SOLUTION: Please help with the problem below. I already know how to create the graph. What I need is to know step by step information on how to shade the particular area called for in the p

Question 114132: Please help with the problem below. I already know how to create the graph.
What I need is to know step by step information on how to shade the particular area called for in the problem. I need help please.
I entered the following information to create the graph
x=4, x=18, y=3, y=18, y=(-3/4)x+15 and y=-x+18
This is how the problem reads.
Draw the appropriate graph for the following.
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 the number of AM/FM radios.
Thanks
JMS

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
I entered the following information to create the graph
x=4, x=18, y=3, y=18, y=(-3/4)x+15 and y=-x+18
This is how the problem reads.
Draw the appropriate graph for the following.
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 the number of AM/FM radios.
--------------
INEQUALITIES:
x+y<=18
15x+20y<=300
----------------
Boundaries:
y = -x+18
y = -(3/4)x+15
x=4, x=18
y=3, y=18
-----------------
To draw the feasible region:
draw vertical lines at x=4 and at x=18
draw horizontal lines at y = 3 and at y = 18
Graph the line y=-x+18 and shade the area below it IN THE FEASIBLE REGION
Graph the line y=(-3/4)x+15 and shade the area below it IN THE F.R.
----------------------
Comment: The feasible region is between x=4 and x=18 and between y=3
and y=18, and under both y=-x+18 and y=(-3/4)x+15
--------------------------

======================
Cheers,
Stan H.
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