A conical vessel is to be prepared out of a circular sheet of metal of unit radius in order that the vessel has maximum value, the sectorial area that must be removed from the sheet is `A_(1)` and the area of the given sheet is `A_(2)`, then `A_(2)/A_(1)` is equal to

Options

A. `3+sqrt6`
B. `7+sqrt5`
C. `2+sqrt5`
D. `3 + sqrt2`

Answer

To solve the problem of finding the ratio \( \frac{A_2}{A_1} \) where \( A_2 \) is the area of the circular sheet of metal and \( A_1 \) is the area of the sector that must be removed to create a conical vessel with maximum volume, we will follow these steps:

### Step 1: Calculate the area of the circular sheet \( A_2 \)

The area \( A_2 \) of a circular sheet with a unit radius is given by the formula for the area of a circle:

\[ A_2 = \pi R^2 \]

Since the radius \( R = 1 \):

\[ A_2 = \pi (1)^2 = \pi \]

### Step 2: Set up the volume of the cone

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

\[ V = \frac{1}{3} \pi r^2 h \]

Where \( r \) is the radius of the base of the cone and \( h \) is the height of the cone.

### Step 3: Relate the radius \( r \), height \( h \), and slant height \( l \)

Using the Pythagorean theorem, we have:

\[ l^2 = r^2 + h^2 \]

Since the slant height \( l \) is equal to the radius of the circular sheet, which is 1, we can write:

\[ 1^2 = r^2 + h^2 \implies r^2 + h^2 = 1 \]

### Step 4: Express \( r^2 \) in terms of \( h \)

From the equation above, we can express \( r^2 \) as:

\[ r^2 = 1 - h^2 \]

### Step 5: Substitute \( r^2 \) into the volume formula

Substituting \( r^2 \) into the volume formula gives:

\[ V = \frac{1}{3} \pi (1 - h^2) h = \frac{\pi}{3} (h - h^3) \]

### Step 6: Differentiate the volume with respect to height \( h \)

To find the maximum volume, we differentiate \( V \) with respect to \( h \):

\[ \frac{dV}{dh} = \frac{\pi}{3} (1 - 3h^2) \]

### Step 7: Set the derivative to zero to find critical points

Setting the derivative equal to zero:

\[ 1 - 3h^2 = 0 \implies 3h^2 = 1 \implies h^2 = \frac{1}{3} \implies h = \frac{1}{\sqrt{3}} \]

### Step 8: Calculate the corresponding radius \( r \)

Substituting \( h \) back into the equation for \( r^2 \):

\[ r^2 = 1 - h^2 = 1 - \frac{1}{3} = \frac{2}{3} \implies r = \sqrt{\frac{2}{3}} \]

### Step 9: Calculate the curved surface area of the cone

The curved surface area \( A_c \) of the cone is given by:

\[ A_c = \pi r l \]

Where \( l = 1 \):

\[ A_c = \pi \left(\sqrt{\frac{2}{3}}\right)(1) = \pi \sqrt{\frac{2}{3}} \]

### Step 10: Calculate the area of the sector \( A_1 \)

The area of the sector that must be removed is given by:

\[ A_1 = A_2 - A_c = \pi - \pi \sqrt{\frac{2}{3}} = \pi \left(1 - \sqrt{\frac{2}{3}}\right) \]

### Step 11: Calculate the ratio \( \frac{A_2}{A_1} \)

Now we can find the ratio:

\[ \frac{A_2}{A_1} = \frac{\pi}{\pi \left(1 - \sqrt{\frac{2}{3}}\right)} = \frac{1}{1 - \sqrt{\frac{2}{3}}} \]

### Step 12: Rationalize the denominator

To rationalize the denominator:

\[ \frac{1}{1 - \sqrt{\frac{2}{3}}} \cdot \frac{1 + \sqrt{\frac{2}{3}}}{1 + \sqrt{\frac{2}{3}}} = \frac{1 + \sqrt{\frac{2}{3}}}{1 - \frac{2}{3}} = \frac{1 + \sqrt{\frac{2}{3}}}{\frac{1}{3}} = 3(1 + \sqrt{\frac{2}{3}}) \]

Finally, simplifying gives:

\[ \frac{A_2}{A_1} = 3 + 3\sqrt{\frac{2}{3}} = 3 + \sqrt{6} \]

### Final Answer

Thus, the ratio \( \frac{A_2}{A_1} \) is:

\[ \frac{A_2}{A_1} = 3 + \sqrt{6} \]