Question

The equation `P = 1 + d/33` gives the pressure, `P`, in atmospheres (atm) at a depth of `d` feet below sea level. For what depths `d` is the pressure at least 1 atm and at most 2 atm?

Answer 1

`d in ["0 feet", "33 feet"]`

Explanation:

For starters, you can find the value of `d` for which the pressure is at least `"1 atm"` by using the definition of an atmosphere.

As you know, the atmosphere is a unit of pressure defined as the pressure exerted by the Earth's atmosphere at sea level.

Since at sea level basically means a depth of `0` feet, you can say that the first value of `d` for which the pressure is at least `"1 atm"` is

`d = "0 feet"`

Now, you know that you must have

`P_1 = 1 + d_1/33 >= "1 atm"`

The depth `d_1` corresponds to a pressure of at least `"1 atm"`

and also

`P_2 = 1 + d_2/33 <= "2 atm"`

The depth `d_2` corresponds to a pressure of at most `"2 atm"`

From the first inequality, you get that

`1 + d_1/33 = 1`

`d_1/33 = 0 implies d_1 = "0 feet" ->` matches what we got by using the definition of an atmosphere!

For the second inequality, you get that

`1 + d_2/33 <= 2`

`d_2/33 <= 1 implies d_2 <= "33 feet"`

Therefore, you can say that the pressure is at least `"1 atm"` and at most `"2 atm"` at depths that vary between `"0 feet"` and `"33 feet"`, respectively.

Notice that you can also set up this as a compound inequality

`1 <= 1 + d/33 <= 2`

Solve this to get

`0 <= d/33 <= 1`

`o<= d <= 33`

Once again, you get `"0 feet"` and `"33 feet"` as the two values.