Question 58822
Here's a word problem that has me stumped, 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 are 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. How do you solve this?
:
Let x = no. of AM radios; Let y = no.of AM/FM 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 hour equation:
15x + 20y =< 300
Arrange it "y=" in order graph it
20y =< 300 - 15x
y =< 300/20 - (15/20)x
y =< 15 - .75x; plotted as the red line
:
"The plant’s capacity is limited to a total of 18 radios"
The quantity equation:
x + y <= 18
y <= 18 - x; plotted as the green line
:
"At least 4 AM Radios"
x => 4
This will be vertical line parallel to y axis, passing thru x =>4. You will have
to draw this in yourself
:
"At least 3 AM/FM Radios"
y => 3, plotted as the black line
:
The graph:
{{{ graph( 300, 200, -4, 20, -4, 20, 15 -.75x, 18-x, 3 ) }}}


:
Note that the feasibility region would be bounded by: 
1. area equal or above the horizontal line going thru y = 3, 
2. the vertical line that you will draw going thru x = 4, or to the right,
3. Equal or below the red and the green lines whichever is lowest.
:
To further explain, here are the 4 corner coodinates of the feasiblity region
(it's inside these four corners)
x = 4; y = 3 (on the vertical line)
:
x = 4; y = 12 (on the vertical line)
:
x = 12; y = 6
:
x = 15; y = 3
:
Does this help?