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?

1 answer

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+15y300

  2. Total Hours Inequality:
    The total hours worked (both babysitting and mowing) cannot exceed 50 hours: x+y50

  3. Non-negativity Constraints:
    Both x and y must be non-negative (she can't work negative hours): x0,y0

To find combinations of x and y that satisfy the equations, we will first rewrite the earnings inequality:

12x+15y300

We can simplify it by dividing the entire inequality by 3:

4x+5y100

Now we need to consider the two inequalities:

  1. 4x+5y100
  2. x+y50

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=1004x5

  2. From the total hours inequality: We can express y as: y=50x

Setting the Equations Equal:

Now, we can set these equal to each other to find the intersection: 1004x5=50x

Multiplying through by 5 to eliminate the fraction: 1004x=2505x

Rearranging gives: 5x4x=250100 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+5y100: 4x100x25,(25,0)

  2. Substituting x=0 into 4x+5y100: 5y100y20,(0,20)

  3. Now check x+y=50: x+y=50,y=50x

Substituting into 4x+5(50x)100: 4x+2505x100 x+250100x150x150

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 \( 20 hours.\)