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.
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.
3 answers
PS
Label questions like this "linear programming" so people who do linear programming will help.
Label questions like this "linear programming" so people who do linear programming will help.
Can I get some questions for how to do capacity