Chapter 5.7, Problem 48E

a.

To determine

To write: a system of four linear inequalities that represents the number x of necklaces and number y of key chains that we can make and graph it.

Answer to Problem 48E

  $\{\begin{array}{l}0.5x+0.25y\le 20\\ 2x+3y\le 120\\ x\ge 0\\ y\ge 0\end{array}$

Explanation of Solution

Since time required in making 1 necklace is 0.5 hours

Then time required in making $x$ necklaces is $0.5x$

Again, since time required in making 1 key chain is 0.25 hours

Then time required in making $y$ key chains is $0.25y$

Since available time is 20 hours

Then possible inequality will be $0.5x+0.25y\le 20$

Similarly,

The other possible inequality can be written as $2x+3y\le 120$

Now the number of necklaces ( $x$ ) and key chains ( $y$ ) can be 0 but never negative

Then, $x\ge 0\text{ and }y\ge 0$

Hence, the required set of inequalities is given by

  $\{\begin{array}{l}0.5x+0.25y\le 20\\ 2x+3y\le 120\\ x\ge 0\\ y\ge 0\end{array}$

And its graph is as follows-

  

The shaded region is the required graph of set of inequalities.

b.

To determine

To write: the vertices (corner points) of solution graph of system of inequalities find in sub-part (a)

Explanation of Solution

It is clear from the graph the vertices (corner points) are (0, 40), (30, 20), (40, 0), (0, 0).

c.

To determine

To find: the revenue(R) for each order pair find in part (b).

Answer to Problem 48E

For (0, 40), R=$320

For (30, 20), R=$460

For (40, 0), R=$400

For (0, 0), R=$0

The vertex (30, 20) results in maximum revenue.

Explanation of Solution

The revenue R is given by equation $R=10x+8y$

For vertex (0, 40) the value of R is given by, $R=10x+8y$

  $\Rightarrow R=10\times 0+8\times 40=0+320=320$

For vertex (30, 20) the value of R is given by, $R=10x+8y$

  $\Rightarrow R=10\times 30+8\times 20=300+160=460$

For vertex (40, 0) the value of R is given by, $R=10x+8y$

  $\Rightarrow R=10\times 40+8\times 0=400+0=400$

For vertex (0, 0) the value of R is given by, $R=10x+8y$

  $\Rightarrow R=10\times 0+8\times 0=0+0=0$

Here the maximum revenue is $460 for order pair (30, 20).