###### 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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