum profit is calculated by...
$MP=300x+375y..............EqMP, maximum profit, both machines


The graph of the above three equations can be found at this URL...

https://www.wyzant.com/resources/files/405142/linear_programming_graph

We will consider the portion of graph in quadrant one (positive
values only for "x" and "y") as any negative values in other quadrants
have no meaning in the context of this problem.

The solution set is the intersection of the the three shaded areas
bounded by points with coordinates (0,0), (0,1575), (500,1200),
and (1500,0). These are the vertices (or corner points) we will

test in EqMP. Plug the values for each of these points into EqMP....

   ($MP)₁=300(0)+375(1575).......substitute point coords (0,1575)
                =0+590,625
∴ ($MP)₁=$590,625

   ($MP)₂=300(500)+375(1200)....substitute point coords (500,1200)
                =150,000+450,000
∴ ($MP)₂=$600,000....................maximum value

   ($MP)₃=300(1500)+375(0)........substitute point coords (1500,0)
                =450,000+0
∴ ($MP)₃=$450,000

 

Disregard point coordinates (0,0) as substitution will produce a

$0 profit value.

So, optimal profit is $600,000 with optimal manufacture of
500 units of machine X and 1200 units of machine Y.

 

(You probably have the solution for this problem by now as it was

initially asked on Nov22, put this post will be available for others

to view the graph and procedure)