SOLUTION: A small firm produces both AM and AM/FM car radios. The AM radio take 15h to produce, and the AM/FM radios take 20h. The number of production hours is limited to 300h per week. The

Question 71877This question is from textbook
: A small firm produces both AM and AM/FM car radios. The AM radio take 15h to produce, and the AM/FM radios take 20h. The number of production hours is limited to 300h per week. The plant's capacity is limited to a tool of 18 radios per week, and existing orders require that at least 4 AM radios and at least 3 AM/FM radios be produced per week. Write a system of inequalities representing this situation. Then, draw a graph of the feasible region given these conditions, in which x is the number of AM radios and y the number of AM/FM radios. This question is from textbook

Found 2 solutions by ankor@dixie-net.com, josmiceli:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A small firm produces both AM and AM/FM car radios. The AM radios take 15 h to produce, and the AM/FM radios take 20 h. The number of production hours is limited to 300 h per week. The plant's capacity is limited to a total of 18 radios per week, and existing orders require that at least 4 AM radios and at least 3 AM/FM radios be produced per week. Write a system of inequalities representing this situation. Then, draw a graph of the feasible region given these conditions, in which X is the number of AM radios and Y is the number of AM/FM radios.
:
Let x = number of AM radios; let y = number of AM/FM radios

The production hour constraint:
15x + 20y =< 300
Arrange in general form to plot the graph
20y =< 300 - 15x
y =< 300/20 - (15/20)x
y =< 15 - .75x, use this to plot the production hr graph (Purple line)
:
Plant's capacity constraint:
x + y =< 18
y =< 18 - x, plot this graph also (Green line)
:
Min AM radio production constraint:
x => 4
This will be a vertical line going thru the x axis at +4, (I can't draw this in)
:
Min AM/FM radio production constaint
y => 3
This will be a horizontal line going thru the y axis at +3, (Dark blue line)
:
Here is the graph as presented by these equations except for x => 4,
:

:
The feasibility region:
1. At or below the purple or green line, whichever is lowest
2. At or above the horizontal line
3. At or to the right of the vertical line which you have to draw in at x = +4
:
How about this, did it all make some sense to you?
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!

divide by 5

I can determine that
from above
I now have 3 inequalities I can plot

The feasible area is the smallest area bounded by
the 3 lines. You can pick values that are just inside
or just outside to test them by putting into the inequalities
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