Word problem - college algebra

Question

A furniture factory makes wooden tables, chairs, and armoires. Each piece of furniture requires three operations: cutting the wood, assembling, and finishing. Each operation requires the number of hours given in the table. The workers in the factory can provide 260 h of cutting, 380 h of assembling, and 560 h of finishing each work week. How many tables, chairs, and armoires should be produced so that all available labor-hours are used? Or is this impossible? (If not possible, enter IMPOSSIBLE.)

T C A

Cutting: 0.5 1 1

Assembling: 0.5 1.5 1

Finishing: 1 1.5 2

Answer 1

Hi Eric, let's do this!

Let the numbers of chairs, tables and armoires produced be C, T, and A, respectively.

Then, from the maximum numbers of hours for each process, we have:

0.5T + C + A = 260 . . . . . . . . (1)

0.5T + 1.5C + A = 380 . . . . . . . . (2)

T + 1.5C + 2A = 560 . . . . . . . .(3)

(2) - (1): 0.5C = 120 => C = 240

Substitute in (1): 0.5T + A = 20 . . . . (4)

Substitute in (2); 0.5T + A = 20 . . . .(5)

Since (4) and (5) are the same equation, there are many solutions.

For example, take A = 0. Then, from (4): 0.5T = 20 => T = 40

So, a solution is (C, T, A) = (240, 40, 0)

Now take A= 10. Then from (4): 0.5T = 10 => T = 20.

So, another solution is (C, T, A) = (240, 20, 10)

And so on.

In fact, for any integer value of A between 0 and 20, inclusive, we have a valid solution.