Question
Lynsay plans to earn at least $300 babysitting and mowing lawns this summer. She will charge $12 per hour for babysitting, and $15 per hour for mowing lawns. She would like to work a maximum of 50 hours this summer.
She wrote the following system of inequalities to determine combinations of jobs she can do. Let x = number of hours spent babysitting, and y = number of hours spent mowing lawns.
What is the smallest number of total hours she can work in order to earn at least $300 without working more than 50 hours?
She wrote the following system of inequalities to determine combinations of jobs she can do. Let x = number of hours spent babysitting, and y = number of hours spent mowing lawns.
What is the smallest number of total hours she can work in order to earn at least $300 without working more than 50 hours?
Answers
GPT-4o mini
To determine the smallest number of total hours Lynsay can work while earning at least $300 without exceeding 50 hours, we can set up the following inequalities based on her situation:
1. **Earnings Inequality**:
The earnings from babysitting at $12 per hour and mowing lawns at $15 per hour must be at least $300:
\[
12x + 15y \geq 300
\]
2. **Total Hours Inequality**:
The total hours worked (both babysitting and mowing) cannot exceed 50 hours:
\[
x + y \leq 50
\]
3. **Non-negativity Constraints**:
Both \(x\) and \(y\) must be non-negative (she can't work negative hours):
\[
x \geq 0, \quad y \geq 0
\]
To find combinations of \(x\) and \(y\) that satisfy the equations, we will first rewrite the earnings inequality:
\[
12x + 15y \geq 300
\]
We can simplify it by dividing the entire inequality by 3:
\[
4x + 5y \geq 100
\]
Now we need to consider the two inequalities:
1. \(4x + 5y \geq 100\)
2. \(x + y \leq 50\)
The goal is to minimize \(x + y\), which represents the total hours worked.
To find the intersection points of the inequalities, we will find the lines \(4x + 5y = 100\) and \(x + y = 50\).
### Finding Intersection Points:
1. **From the earnings inequality**:
We can express \(y\) as:
\[
y = \frac{100 - 4x}{5}
\]
2. **From the total hours inequality**:
We can express \(y\) as:
\[
y = 50 - x
\]
### Setting the Equations Equal:
Now, we can set these equal to each other to find the intersection:
\[
\frac{100 - 4x}{5} = 50 - x
\]
Multiplying through by 5 to eliminate the fraction:
\[
100 - 4x = 250 - 5x
\]
Rearranging gives:
\[
5x - 4x = 250 - 100
\]
\[
x = 150
\]
However, this value does not make sense in our hour bounds. We might have miscalculated there. Let's solve stepwise for points of intersection:
To find the feasible solutions, we can check values:
1. Substituting \(y = 0\) into \(4x + 5y \geq 100\):
\[
4x \geq 100 \implies x \geq 25 \, \Rightarrow (25, 0)
\]
2. Substituting \(x = 0\) into \(4x + 5y \geq 100\):
\[
5y \geq 100 \implies y \geq 20 \, \Rightarrow (0, 20)
\]
3. Now check \(x + y = 50\):
\[
x + y = 50 \, \Rightarrow y = 50 - x
\]
Substituting into \(4x + 5(50 - x) \geq 100\):
\[
4x + 250 - 5x \geq 100
\]
\[
-x + 250 \geq 100 \implies -x \geq -150 \implies x \leq 150
\]
Let's check the two valid points to find minimum \(x+y\):
So we checkon \( (25, 0) \) and \( (0, 20) \):
1. For \(x+y\):
\[
25 + 0 = 25
\]
2. For \(0 + 20 = 20 \)
Thus the feasible region points gives us minimum:
### Final Validation:
Checking at \(x=0/y=20\)
- Hours: \(20\) hours (minimum selection)
Hence the minimum number of total hours she can work to earn at least $300 while not exceeding 50 hours is \( **20 hours**.\)
1. **Earnings Inequality**:
The earnings from babysitting at $12 per hour and mowing lawns at $15 per hour must be at least $300:
\[
12x + 15y \geq 300
\]
2. **Total Hours Inequality**:
The total hours worked (both babysitting and mowing) cannot exceed 50 hours:
\[
x + y \leq 50
\]
3. **Non-negativity Constraints**:
Both \(x\) and \(y\) must be non-negative (she can't work negative hours):
\[
x \geq 0, \quad y \geq 0
\]
To find combinations of \(x\) and \(y\) that satisfy the equations, we will first rewrite the earnings inequality:
\[
12x + 15y \geq 300
\]
We can simplify it by dividing the entire inequality by 3:
\[
4x + 5y \geq 100
\]
Now we need to consider the two inequalities:
1. \(4x + 5y \geq 100\)
2. \(x + y \leq 50\)
The goal is to minimize \(x + y\), which represents the total hours worked.
To find the intersection points of the inequalities, we will find the lines \(4x + 5y = 100\) and \(x + y = 50\).
### Finding Intersection Points:
1. **From the earnings inequality**:
We can express \(y\) as:
\[
y = \frac{100 - 4x}{5}
\]
2. **From the total hours inequality**:
We can express \(y\) as:
\[
y = 50 - x
\]
### Setting the Equations Equal:
Now, we can set these equal to each other to find the intersection:
\[
\frac{100 - 4x}{5} = 50 - x
\]
Multiplying through by 5 to eliminate the fraction:
\[
100 - 4x = 250 - 5x
\]
Rearranging gives:
\[
5x - 4x = 250 - 100
\]
\[
x = 150
\]
However, this value does not make sense in our hour bounds. We might have miscalculated there. Let's solve stepwise for points of intersection:
To find the feasible solutions, we can check values:
1. Substituting \(y = 0\) into \(4x + 5y \geq 100\):
\[
4x \geq 100 \implies x \geq 25 \, \Rightarrow (25, 0)
\]
2. Substituting \(x = 0\) into \(4x + 5y \geq 100\):
\[
5y \geq 100 \implies y \geq 20 \, \Rightarrow (0, 20)
\]
3. Now check \(x + y = 50\):
\[
x + y = 50 \, \Rightarrow y = 50 - x
\]
Substituting into \(4x + 5(50 - x) \geq 100\):
\[
4x + 250 - 5x \geq 100
\]
\[
-x + 250 \geq 100 \implies -x \geq -150 \implies x \leq 150
\]
Let's check the two valid points to find minimum \(x+y\):
So we checkon \( (25, 0) \) and \( (0, 20) \):
1. For \(x+y\):
\[
25 + 0 = 25
\]
2. For \(0 + 20 = 20 \)
Thus the feasible region points gives us minimum:
### Final Validation:
Checking at \(x=0/y=20\)
- Hours: \(20\) hours (minimum selection)
Hence the minimum number of total hours she can work to earn at least $300 while not exceeding 50 hours is \( **20 hours**.\)