word problem?

Question

a small firm poduces both am and am/fm car radios. the am radios take 15h to produce and the am/fm take 20h. The number of production hours is limited to 300h per week. The plant's capacity is limited to a total of 18 radios per radios be produced per week. write a system of inequalities respresenting 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.

Could someone draw this graph for me i do not have the tools to do this please

Answer 1

Knowing
x = number of am radios
y = number of fm radios

We have 4 limitations:

1. Amount of radio productions per week limitation

x + y <= 18
y <= 18 - x [1]

2. Amount of time for radio production in a week

15x + 20y <= 300
20 y <= 300 - 15x
y <= 300/20 -15x/20
y <= 15 - 0,75 x [2]

3. We also know we're talking of real objects, so we only can think in units above 0

x >= 0 [3]
y >= 0 [4]

The area should be drawn as follows

[1] A line between (18,0) and (0,18), selecting the area under the line (considering the line too)
[2] A line between (0,15) and (20,0), selecting the area under the line (considering the line too)
[3] The area over the X axis (considering the axis)
[4] The area at the right of the Y axis (considerind the axis)

Then, the intersection of [1] and [2] can be get as follows:

[1]=[2]

18 - x = 15 - 0,75x
3 - x = - 0,75x
3 = x - 0,75x
3 = 0,25x
x = 3 / 0,25
x = 12

Replacing x = 12 in [1]

y = 18 - 12
y = 6

Then [1] and [2] are intersected in P (12,6)

The feasible region should be where all the areas get intersected. In other words, the polygon formed by the following coordinates:

(0,0);(18,0);(12,6):(0,15); considering all the edges

Answer 2

Let's call the am radios A and the am/fm radios F.

15A + 20F <= 300
A + F <= 18

For the graph,
20y = -15x + 300 or y = (-3/4)x + 15
y = -x + 18

It's best to draw a graph using two colors. For the second line, which is the simplest, draw a line that starts at (0,18) and goes to (18,0). Shade the area under this graph one color, say yellow.

For the second equation, start the line at (15,0) and ends at (0,20). Color everything under that pale blue.

Where the graph is pale green, production is OK.

I can do it in Visio if you want and post it somewhere, but it'll take a bit.