She has to do at least 8.6 hours walking (and then 7.5 hours dog walking) for the minimum of $80 earning
Keilantra is working two summer jobs, making $12 per hour lifeguarding and $6 per hour walking dogs. Last week Keilantra worked a total of 11 hours and earned a total of $108.
x = number hours dog walking
y = number hours Lifeguarding
x + y <= 11
12x + 6y >= 80
The solution is the common area below the first line and above the second line.
The first line is above the second line (due to the y-intercept of 11 vs. 8) until the crossing point.
for the crossing point we use the regular equations :
from the first we get
x = 11 - y
using this in the second
12×(11 - y) + 6y = 80
132 - 12y + 6y = 80
6y = 52
y = 52/6 = 26/3 = 8.6
x = 11 - y = 11 - 8.6 = 2.4
So, she has to do at least 8.6 hours lifeguarding (and then 7.5 hours dog walking) for the minimum of $80 earning, all the way up to the maximum of 11 hours lifeguarding (and 0 dog walking).
Hence the answer is she has to do at least 8.6 hours lifeguarding (and then 7.5 hours dog walking) for the minimum of $80 earning.
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