Solving systems of inequalities

Question

A store has two types of animal feed available. Type A contains 5

 pounds of oats and 7

 pounds of corn per bag. Type B contains 6

 pounds of oats and 4

 pounds of corn per bag. A farmer buys both types and combines them so that the resulting mixture has at least 150

 pounds of oats and at least 140

 pounds of corn. The store only has 20

 bags of type A feed and 25

 bags of type B feed in stock. Let x

 be the number of type A bags purchased. Let y

 be the number of type B bags purchased. Shade the region corresponding to all values of x

 and y

 that satisfy these requirements

Answer 1

To graph this problem, you can start by determining the parameters. From the problem,

For oats,

5A + 6B >= 150

For corn,

7A + 4B >= 140

With the limits

A<=20

B<=25

Which can be drawn as straight boundary lines on the graph

An additional boundary would be made by determining what combinations would get us the minimum 150lb of oats and 140lb of corn

If purchasing the maximum 20 bags of A,

To fulfill oats:

5(20) + 6B = 150

6B = 150 - 100

6B = 50

B = 8.3

To fulfill corn:

7(20) + 4B = 140

4B = 140 - 140

B=0

So we would take the greater number, B=8.3

If producing the maximum 25 bags of B,

To fulfill oats:

5A + 6(25) = 150

5A = 150 - 150

A = 0

To fulfill corn:

7A + 4(25) = 140

7A = 140 - 100

7A = 40

A = 5.7

So we take the greater number, A = 5.7

Thus, we can draw a boundary line from (20, 8.3) to (5.7, 25), or B = (-1.2)A + 32.3

The area above that line and below B=25 and to the left of A=20 is the area of interest