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?

1 Answer
Jan 7, 2017

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.