Owen would like to make a small income as an artist. Owen asked his friend Emily for advice about what
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.
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.
a. Let's define the variables:
- Let x be the number of pastels Owen makes.
- Let y be the number of watercolors Owen makes.
Now, let's write the inequalities for the given constraints:
1. The total cost of materials used cannot exceed $180:
5x + 15y ≤ 180
2. Owen can make at most 16 pictures:
x + y ≤ 16
3. Both x and y must be non-negative:
x ≥ 0
y ≥ 0
Therefore, the system of inequalities representing this situation is:
5x + 15y ≤ 180
x + y ≤ 16
x ≥ 0
y ≥ 0
b. The inequalities in the system represent all the constraints that Owen needs to follow in order to make a reasonable profit while considering the given conditions. By solving this system, we can find different combinations of pastels and watercolors that fulfill these conditions.
d. To find three different combinations of watercolors and pastels that would earn Owen a profit of exactly $1,000, we need to solve the system of inequalities where the profit expressions are equated to $1,000.
Let's set up the equations for the profit:
- For pastels: Profit from each pastel = $40
Therefore, the profit from x pastels = 40x
- For watercolors: Profit from each watercolor = $100
Therefore, the profit from y watercolors = 100y
Now, we need to find three different combinations of x and y such that the total profit from pastels and watercolors adds up to exactly $1,000.
Using the equations: 40x + 100y = 1000
We can find some possible combinations by substituting different values for x and solving for y.
Possible combinations are:
1. Let x = 5 and y = 6
In this case, 40(5) + 100(6) = 200 + 600 = $800.
2. Let x = 4 and y = 8
In this case, 40(4) + 100(8) = 160 + 800 = $960.
3. Let x = 2 and y = 10
In this case, 40(2) + 100(10) = 80 + 1000 = $1,080.
Note: It is not possible to find an exact combination that results in a profit of exactly $1,000. However, the three combinations mentioned above are the closest to this target profit.