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
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.

PS

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

Can I get some questions for how to do capacity

To represent this situation with a system of inequalities, let's assign variables to the given information:

Let x be the number of AM radios produced per week.
Let y be the number of AM/FM radios produced per week.

Since the AM radios take 15 hours to produce and the AM/FM radios take 20 hours, the total number of production hours is limited to 300 hours per week. We can represent this constraint with the following inequality:

15x + 20y ≤ 300

The plant's capacity is limited to a total of 18 radios per week, so we can express this constraint as:

x + y ≤ 18

Furthermore, existing orders require that at least 4 AM radios and at least 3 AM/FM radios be produced per week. This gives us the following constraints:

x ≥ 4
y ≥ 3

Combining all these inequalities, we can write the system of inequalities as:

15x + 20y ≤ 300
x + y ≤ 18
x ≥ 4
y ≥ 3

To graph the feasible region for this system of inequalities, we will first graph the lines corresponding to the equalities in each inequality using the coordinates (x, y). Then, we will shade the region that satisfies all inequalities.

The feasible region describes the area where all the constraints are satisfied. In this case, it will be the region below the first line, to the left of the second line, and above the x-axis and y-axis.

Here's a rough sketch of the graph:

|
18 |________________
| /
| /
| /
| ________/
| |
|-----|----------|
0 3.75 7.50

The shaded region represents the feasible region satisfying all the given conditions. It can be visualized as a triangular region in the first quadrant of the graph.

Note: The graph is just a visual representation to help illustrate the feasible region. For accurate calculations and a more precise graph, consider using graphing software or a graphing calculator.