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.
* r^2