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

Answer

To solve the problem, we need to find the ratio \( \frac{A_2}{A_1} \), where \( A_2 \) is the area of the circular sheet of metal and \( A_1 \) is the sectorial area that must be removed to form a conical vessel with maximum volume.

### Step 1: Define the areas

1. **Area of the circular sheet \( A_2 \)**:
The area \( A_2 \) of a circle with radius 1 is given by:
\[ A_2 = \pi r^2 = \pi (1^2) = \pi \]

2. **Sectorial area \( A_1 \)**:
The sectorial area \( A_1 \) that must be removed is related to the angle \( \theta \) of the sector. The formula for the area of a sector is:
\[ A_1 = \frac{1}{2} r^2 \theta \]
Since the radius \( r \) is 1, we have:
\[ A_1 = \frac{1}{2} (1^2) \theta = \frac{1}{2} \theta \]

### Step 2: Relate the angle \( \theta \) to the cone's dimensions

To form the cone, we need to relate the angle \( \theta \) to the radius \( r_c \) of the cone's base and its height \( h \).

Using the Pythagorean theorem in the right triangle formed by the radius of the cone, the height, and the slant height (which is the radius of the original circle, equal to 1):
\[ 1^2 = r_c^2 + h^2 \implies r_c^2 = 1 - h^2 \]

### Step 3: Find the volume of the cone

The volume \( V \) of the cone is given by:
\[ V = \frac{1}{3} \pi r_c^2 h = \frac{1}{3} \pi (1 - h^2) h \]

### Step 4: Maximize the volume

To find the maximum volume, we differentiate \( V \) with respect to \( h \) and set the derivative to zero:
\[ \frac{dV}{dh} = \frac{\pi}{3} (1 - 3h^2) = 0 \]
Solving for \( h \):
\[ 1 - 3h^2 = 0 \implies h^2 = \frac{1}{3} \implies h = \frac{1}{\sqrt{3}} \]

### Step 5: Find \( r_c \)

Substituting \( h \) back to find \( r_c \):
\[ r_c^2 = 1 - h^2 = 1 - \frac{1}{3} = \frac{2}{3} \implies r_c = \sqrt{\frac{2}{3}} \]

### Step 6: Find the angle \( \theta \)

The arc length \( AB \) of the sector is equal to the circumference of the base of the cone:
\[ \text{Arc length} = 2\pi r_c = 2\pi \sqrt{\frac{2}{3}} \]
Since the arc length is also given by \( r \theta \) where \( r = 1 \):
\[ \theta = 2\pi \sqrt{\frac{2}{3}} \]

### Step 7: Calculate \( A_1 \)

Substituting \( \theta \) into the formula for \( A_1 \):
\[ A_1 = \frac{1}{2} \theta = \frac{1}{2} \left( 2\pi \sqrt{\frac{2}{3}} \right) = \pi \sqrt{\frac{2}{3}} \]

### Step 8: Find the ratio \( \frac{A_2}{A_1} \)

Now we can find the ratio:
\[ \frac{A_2}{A_1} = \frac{\pi}{\pi \sqrt{\frac{2}{3}}} = \frac{1}{\sqrt{\frac{2}{3}}} = \sqrt{\frac{3}{2}} = \frac{\sqrt{3}}{\sqrt{2}} \]

Thus, the final answer is:
\[ \frac{A_2}{A_1} = \frac{\sqrt{3}}{\sqrt{2}} \]