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

Answer:

#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.