Question

Let’s think about another type of scenario. What if you were told that a bracelet requires 10 beads and 10 minutes to make while a necklace requires 20 beads and takes 40 minutes to make. The craftsman has 1000 beads to work with and he has 1600 minutes in which to work. If a bracelet costs $5 and a necklace costs $7.50, what is the maximum revenue that the craftsman can take in?

There are four inequalities in this situation. Let b be the number of bracelets made and n be the number of necklaces made. The system of inequalities for this situation is:

Answer 1

Hi, your question appears to be unclear. However, I inferred this to be a linear programming problem.

Answer:

$500

Explanation:

To begin we need to state the system of inequalities for this situation (constraints):

• For the number of available beads:
  `10b + 20n \leq 1000` where b represents the number of bracelets made, and n represents the number of necklaces made.

• For the available time to make the jewelry:
  `10b+40n\leq 1600`

• 
  `b\geq 0`

• 
  `n\geq 0`

From the question, we note we are required to find the maximum revenue that the craftsman can take in. In other words, the optimization equation is
`5b+7.5n` = maximum revenue (taking note that a bracelet costs $5 and a necklace costs $7.50)

Next, we are to plot the inequalities on a graph to determine the feasible region: Doing so should give us these main vertices: (0,0) (0,40) (40,30 (100,0).

By substituting the vertices into the optimization equation (replacing b, and n) we can determine which quantity gives the maximum revenue:

For (0,40) ⇒ 5(0) + 7.5(0) = $0

For (0,40) ⇒ 5(0) + 7.5 (40) = $300

For (40, 30) ⇒ 5(40) + 7.5 (30) = $425

For (100, 0) ⇒ 5(100)+7.50(0) = $500

We notice that at point (100, 0) we have a maximum revenue of $500.