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.
b. What is the optimization equation?
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. (3 marks)
e. Now suppose Owen wanted to earn only $500 in profit. Find three different combinations of
watercolors and pastels that will earn Owen a profit of exactly $500. Using a different-coloured
pencil, add those points to your graph.

Answer 1

a. Let:
x = the number of pastels Owen makes
y = the number of watercolors Owen makes

The system of inequalities is:
5x + 15y ≤ 180 (constraint for available materials)
x + y ≤ 16 (constraint for number of pictures)

Additionally, the constraints must be non-negative:
x ≥ 0
y ≥ 0

b. The optimization equation is:
Profit = 40x + 100y

d. To find three different combinations of watercolors and pastels that would earn Owen a profit of exactly $1,000, we must solve the system of inequalities along with the optimization equation:

5x + 15y ≤ 180
x + y ≤ 16
x ≥ 0
y ≥ 0
Profit = 40x + 100y = 1000

One possible solution is:
x = 4
y = 8

Another possible solution is:
x = 2
y = 10

And one more possible solution is:
x = 6
y = 7

e. To find three different combinations of watercolors and pastels that will earn Owen a profit of exactly $500, we need to solve the same system of inequalities and optimization equation, but change the profit value to 500:

5x + 15y ≤ 180
x + y ≤ 16
x ≥ 0
y ≥ 0
Profit = 40x + 100y = 500

One possible solution is:
x = 3
y = 6

Another possible solution is:
x = 0
y = 5

And one more possible solution is:
x = 6
y = 2

These points can be added to the graph as requested.