posted by amy .
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.
This seems to be linear programming night
let number of AMs be x
let number of FMs be y
15 x + 20 y </= 300 hour limit line
x+y </= 18 number limit line
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)
(4,48) 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.
Label questions like this "linear programming" so people who do linear programming will help.
Can I get some questions for how to do capacity