Example3.1.23

When a stopwatch started, the pressure inside a gas container was \(4.2\) atm (standard atmospheric pressure). As the container was heated, the pressure increased by \(0.7\) atm per minute. The maximum pressure the container can handle was \(21.7\) atm. Heating must be stopped once the pressure reaches \(21.7\) atm. In what time interval was the container safe?

The pressure increases by \(0.7\) atm per minute, so it increases by \(0.7m\) after \(m\) minutes. Counting in the original pressure of \(4.2\) atm, pressure in the container can be modeled by \(0.7m+4.2\text{,}\) where \(m\) is the number of minutes since the stop watch started.

The container is safe when the pressure is \(21.7\) atm or lower. We can write and solve this inequality:

\begin{align*} 0.7m+4.2\amp\leq21.7\\ 0.7m+4.2\subtractright{4.2}\amp\leq21.7\subtractright{4.2}\\ 0.7m\amp\leq17.5\\ \divideunder{0.7m}{0.7}\amp\leq\divideunder{17.5}{0.7}\\ m\amp\leq25 \end{align*}

In summary, the container was safe as long as \(m\leq25\text{.}\) Assuming that \(m\) also must be greater than or equal to zero, this means \(0\leq m\leq 25\text{.}\) We can write this as the time interval as \([0,25]\text{.}\) Thus the container was safe between 0 minutes and 25 minutes.

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