|
Question 58444This question is from textbook Beginning Algebra
: Could someone help me on this problem please?
I need to draw the appropriate graph for the following problem:
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/PM radios be produced per week. Write a system of inequalities representing this situation. Then, I need a raph of the feasble region given these conditions, in which x is the number of AM radios and y the number of AM/FM radios.
Thanks so much,
Lisa
This question is from textbook Beginning Algebra
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! I need to draw the appropriate graph for the following problem:
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/PM radios be produced per week. Write a system of inequalities representing this situation. Then, I need a raph of the feasble region given these conditions, in which x is the number of AM radios and y the number of AM/FM radios.
---------------
Let # of AM radios be A and # of FM radios be F.
1st: A>=4
2nd: F>=3
3rd: A+F<=18
4th: 15A+20F<=300
-----------------
Draw a coordinate system with "A" as the vertical axis and "F" as the
horizontal axis.
Using 1st and 2nd, sketch a vertical line at F=3 and a horizontal line at A=4
Using 3rd, sketch the line A=-F+18 with points (3,15) and (14,4)
Using 4th, sketch the line A=(-4/3)F+20 with points (3,16) and (12,4)
Find the intersection of 3rd and 4th:
-F+18=(-4/3)F+20
(1/3)F=2
F=6 ; then A=-6+18=12
The intersection pt. is (6,12)
If you shade in the inequality areas you will see you
have vertices at (3,15),(6,12),(12,4),and (3,4)
Cheers,
Stan H.
|
|
|
| |
