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Let's break down and solve the problems step by step.

### 4.1.1: Show that the height of the cylindrical tank is \( h = \frac{75}{\pi r} - r \)

The total surface area of the cylindrical tank includes the lateral surface area and the two circular ends. For a cylinder:

- **Lateral surface area** = \( 2\pi r h \)
- **Area of the two circular ends** = \( 2 \times \pi r^2 = 2\pi r^2 \)
  
The total surface area is given as 150 m², so we can write the equation:

\[
2\pi r h + 2\pi r^2 = 150
\]

Dividing both sides by \( 2\pi r \) to isolate \( h \):

\[
h + r = \frac{75}{\pi r}
\]

Finally, subtract \( r \) from both sides:

\[
h = \frac{75}{\pi r} - r
\]

This shows the result as required.

---

### 4.1.2: Show that the volume of the cylindrical tank is \( V = 75r - \pi r^3 \)

The volume \( V \) of a cylinder is given by the formula:

\[
V = \pi r^2 h
\]

We already have the expression for \( h \) from 4.1.1 as \( h = \frac{75}{\pi r} - r \). Substituting this into the volume formula:

\[
V = \pi r^2 \left( \frac{75}{\pi r} - r \right)
\]

Simplifying the expression:

\[
V = \pi r^2 \left( \frac{75}{\pi r} \right) - \pi r^2 \cdot r
\]

\[
V = 75r - \pi r^3
\]

This confirms the required volume expression.

---

### 4.1.3: Calculate the maximum volume of the cylindrical tank

To find the maximum volume, we need to differentiate the volume function \( V = 75r - \pi r^3 \) with respect to \( r \), and then set the derivative equal to zero.

The derivative of \( V \) with respect to \( r \) is:

\[
\frac{dV}{dr} = 75 - 3\pi r^2
\]

Setting this equal to zero to find the critical points:

\[
75 - 3\pi r^2 = 0
\]

Solving for \( r \):

\[
3\pi r^2 = 75
\]
\[
r^2 = \frac{75}{3\pi}
\]
\[
r^2 = \frac{25}{\pi}
\]
\[
r = \frac{5}{\sqrt{\pi}}
\]

To confirm this is a maximum, we can check the second derivative:

\[
\frac{d^2V}{dr^2} = -6\pi r
\]

Since \( \frac{d^2V}{dr^2} < 0 \), the volume is maximized when \( r = \frac{5}{\sqrt{\pi}} \).

Now substitute this value of \( r \) back into the volume equation to find the maximum volume.

---

### 4.1.4: Show that the height of the rectangular prism tank with a square base is closed top and bottom

The total surface area for the rectangular prism with a square base, assuming length \( L = B = x \), is given by:

\[
\text{Surface Area} = 2x^2 + 4xh
\]

The total surface area is 150 m², so we can write the equation:

\[
2x^2 + 4xh = 150
\]

Solving for \( h \):

\[
4xh = 150 - 2x^2
\]
\[
h = \frac{150 - 2x^2}{4x}
\]

This is the required height of the rectangular prism tank.

---

Let me know if you want more detailed steps or further clarifications.

### Related Questions:
1. How do we apply calculus to find maximum and minimum values in geometry problems?
2. Why do we differentiate the volume formula to find the maximum volume?
3. How do the surface area and volume formulas change if the tank had a different shape, like a cone?
4. Can we generalize the method for maximizing the volume for any surface area constraint?
5. How does the relationship between the surface area and volume affect the efficiency of the tank's design?

### Tip:
Always remember to check if a critical point is a maximum or a minimum by evaluating the second derivative after setting the first derivative equal to zero.

A company has building material which can cover a total surface area of 150m² to build a water tank. They can choose from a cylindrical or rectangular prism tank. Tasks include showing the height of the cylindrical tank, calculating the volume, finding the maximum volume, and determining the height of the rectangular prism tank.

Learn how to calculate and maximize the volume of cylindrical and rectangular water tanks with given surface area constraints. This problem covers geometry concepts such as surface area, volume, and the application of differentiation for optimization. Suitable for Grade 10-12 students.

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