Marissa is selling paintings for $12 each and bracelets for $6 each. Her goal is to sell at least $500 in products, and she must sell at least 40 bracelets. Which of the following combinations will satisfy these constraints?
Let's use the variables x and y to represent the number of paintings and bracelets Marissa sells, respectively.
We want to find a combination of x and y that satisfies the following constraints:
- The total amount Marissa makes must be at least $500:
- 12x + 6y ≥ 500
- Marissa must sell at least 40 bracelets:
- y ≥ 40
Let's use these two inequalities to eliminate any answer choices that don't work:
A) 30 paintings and 30 bracelets
- Revenue: 12(30) + 6(30) = $540
- Does not satisfy the first constraint
B) 40 paintings and 40 bracelets
- Revenue: 12(40) + 6(40) = $960
- Satisfies both constraints
C) 50 paintings and 20 bracelets
- Revenue: 12(50) + 6(20) = $720
- Satisfies the first constraint but does not satisfy the second
D) 20 paintings and 60 bracelets
- Revenue: 12(20) + 6(60) = $720
- Satisfies the first constraint but does not satisfy the second
Therefore, the only combination that satisfies both constraints is B) 40 paintings and 40 bracelets.
Let's represent the number of paintings as "p" and the number of bracelets as "b".
We know that Marissa is selling paintings for $12 each and bracelets for $6 each.
Therefore, the equation for the total sales can be written as:
Total sales = (number of paintings * price per painting) + (number of bracelets * price per bracelet)
Total sales = (p * 12) + (b * 6)
Marissa's goal is to sell at least $500 in products. So we have the inequality:
(p * 12) + (b * 6) >= 500
Also, she must sell at least 40 bracelets. So we have the inequality:
b >= 40
We need to find the combinations of p and b that satisfy these constraints.
We can start by finding the possible values for b. Since b must be at least 40, the possible values for b are 40, 41, 42, ...
Now, let's substitute the values of b into the first inequality and solve for p:
(12 * p) + (6 * b) >= 500
Case 1: b = 40
(12 * p) + (6 * 40) >= 500
12p + 240 >= 500
12p >= 260
p >= 21.67
Since the number of paintings must be a whole number, the possible values for p are 22, 23, ...
Case 2: b = 41
(12 * p) + (6 * 41) >= 500
12p + 246 >= 500
12p >= 254
p >= 21.17
The possible values for p are 22, 23, ...
And we continue this process for each value of b.
Based on these calculations, the following combinations will satisfy the given constraints:
Option A: (p = 22, b = 41)
Option B: (p = 22, b = 42)
Option C: (p = 23, b = 42)
Option D: (p = 23, b = 43)
Option E: (p = 24, b = 43)
...
Therefore, there are multiple combinations that satisfy the given constraints.