Question

1. chiang is filling a 50 ft³ container with water at a rate of 0.5 ft³/min. interpret the key features for this situation.

Explanation:

Step1: Identify the type of problem

This is a problem about filling a container with water, which involves rates and volumes. We can model it using a linear function. Let \( t \) be the time in minutes and \( V \) be the volume of water in cubic feet. The rate of filling is \( 0.5\space ft^3/min \), so the volume as a function of time is \( V(t)=0.5t \), with the constraint that \( V(t)\leq50 \) (since the container has a volume of \( 50\space ft^3 \)).

Step2: Analyze the key features

• Initial value (y - intercept): When \( t = 0 \) (at the start), the volume of water \( V(0)=0.5\times0 = 0\space ft^3 \). So the initial volume of water in the container is \( 0\space ft^3 \).
• Rate of change (slope): The rate at which the container is being filled is \( 0.5\space ft^3/min \). This is the slope of the linear function \( V(t)=0.5t \), which means the volume of water in the container increases by \( 0.5\space cubic\space feet \) for each minute that passes.
• Maximum time to fill the container: We want to find when the container is full, i.e., when \( V(t)=50 \). Solving \( 0.5t = 50 \) for \( t \), we divide both sides by \( 0.5 \): \( t=\frac{50}{0.5}=100 \) minutes. So it will take 100 minutes to fill the 50 - cubic - foot container at a rate of \( 0.5\space ft^3/min \). Also, the domain of the function (time) is \( 0\leq t\leq100 \) and the range (volume) is \( 0\leq V\leq50 \).

Answer:

• Initial volume: \( 0\space ft^3 \) (at \( t = 0 \)).
• Filling rate: \( 0.5\space ft^3 \) per minute (slope of the volume - time function).
• Time to fill: 100 minutes (when \( V = 50\space ft^3 \), \( t=\frac{50}{0.5}=100 \) min).
• Domain (time): \( 0\leq t\leq100 \) minutes.
• Range (volume): \( 0\leq V\leq50\space ft^3 \).