SOLUTION: Mark is a student, and he can work for at most 20 hours a week. He needs to earn at least $75 to cover his weekly expenses. His dog-walking job pays $5 per hour and his job as a ca

Question 1010905: Mark is a student, and he can work for at most 20 hours a week. He needs to earn at least $75 to cover his weekly expenses. His dog-walking job pays $5 per hour and his job as a car wash assistant pays $4 per hour. Write a system of inequalities to model the situation, and graph the inequalities.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
x = number of hours as dog walker.
y = number of hours as car wash assistant.

x + y <= 20

5x + 4y >= 75

solve for y in the first equation to get y <= 20 - x

solve for y in the second equation to get y >= (75-5x)/4

you will graph the equality portion of these equations and then you will comply with the inequality portions of these equations.

graphing both y = 20 - x and y = (75-5x)/4, you will get what is shown below:

$$$

the line that intersects the x-axis at 15 is the revenue equation.

the line that intersects the x-axis at 20 is the hours equation.

the space between these 2 lines and to the right of the y-axis and above the x-axis is the feasible region.

this region allows him to make more than 75 dollars and keeps the total hours less than 20.

the value of the sum of x and y at each coordinate point has to be less than or equal to 20.

the value of the sum of 5x and 4y at each coordinate point has to be greater than 75 dollars or he will not have covered the minimum amount he needs to break even.

for example:

at the coordinate point of (x,y) = (5,8):
5+8 = 13 which is less than 20 so that's ok.
5*5 + 4*8 = 25 + 32 = 57 which is less than 75 so that's not ok.
he needs 75 dollars to break even.
as you can see, the point (5,8) is not in the feasible region so it would not satisfy all the requirements.

at the coordinate point of (x,y) = 15,8):
15 + 8 = 23 which is more than 20 so that is not ok.
the revenue is good but he worked more than 23 hours which is not allowed.
as you can see, the point (15,8) is also not in the feasible region so it would not satisfy all the requirements.

at the coordinate point of (x,y) = 10,8):
10 + 8 = 18 hours which is less than 20 so that's ok.
5*10 + 4*8 = 50 + 32 = 82 which is more than 75 so that's ok since he made more than 75 dollars and has a little left over.

if you look at the corner points of the feasible region, you should be able to find the maximum revenue he can make.

the corner points are at:

(0,20)
(0,18.75)
(15,0)
(20,0)

at (0,20), he makes 0 * 5 + 20 * 4 = 80 dollars.
at (0,18.75) he makes less than that.
at (15,0) he makes 15 * 5 + 0 * 4 = 75 dollars.
at (20,0) he makes 20 * 5 + 0 * 4 = 100 dollars.

he can make the most money when x = 20 and y = 0

that means he works all 20 hours at walking the dog and 0 hours at assisting the car washer.

your system of inequalities is:

x + y <= 20

5x + 4y >= 75

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