Of all the closed cylinderical cans (right - circular), which enclose a given volume of :
(i) 100 cubic centimeters
(ii) `128pi` cubic centimeters,
find the dimensions of the can, which has the minimum surface area.

Answer

To solve the problem of finding the dimensions of a closed cylindrical can that encloses a given volume with minimum surface area, we will break down the solution into steps for both parts of the question.

### Part (i): Volume = 100 cubic centimeters

1. **Write the formula for the volume of a cylinder**:
\[ V = \pi r^2 h \]
Given that \( V = 100 \), we have:
\[ \pi r^2 h = 100 \]

2. **Express height \( h \) in terms of radius \( r \)**:
\[ h = \frac{100}{\pi r^2} \]

3. **Write the formula for the surface area \( A \) of the cylinder**:
\[ A = 2\pi rh + 2\pi r^2 \]
Substitute \( h \) from step 2 into the surface area formula:
\[ A = 2\pi r \left(\frac{100}{\pi r^2}\right) + 2\pi r^2 \]
Simplifying this gives:
\[ A = \frac{200}{r} + 2\pi r^2 \]

4. **Differentiate the surface area \( A \) with respect to \( r \)**:
\[ \frac{dA}{dr} = -\frac{200}{r^2} + 4\pi r \]

5. **Set the derivative equal to zero to find critical points**:
\[ -\frac{200}{r^2} + 4\pi r = 0 \]
Rearranging gives:
\[ 4\pi r^3 = 200 \]
\[ r^3 = \frac{200}{4\pi} = \frac{50}{\pi} \]
\[ r = \left(\frac{50}{\pi}\right)^{1/3} \]

6. **Find the height \( h \)** using the value of \( r \)**:
Substitute \( r \) back into the equation for \( h \):
\[ h = \frac{100}{\pi \left(\left(\frac{50}{\pi}\right)^{1/3}\right)^2} \]
Simplifying gives:
\[ h = \frac{100}{\pi \cdot \left(\frac{50^{2/3}}{\pi^{2/3}}\right)} = \frac{100 \cdot \pi^{2/3}}{50^{2/3}} \]

### Part (ii): Volume = \( 128\pi \) cubic centimeters

1. **Write the formula for the volume of a cylinder**:
\[ V = \pi r^2 h \]
Given that \( V = 128\pi \), we have:
\[ \pi r^2 h = 128\pi \]

2. **Express height \( h \) in terms of radius \( r \)**:
\[ h = \frac{128}{r^2} \]

3. **Write the formula for the surface area \( A \) of the cylinder**:
\[ A = 2\pi rh + 2\pi r^2 \]
Substitute \( h \) from step 2 into the surface area formula:
\[ A = 2\pi r \left(\frac{128}{r^2}\right) + 2\pi r^2 \]
Simplifying this gives:
\[ A = \frac{256}{r} + 2\pi r^2 \]

4. **Differentiate the surface area \( A \) with respect to \( r \)**:
\[ \frac{dA}{dr} = -\frac{256}{r^2} + 4\pi r \]

5. **Set the derivative equal to zero to find critical points**:
\[ -\frac{256}{r^2} + 4\pi r = 0 \]
Rearranging gives:
\[ 4\pi r^3 = 256 \]
\[ r^3 = \frac{256}{4\pi} = \frac{64}{\pi} \]
\[ r = \left(\frac{64}{\pi}\right)^{1/3} \]

6. **Find the height \( h \)** using the value of \( r \)**:
Substitute \( r \) back into the equation for \( h \):
\[ h = \frac{128}{\left(\left(\frac{64}{\pi}\right)^{1/3}\right)^2} \]
Simplifying gives:
\[ h = \frac{128 \cdot \pi^{2/3}}{64^{2/3}} = 2 \cdot \pi^{2/3} \]

### Summary of Results:
- For **Part (i)**:
- Radius \( r = \left(\frac{50}{\pi}\right)^{1/3} \)
- Height \( h = \frac{100 \cdot \pi^{2/3}}{50^{2/3}} \)

- For **Part (ii)**:
- Radius \( r = \left(\frac{64}{\pi}\right)^{1/3} \)
- Height \( h = 2 \cdot \pi^{2/3} \)