Let's go through this problem step-by-step to make it easier to understand.

Problem Breakdown

The question involves a lake where:

1. Fish (baars) can live if the water temperature is below 25°C and there is at least 2 mg of oxygen per liter of water.
2. At the start of a hot summer on July 1 (t = 0), the temperature is 25°C at the surface (depth $d = 0$) and gets deeper as days progress.
3. The depth at which the temperature reaches 25°C increases by 10 cm every three days (or approximately 1/30 meters per day).
4. Oxygen levels also change, with a low-oxygen layer expanding from the bottom upwards by 5 cm each day (1/20 meters per day).

The variables in this problem are:

• $d_T(t)$: Depth at which temperature reaches 25°C after $t$ days.
• $d_Z(t)$: Depth at which there is sufficient oxygen for the fish after $t$ days.

Step-by-Step Explanation

Part (a): Formula for $d_T(t)$

Since the temperature boundary (25°C) starts at the surface on July 1 and moves downward at a rate of 0.1 meters every 3 days, we calculate the daily depth increase: $d_T = \frac{0.1}{3} = \frac{1}{30} \text{ meters per day}$ So the formula for $d_T(t)$ after $t$ days is: $d_T(t) = \frac{1}{30}t$

Part (b): Formula for $d_Z(t)$

The sufficient oxygen depth starts from the lake bottom (4 meters) and increases upward by 0.05 meters (5 cm) per day. This means: $d_Z = -\frac{0.05}{1} = -\frac{1}{20} \text{ meters per day}$ Thus, the formula for $d_Z(t)$ after $t$ days is: $d_Z(t) = -\frac{1}{20}t + 4$

Part (c): Graph Interpretation

The graph plots $d_T(t)$ and $d_Z(t)$ over time. By plotting these, we observe how the depth levels change.

• For $d_T(t) = \frac{1}{30}t$: This line starts at 0 and increases with a slope of $\frac{1}{30}$.
• For $d_Z(t) = -\frac{1}{20}t + 4$: This line starts at 4 meters and decreases with a slope of $-\frac{1}{20}$.

Finding the Intersection

To find when the fish can no longer survive in the lake, set $d_T(t) = d_Z(t)$: $\frac{1}{30}t = -\frac{1}{20}t + 4$ Multiplying through by 60 to clear the fractions: $2t + 3t = 240$ $5t = 240$ $t = 48$ Thus, after 48 days (August 18), the fish can no longer survive because the temperature and oxygen levels intersect at the same depth.

Summary of Key Points

1. $d_T(t) = \frac{1}{30}t$
2. $d_Z(t) = -\frac{1}{20}t + 4$
3. Intersection occurs at $t = 48$ days, which is August 18.

Would you like further clarification on any specific part, or have questions on how to plot these functions?

Here are five questions to consider:

1. Why does the rate of depth increase differ for temperature and oxygen levels?
2. What would happen if the rates for $d_T(t)$ and $d_Z(t)$ were equal?
3. How could we adjust the formulas if the rates of depth change were altered?
4. What other factors might affect the survival of fish in the lake?
5. Could similar formulas apply to other environmental scenarios?

Tip: Always check units and rates when working with real-life applications, as small conversions (like cm to meters) are common sources of error.