Jose runs a factory that makes stereo tuners. Each S100 takes 8 ounces of plastic and 4 ounces of metal. Each FS20 requires 4 ounces of plastic and 6 ounces of metal. The factory has 312 ounces of plastic, 372 ounces of metal available, with a maximum of 20 S100 that can be built each week. If each S100 generates $7 in profit, and each FS20 generates $13, how many of each of the stereo tuners should Jose have the factory make each week to make the most profit

Since each S100 takes 8 ounces of plastic and 4 ounces of metal. Each FS20 requires 4 ounces of plastic and 6 ounces of metal. And the factory has 312 ounces of plastic, 372 ounces of metal available, then,

For plastic

8 ounces + 4 ounces = 12 ounces

The number of stereo tuners it can produce will be

312/12 = 26 stereo tuners

For metal

4 ounces + 6 ounces = 10 ounces

The number of stereo tuners it can produce will be

372/10 = 37.2 = 37 approximately

Since FS20 generate more profit than S100, let assume that Jose produces 50 FS20 by consuming

4 × 50 = 200 ounces of plastic

6 × 50 = 300 ounces of metal

The remaining plastic will be

312 - 200 = 112

The remaining plastic will be

372 - 300 = 72

Divide 112 by 8 in order to make S100

112/8 = 14

Also 72/4 = 18.

Therefore, Jose should have the factory make 50 FS20 stereo tuners and 14 S100 stereo tuners each week to make the most profit