Can You Balance Two Jobs and Meet Your Weekly Financial Goals?

[rebo1984]

You can work a total of no more than 41 hours each week at your two
jobs. Housecleaning pays \$6 per hour and your sales job pays \$9 per hour. You
need to earn at least \$252 each week to pay your bills.

Are the answers:

x+y\le 41

6x+9y\ge252

How would you solve this?

Thanks

 

[MarkFL]

To solve a system of linear inequalities, such as the system we have here:

\(\displaystyle x+y\le41\)

\(\displaystyle 2x+3y\ge84\) (I divided through by 3 to make the numbers smaller)

We observe that the variables represent the number of hours worked, and so we are only interested in non-negative values, so our solution set will be in quadrant I, or along the positive axes.

We begin with the first inequality and consider the equation:

\(\displaystyle x+y=41\)

This line will be the boundary of the solution set of the inequality. To plot the graphs of a line, I like to arrange it in slope-intercept form $y=mx+b$:

\(\displaystyle y=-x+41\)

We immediately see the point $(0,41)$ is on the line. Plot that point. Now, since the slope is -1, we may go one unit to the right and one unit down to the point $(1,40)$. Plot that point, and since the inequality is a weak one, we plot the solid line passing through the two points we plotted.

Now, to see which side of the line the solution set exists, we check a point not on the line to see if it satisfies the inequality or not. It if satisfies the inequality, then we know the solution set is on the same side of the line as our test point. Let's use the origin $(0,0)$...

\(\displaystyle 0+0\le41\)

\(\displaystyle 0\le41\quad\checkmark\)

This point satisfies the inequality, so we shade underneath the line in the first quadrant up to and including the positive axes:

[DESMOS=-3.332076962397118,46.32304769391533,-4.665236823966467,44.98988783234599]x+y\le41\left\{0\le x\right\}\left\{0\le y\right\}[/DESMOS]

Okay for the second inequality, we write the equation:

\(\displaystyle 2x+3y=84\)

Write in slope-intercept form:

\(\displaystyle y=-\frac{2}{3}x+28\)

Plot the $y$-intercept $(0,28)$, then move three units to the right and two units down to the point $(2,25)$ and plot that point, then connect those points with a solid line (weak inequality) and extend them to the positive axes. We'll use the origin again as our test point:

\(\displaystyle 2(0)+3(0)\ge84\)

\(\displaystyle 0\ge84\quad\xcancel{\checkmark}\)

This point does not satisfy the inequality, so we shade above the line:

[DESMOS=-2.1314516419265574,44.60278332872047,-9.23991795329183,37.49431701735519]2x+3y\ge84\left\{0\le x\right\}\left\{0\le y\right\}[/DESMOS]

We are interested in the set of points that satisfy both inequalities simultaneously, so our plot becomes:

[DESMOS=-1,40,-1,42]0\le x\left\{2x+3y\ge84\right\}\left\{x+y\le41\right\}[/DESMOS]