Answer:
Let x be the number of $10 CDs and y be the number of $15 CDs. Then, the system of inequalities representing this situation is:
10x + 15y ≥ 30 (the customer wants to spend at least $30)
10x + 15y ≤ 60 (the customer wants to spend at most $60)
x ≥ 0 (the customer can't buy a negative number of CDs)
y ≥ 0 (the customer can't buy a negative number of CDs)
Simplifying the first two inequalities:
2x + 3y ≥ 6
2x + 3y ≤ 12
To graph the solution region, we first graph the boundary lines:
2x + 3y = 6 (or y = (-2/3)x + 2)
2x + 3y = 12 (or y = (-2/3)x + 4)
Then, we shade the region that satisfies both inequalities (the region below the upper line and above the lower line), as well as the regions to the right and above the x and y axes:
Let x be the number of $10 CDs and y be the number of $15 CDs. Then, the system of inequalities representing this situation is:
10x + 15y ≥ 30 (the customer wants to spend at least $30)
10x + 15y ≤ 60 (the customer wants to spend at most $60)
x ≥ 0 (the customer can't buy a negative number of CDs)
y ≥ 0 (the customer can't buy a negative number of CDs)
Simplifying the first two inequalities:
2x + 3y ≥ 6
2x + 3y ≤ 12
To graph the solution region, we first graph the boundary lines:
2x + 3y = 6 (or y = (-2/3)x + 2)
2x + 3y = 12 (or y = (-2/3)x + 4)
Then, we shade the region that satisfies both inequalities (the region below the upper line and above the lower line), as well as the regions to the right and above the x and y axes:
inequality graph
As an example of one possible production situation within these restrictions, let's say the customer buys 2 $10 CDs and 2 $15 CDs. This would cost $50, which falls within the customer's desired spending range of $30 to $60. Plugging in x = 2 and y = 2 into the inequalities:
10x + 15y = 10(2) + 15(2) = 50, which satisfies 10x + 15y ≥ 30 and 10x + 15y ≤ 60
x = 2 ≥ 0
y = 2 ≥ 0
Therefore, (2,2) is a valid solution within the given restrictions.
Explanation: