Finite Math is rocket science to me: PLEASE HELP! see problem below...?

Question

A small furniture company finishes two kinds of tables, A and B. Each table must be sanded, stained, and varnished. Table A requires 10 minutes of sanding, 6 minutes of staining, and 9 minutes of varnishing. Table B requires 5 minutes of sanding, 12 minutes of staining, and 9 minutes of varnishing. The profit is $5 on each A-type table and $3 on each B-type table. There is labor available for 450 minutes of sanding, 480 minutes of staining, and 450 minutes of varnishing each day. Formulate this problem as a linear programming problem and determine how many of each type of table should be made each day in order to maximize profits.

Answer 1

If you let a be the number of tables of type A and b be the number of type B, then the sanding requires that:
10a + 5b <= 450, or
2a + b <= 90

The staining requires:
6a + 12b <= 480, or
a + 2b <= 80

And the varnishing that:
9a + 9b <= 450, or
a + b <= 50

The total profit is 5a + 3b.

So you're trying to maximize 5a + 3b under the conditions:
1) 2a + b <= 90
2) a + 2b <= 80
3) a + b <= 50
4) a >= 0
5) b >= 0

(the last two come from the fact that you can't make a negative number of either type of table to gain resources and lose money)

With a problem of this magnitude, the easiest way to solve it is to graph it and then do a bit of algebra.

If you graph the basic lines, you'll discover that the allowable region's boundary is a polygon bounded by the intersections of 2 and 4, 2 and 3, 1 and 3, 1 and 5, and of course 4 and 5 (making no tables is not going to maximize your profit, though).

Intersection of 2 and 4:
a + 2b = 80 & a = 0
0 + 2b = 80
2b = 80
b = 40
(a, b) = (0, 40)

Intersection of 2 and 3:
a + 2b = 80 & a + b = 50
a = 50 - b
(50 - b) + 2b = 80
50 + b = 80
b = 30
a = 50 - 30
a = 20
(a, b) = (20, 30)

Intersection of 1 and 3:
2a + b = 90 && a + b = 50
b = 50 - a
2a + (50 - a) = 90
a + 50 = 90
a = 40
b = 50 - 40
b = 10
(a, b) = (40, 10)

Intersection of 1 and 5:
2a + b = 90 & b = 0
2a + 0 = 90
a = 45
(a, b) = (45, 0)

(We can ignore a = 0, b = 0, the intersection of 4 and 5 unless none of the others turns a positive profit).

Now we have the following:
a b Profit(5a + 3b)
0 40 120
20 30 190
40 10 230
45 0 225

Therefore, the best solution is to make 40 tables of type A and 10 tables of type B, thus realizing a profit of $230.