An open cylindrical can has to be made with `100m^(2)` of tin. If its volume is maximum, then the ratio of its base radius and the height is

Options

A. `1:1`
B. `2:1`
C. `1:3`
D. `sqrt2:1`

Answer

To solve the problem of maximizing the volume of an open cylindrical can made with a surface area of \(100 \, m^2\), we can follow these steps:

### Step 1: Define the variables and the surface area equation
Let:
- \( r \) = radius of the base of the cylinder
- \( h \) = height of the cylinder

The surface area \( S \) of the open cylinder (which includes the curved surface area and the base) is given by:
\[ S = 2\pi rh + \pi r^2 \]
Given that the surface area is \(100 \, m^2\), we can write:
\[ 2\pi rh + \pi r^2 = 100 \]

### Step 2: Express height \( h \) in terms of radius \( r \)
From the surface area equation, we can isolate \( h \):
\[ 2\pi rh = 100 - \pi r^2 \]
\[ h = \frac{100 - \pi r^2}{2\pi r} \]

### Step 3: Write the volume equation
The volume \( V \) of the cylinder is given by:
\[ V = \pi r^2 h \]
Substituting the expression for \( h \) from Step 2:
\[ V = \pi r^2 \left(\frac{100 - \pi r^2}{2\pi r}\right) \]
\[ V = \frac{r(100 - \pi r^2)}{2} \]
\[ V = \frac{100r - \pi r^3}{2} \]

### Step 4: Differentiate the volume function
To find the maximum volume, we differentiate \( V \) with respect to \( r \) and set the derivative to zero:
\[ \frac{dV}{dr} = \frac{1}{2}(100 - 3\pi r^2) \]
Setting the derivative equal to zero:
\[ 100 - 3\pi r^2 = 0 \]
\[ 3\pi r^2 = 100 \]
\[ r^2 = \frac{100}{3\pi} \]
\[ r = \sqrt{\frac{100}{3\pi}} = \frac{10}{\sqrt{3\pi}} \]

### Step 5: Find the corresponding height \( h \)
Substituting \( r \) back into the expression for \( h \):
\[ h = \frac{100 - \pi \left(\frac{100}{3\pi}\right)}{2\pi \left(\frac{10}{\sqrt{3\pi}}\right)} \]
\[ h = \frac{100 - \frac{100}{3}}{2\pi \left(\frac{10}{\sqrt{3\pi}}\right)} \]
\[ h = \frac{\frac{200}{3}}{2\pi \left(\frac{10}{\sqrt{3\pi}}\right)} \]
\[ h = \frac{10\sqrt{3\pi}}{3} \]

### Step 6: Find the ratio of \( r \) to \( h \)
Now, we need to find the ratio \( \frac{h}{r} \):
\[ \frac{h}{r} = \frac{\frac{10\sqrt{3\pi}}{3}}{\frac{10}{\sqrt{3\pi}}} \]
\[ \frac{h}{r} = \frac{10\sqrt{3\pi}}{3} \cdot \frac{\sqrt{3\pi}}{10} \]
\[ \frac{h}{r} = \frac{3}{3} = 1 \]

### Conclusion
Thus, the ratio of the base radius \( r \) to the height \( h \) is:
\[ \frac{h}{r} = 1 \quad \text{or} \quad h : r = 1 : 1 \]