Question 1208741
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x = number of regular gadgets
y = number of premium gadgets


Let's make a table to find out how long each process takes.
<table border = "1" cellpadding = "5"><tr><td></td><td>Regular</td><td>Premium</td><td>Total</td></tr><tr><td>Assembly</td><td>x</td><td>4y</td><td>x+4y</td></tr><tr><td>Finishing</td><td>3x</td><td>5y</td><td>3x+5y</td></tr></table>
Based on the results in the last column, we can set up these constraints
{{{system(x+4y<=12,3x+5y<=12)}}}
This is because assembly and finishing have at most 12 hours for each operation.
"At most 12 hours" is the same as saying "12 hours or fewer".


Another constraint is x+y <= 100 because the factory can make at most 100 objects per day.
{{{system(x+4y<=12,3x+5y<=12,x+y<=100)}}}


Then,
{{{system(x+4y<=12,3x+5y<=12,x+y<=100,x>=0,y>=0)}}}
The last two constraints are there to ensure x,y are never negative. 
Keep in mind that x,y are also integers.


That completes the system of inequalities for the constraints.
If you were to graph each of them on the same xy axis, then you'd get a mess of regions that overlap. The region where all areas overlap is what you're looking for.
That's often a standard approach that many teachers and textbooks will follow.


But perhaps a more efficient and cleaner graph would be to look at the opposite of each inequality mentioned.
Eg: The opposite of {{{x>=0}}} is {{{x < 0}}}
Graph the opposite of each and look where none of the shaded regions overlap. Look at the blank white space. This space is where the original system would overlap.


I'm following a process that tutor Theo is using in this similar question
<a href="https://www.algebra.com/algebra/homework/Linear-equations/Linear-equations.faq.question.1187223.html">https://www.algebra.com/algebra/homework/Linear-equations/Linear-equations.faq.question.1187223.html</a>


For this problem, the final shaded region is marked as an X shown below.
{{{
drawing(500,500,-6,6,-6,6,
graph(500,500,-6,6,-6,6,-1000),

circle(0,0,0.08),circle(0,0,0.1),circle(0,0,0.12),circle(0,0,0.14),circle(4,0,0.08),circle(4,0,0.1),circle(4,0,0.12),circle(4,0,0.14),circle(0,2.4,0.08),circle(0,2.4,0.1),circle(0,2.4,0.12),circle(0,2.4,0.14),

line(-11,9,9,-3),line(-12,6,12,0),
locate(1,1,"X"),

locate(0+0.2,0-0.2,"A"),locate(4,0+0.6,"B"),locate(0+0.2,2.4+0.4,"C"),
locate(4,-1,3x+5y=12),
locate(4,2.5,x+4y=12)
)
}}}
The final shaded region is in the upper right quadrant and below the line 3x+5y=12.
If you pick a point in region X, then that point satisfies all of the original inequalities.


The shaded region is a triangle with the vertices 
A = (0, 0)
B = (4, 0)
C = (0, 2.4)
Vertex B is found by solving the system of equations {{{system(3x+5y=12,y=0)}}} In other words, plug in y = 0 to get x = 4


Vertex C is found by solving the system of equations {{{system(3x+5y=12,x=0)}}} Plug in x = 0 to get y = 12/5 = 2.4
We don't have to worry about the lines x+4y=12 and x+y=100



Regular gadgets give a profit of $20 while premium gives a profit of $30.
The profit function is P = 20x+30y
The goal is to find the max value of P based on the constraints.


It turns out P is maxed when we're at a vertex.
Plug in each (x,y) coordinate to find various values of P.
<table border = "1" cellpadding = "5"><tr><td>Vertex</td><td>(x,y)</td><td>P</td></tr><tr><td>A</td><td>(0,0)</td><td>20x+30y = 20*0+30*0 = 0</td></tr><tr><td>B</td><td>(4,0)</td><td>20x+30y = 20*4+30*0 = <font color=red size=4>80</font></td></tr><tr><td>C</td><td>(0,2.4)</td><td>20x+30y = 20*0+30*2.4 = 72</td></tr></table>
The <font color=red>max profit is P = 80</font>. It happens when you make <font color=red>x = 4 regular gadgets and y = 0 premium gadgets.</font> This corresponds to vertex B.
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