Owen would like to make a small income as an artist. Owen asked his friend Emily for advice about what

Question

combination of pictures to make. She suggested that he determine a reasonable profit for that month’s work
and then paint what he needs in order to earn that amount of profit.
• Each pastel requires $5 in materials and earns a profit of $40 for Owen.
• Each watercolor requires $15 in materials and earns a profit of $100 for Owen.
• Owen has $180 to spend on materials.
• Owen can make at most 16 pictures.
a. State the system of inequalities that represents this situation. Remember to define your variables
and include any non-negative constraints that are required.
d. Suppose Owen decided $1,000 would be a reasonable profit. Find three different combinations of watercolors and pastels that would earn Owen a profit of exactly $1,000.

Answer 1

Let x be the number of pastels Owen makes.
Let y be the number of watercolors Owen makes.

The materials cost for each pastel is $5, so the total cost for all pastels is 5x.
The profit for each pastel is $40, so the total profit for all pastels is 40x.

The materials cost for each watercolor is $15, so the total cost for all watercolors is 15y.
The profit for each watercolor is $100, so the total profit for all watercolors is 100y.

We are given that Owen has $180 to spend on materials, so the cost constraint is:
5x + 15y ≤ 180

We are also given that Owen can make at most 16 pictures, so the quantity constraint is:
x + y ≤ 16

Finally, Owen wants to earn a profit of $1,000, so the profit constraint is:
40x + 100y = 1000

Therefore, the system of inequalities representing this situation is:
5x + 15y ≤ 180
x + y ≤ 16
40x + 100y = 1000

To find three different combinations of watercolors and pastels that would earn Owen a profit of exactly $1,000, we can solve the system of equations:
40x + 100y = 1000

We can set x = 1 and solve for y:
40(1) + 100y = 1000
40 + 100y = 1000
100y = 960
y = 9.6

Since the number of watercolors must be a whole number, we can let y = 10 and solve for x:
40x + 100(10) = 1000
40x + 1000 = 1000
40x = 0
x = 0

So one combination would be 0 pastels and 10 watercolors.

We can set x = 2 and solve for y:
40(2) + 100y = 1000
80 + 100y = 1000
100y = 920
y = 9.2

Since the number of watercolors must be a whole number, we can let y = 9 and solve for x:
40x + 100(9) = 1000
40x + 900 = 1000
40x = 100
x = 2.5

Since the number of pastels must be a whole number, we can let x = 2 and solve for y:
40(2) + 100y = 1000
80 + 100y = 1000
100y = 920
y = 9.2

So another combination would be 2 pastels and 9 watercolors.

We can set x = 3 and solve for y:
40(3) + 100y = 1000
120 + 100y = 1000
100y = 880
y = 8.8

Since the number of watercolors must be a whole number, we can let y = 9 and solve for x:
40x + 100(9) = 1000
40x + 900 = 1000
40x = 100
x = 2.5

Since the number of pastels must be a whole number, we can let x = 2 and solve for y:
40(2) + 100y = 1000
80 + 100y = 1000
100y = 920
y = 9.2

So another combination would be 2 pastels and 9 watercolors.