SOLUTION: A furniture manufacturer makes wooden tables and chairs. The production process involves two types of labor: carpentry and finishing. A table requires 2 hours of carpentry and 1 ho

Algebra ->  Sequences-and-series -> SOLUTION: A furniture manufacturer makes wooden tables and chairs. The production process involves two types of labor: carpentry and finishing. A table requires 2 hours of carpentry and 1 ho      Log On


   



Question 743379: A furniture manufacturer makes wooden tables and chairs. The production process involves two types of labor: carpentry and finishing. A table requires 2 hours of carpentry and 1 hour of finishing, and a chair requires 3 hours of carpentry and 1/2 hour of finishing. The profit is $35 per table and $20 per chair. The manufacturer's employees can supply a maximum of 108 hours of carpentry work and 20 hours of finishing work per day. How many tables and chairs should be made each day to maximize the profit?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
x= number of tables made per day
y= number of chairs made per day
Two obvious constraints are
x%3E=0 and y%3E=0
The amount of carpentry work required per day would be
2x%2B3y%3C=108.
The amount of finishing work required per day would be
x%2B0.5y%3C=20.
The inequalities above represent our design space or feasibility region (or whatever your teacher calls it. That is the region of the x-y plane where the factory will be working, hopefully at the point of maximum profit.
We can graph that region to visualize the problem (and because the teacher wants us to do it).
The feasibility region is the quadrilateral OABC, bounded by the lines
x=0 (the y-axis)
y=0 (the x-axis)
2x%2B3y=108 (the blue line) and
x%2B0.5y=20 (the green line).
To plot the blue and green line, I just found the x- and y-intercepts for each one.
For 2x%2B3y=108:
when x=0, 2%2A0%2B3y=108 --> 3y=108 --> y=108%2F3 --> y=36 gives us point A(0,36).
When y=0, 2x%2B3%2A0=108 --> 2x=108 --> x=108%2F2 --> x=54 gives us point E(54,0)
For x%2B0.5y=20:
when x=0, 0%2B0.5y=20 --> 0.5y=20 --> y=20%2F0.5 --> y=40 gives us point D(0,40).
When y=0, x%2B0.5%2A0=20 --> x=20 gives us point C(20,0).
The intersection of the 2 lines is the point that satisfies 2x%2B3y=108 and
x%2B0.5y=20 <--> 2x%2By=40. It is the solution to
system%282x%2B3y=108%2C2x%2By=40%29 --> 3y-y=108-40 --> 2y=68 --> y=34 and
with system%28y=34%2C2x%2By=40%29 --> 2x%2B34=40 --> 2x=40-34 --> 2x=6 --> x=4.
That intersection is the point C(3,34).

The profit, as a function of x and y is the linear function
P%28x%2Cy%29=35x%2B20y
The profit maximum will occur either at a vertex of that feasibility quadrilateral, or all along an edge. All we need to do is calculate P at all the vertices and see which vertex wins. If there is a tie, it will be between adjacent vertices, and in that case, the maximum happens all along the edge connecting those two vertices.
P%280%2C0%29=0
P%2820%2C0%29=700
P%280%2C36%29=720
P%283%2C34%29=35%2A3%2B20%2A34=105%2B680=785
There should be highlight%283%29 tables and highlight%2834%29 chairs made each day to maximize profit, and that would yield a profit of $785.