Q. 405 - Intermediate Algebra

Question

Solve Applications of Systems of Inequalities
In the following exercises, translate to a system of inequalities and solve.

Roxana makes bracelets and necklaces and sells them at the farmers’ market. She sells the bracelets for $\backslash (12$ each and the necklaces for $\backslash )18$ each. At the market next weekend she will have room to display no more than $40$ pieces, and she needs to sell at least $\$500$ worth in order to earn a profit.

(a) Write a system of inequalities to model this
situation.
(b) Graph the system.
(c) Should she display $26$ bracelets and $14$ necklaces?
(d) Should she display $39$ bracelets and $1$ necklace?

Step-by-Step Solution

Answer

(a) The system of inequalities to model the situation is $\left\{\begin{array}{l}x+y\le 40\\ 12x+18y\ge 500\end{array}\right.$

(b) The graph of the inequality is the overlapped region of both the inequalities:

(c) Yes, she should display $26$ bracelets and $14$ necklaces.

(d) No, she should not display $39$ bracelets and $1$ necklace.

Part (a) Step 1. Given

She sells bracelets for each $\$12$ and necklaces for each $\$18$.

She will have to display no more than $40$ pieces.

She needs to sell for at least $\$500$

To write a system of inequalities to model this
situation.

Part (a) Step 2. Form linear inequality

Let $x$ denote the number of bracelet and

$y$ denote the number of necklace.

She will have to display no more than $40$ pieces.

So, $x+y\le 40$

And She needs to sell for at least $\$500$ and the price of each bracelet is $\$12$ and that of necklace is $\$18$

$12x+18y\ge 500$

Part (b) Step 1. Graph the first inequality

Graph the  inequality $x+y=40$ 

The boundary line is a solid line since the inequality is $\le$.

Test $(0,0)$ and it is true so shade the region that contains the test point.

Part (b) Step 2. Graph the second inequality

Graph the equation $12x+18y=500$

The boundary line is solid line sine the inequality sign is $\ge$.

Test $(0,0)$ and it is false so shade the region that does not contain the test point.

Part (b) Step 3. Check the inequality

The solution of the inequality is the double shaded region.

Choose any point from the double shaded region and check the inequality

Choose $(0,40)$

Check the first inequality

$x+y\le 40$

$0+40\le 40$ is true.

Check the second inequality:

        $12x+18y\ge 500$

$12\left(0\right)+18\left(40\right)\ge 500$

                 $720\ge 500$ is true.

So the overlapped region is the solution to the system.

Part (c) Step 1. Given

Can she display $26$ bracelets and $14$ neclaces

Part (c) Step 2. Substitute the points in inequalities

Substitute $x=26$ and

                   $y=14$ in both the inequalities:

    $x+y\le 40$

$26+14\le 40$

        $40\le 40$ is true.

Then $12x+18y\ge 500$

$12\left(26\right)+18\left(14\right)\ge 500$

                   $564\ge 500$ is true.

Since both the inequalities are true, she can display $26$ bracelets and $14$ necklaces.

Pat (d) Step 1. Given

Can she display $39$ bracelets and $1$ necklace

Part (d) Step 2. Substitute the points in inequalities

Substitute $x=39$ and

                  $y=1$ in both the inequality,

  $x+y\le 40$

$39+1\le 40$

      $40\le 40$ is true.

And, $12x+18y\ge 500$

$12\left(39\right)+18\left(1\right)\ge 500$

                 $486\ge 500$ is false.

Since one of the inequality is true and the other one is false, she cannot display $39$ bracelets and $1$ necklace.