# 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?

Jan 7, 2017

$d \in \left[\text{0 feet", "33 feet}\right]$

#### Explanation:

For starters, you can find the value of $d$ for which the pressure is at least $\text{1 atm}$ 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 $\text{1 atm}$ is

$d = \text{0 feet}$

Now, you know that you must have

${P}_{1} = 1 + {d}_{1} / 33 \ge \text{1 atm}$

The depth ${d}_{1}$ corresponds to a pressure of at least $\text{1 atm}$

and also

${P}_{2} = 1 + {d}_{2} / 33 \le \text{2 atm}$

The depth ${d}_{2}$ corresponds to a pressure of at most $\text{2 atm}$

From the first inequality, you get that

$1 + {d}_{1} / 33 = 1$

${d}_{1} / 33 = 0 \implies {d}_{1} = \text{0 feet} \to$ matches what we got by using the definition of an atmosphere!

For the second inequality, you get that

$1 + {d}_{2} / 33 \le 2$

${d}_{2} / 33 \le 1 \implies {d}_{2} \le \text{33 feet}$

Therefore, you can say that the pressure is at least $\text{1 atm}$ and at most $\text{2 atm}$ at depths that vary between $\text{0 feet}$ and $\text{33 feet}$, respectively.

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

$1 \le 1 + \frac{d}{33} \le 2$

Solve this to get

$0 \le \frac{d}{33} \le 1$

$o \le d \le 33$

Once again, you get $\text{0 feet}$ and $\text{33 feet}$ as the two values.