Posted by **amy** on Tuesday, January 22, 2008 at 7:11pm.

A small firm produces AM and AM/FM car radios. The AM radios take 15 hours to produce, and the AM/FM radios take 20 hours. The number of production hours is limited to 300 hours 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 a 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.

- math -
**Damon**, Tuesday, January 22, 2008 at 8:51pm
This seems to be linear programming night

let number of AMs be x

let number of FMs be y

then constraints:

15 x + 20 y </= 300 hour limit line

x+y </= 18 number limit line

x>/= 4

y>/= 3

graph

the hour limit line hits the y axis at (0,15) and the x axis at (20,0)

the number limit line hits the y axis at (0,18) and the x axis at (18,0)

we have a vertical line at x = 4

and a horizontal line at y =3

now find intersections (vertices)

(4,3)

(4,48) intersection of x = 4 with hours limit line

(12,6) intersection of hours line and numbers line

(15,3) intersection of numbers line with horizontal line at y = 3

Draw straight lines around the vertices and you have your region inside.

- Linear programming -
**Damon**, Tuesday, January 22, 2008 at 8:54pm
PS

Label questions like this "linear programming" so people who do linear programming will help.

- math -
**Anonymous**, Monday, January 28, 2008 at 5:25pm
Can I get some questions for how to do capacity

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