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

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

Step-by-step explanation:

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

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}