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.

Question

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?

Answer 1

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 \( 20 hours.\)