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.
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-colored pencil, add those points to your graph.

Answer 1

Let's set up a system of equations to solve this problem.

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

The profit equation is given by:
Profit = 40x + 100y

The material cost equation is given by:
Material cost = 5x + 15y

We have the following additional constraints:
1. Owen has $180 to spend on materials, so we have the equation:
5x + 15y ≤ 180

2. Owen can make at most 16 pictures, so we have the equation:
x + y ≤ 16

We want to find three different combinations of watercolors and pastels that will earn Owen a profit of exactly $500, which can be represented by the equation:
40x + 100y = 500

Now, let's solve this system of equations and find the different combinations.

First, we'll graph the feasible region by plotting the inequality constraints:

The inequality 5x + 15y ≤ 180 can be rewritten as:
x + 3y ≤ 36

Plotting this equation on a graph, we get a line with slope -1/3 passing through the points (0, 12) and (36, 0).

The inequality x + y ≤ 16 represents a line with slope -1 passing through the points (0, 16) and (16, 0).

To find the points of intersection, we solve the system of equations:
x + 3y = 36
x + y = 16

Subtracting the second equation from the first, we get:
2y = 20
y = 10

Substituting this value of y into the second equation, we get:
x + 10 = 16
x = 6

The points of intersection are (6, 10).

Now, let's add the points (0, 12), (36, 0), (0, 16), (16, 0), and (6, 10) to the graph.

Next, let's find three different combinations of watercolors and pastels that will earn Owen a profit of exactly $500.

We can substitute the value of y = 10 into the profit equation:
40x + 100(10) = 500
40x + 1000 = 500
40x = 500 - 1000
40x = -500
x = -12.5

Since we can't have a negative number of pictures, this combination is not feasible.

Let's try another combination.

Let's set y = 16 and solve the profit equation:
40x + 100(16) = 500
40x + 1600 = 500
40x = 500 - 1600
40x = -1100
x = -27.5

Again, we can't have a negative number of pictures, so this combination is not feasible.

Finally, let's set y = 6 and solve the profit equation:
40x + 100(6) = 500
40x + 600 = 500
40x = 500 - 600
40x = -100
x = -2.5

This combination is also not feasible as we can't have a negative number of pictures.

Therefore, there are no feasible combinations of watercolors and pastels that will earn Owen a profit of exactly $500.