Question

Suppose that a factory manufactures only tables and chairs and that the profit on one chair is $15 and on the table is $20. Each chair requires 1 large piece and 2 small pieces. Each table requires 2 large pieces and 2 small pieces (see Figure 1 below). Finally, suppose that you have only 6 large pieces and 8 small pieces. How many chairs and how many tables should you build to maximize profit? Write a system of equations using inequalities and constraints

Answer 1

Answer:

The data we have is:

Profit per chair = $15

Profit per table = $20

Chair needs: 1 large pice, 2 small pieces.

Table needs: 2 large pieces, 2 small pieces.

You have 6 large pieces and 8 small pieces, and you want to maximize the profit.

Let's define:

T = # of tables

C = # of chairs.

Total profit:

P = C*$15 + M*$20

The number of chairs that you can make is given by:

0 ≤ C ≤ 4

If you make 4 chairs, the profit is:

P = 4*$15 = $60

And there will be 2 large pieces left.

Now for the tables.

0 ≤ T ≤ 3

If you make 4 tables, the profit will be:

P = 3*$20 = $60 (same as before)

And there will be two small pieces left.

Then we want to use a medium number of tables and chairs, we can not use the maximum for any of those, the first thing we should try is:

we ideally would want to use all the pieces that we have.

8 small, 6 large:

Then we can make:

2 tables: we use 4 large pieces and 4 small pieces.

T = 2

And we have left 2 large pieces and 4 small pieces.

The leftover is enough to make two chairs.

C = 2.

In this case, where we used all our materials, the profit will be:

P = 2*$15 + 2*$20 = $30 + $40 = $70

The profit is maximized when we make 2 tables and two chairs.