SOLUTION: Anna has two after school jobs, babysitting and working in a local restaurant. She can spend at most 18 hours a week between these two jobs. She always babysits for her neighb

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Question 990328: Anna has two after school jobs, babysitting and working in a local restaurant. She
can spend at most 18 hours a week between these two jobs.
She always babysits for her neighbours for three hours after school on Thursday and
for four hours on Saturday night, and in order to keep her employment at the
restaurant she must work there for at least three hours a week.
Anna can do some of her homework whilst she is babysitting, so prefers to spend at
least twice as much time doing this as working at the restaurant.
Set up this scenario as a linear programming problem.

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
We start by defining our variables:
x= hours spent babysitting in a week, and
y= hours working at the restaurant in a week.
We know, of course, that x and y cannot be negative numbers.
Next, we translate the worded information given into equations or inequalities.
"She can spend at most 18 hours a week between these two jobs" translates as
x%2By%3C=18 .
"She always babysits for her neighbours for three hours after school on Thursday and for four hours on Saturday night" means that she babysits at least 3%2B4=7 hours each week.
That translates as
x%3E=7 .
Also, "in order to keep her employment at the
restaurant she must work there for at least three hours a week" translates as
y%3E=3 .
"Anna can do some of her homework whilst she is babysitting, so prefers to spend at least twice as much time doing this as working at the restaurant" translates as
x%3E=2y .
The four inequalities above are part of our constraints.
Graphed separately those 4 inequalities look like this:

Taking the four inequalities together we have as feasible region the triangle ABC below:


We also know, of course, that x and y cannot be negative numbers,
or in other words, that we were working in the first quadrant,
so we could have added x%3E=0 and y%3E=0 as constraints for a total of
system%28x%3E=0%2Cy%3E=0%2Cx%2By%3C=18%2Cy%3E=3%2Cx%3E=7%2Cx=2y%29 .
However, once we graph the feasible region, we realize that it is triangle ABC,
which is completely specified with just
system%28x%2By%3C=18%2Cx%3E=7%2Cx%3E=2y%29 .

The coordinates for points A, B, and C can be fund by solving
For A: system%28x=7%2Cx=2y%29--->system%28x=7%2Cy=3.5%29
For B: system%28x%2By=18%2Cx=2y%29--->system%28x=12%2Cy=6%29
For C: system%28x%2By=18%2Cx=7%29--->system%28x=7%2Cy=11%29