A conical vessel is to be prepared out of a circular sheet of gold of unit radius. How much sectorial area is to be removed from the sheet so that the vessel has maximum volume?

Answer

To solve the problem of determining how much sectorial area is to be removed from a circular sheet of gold of unit radius to maximize the volume of a conical vessel, we can follow these steps:

### Step 1: Define the Variables
Let:
- The radius of the cone be \( r \).
- The height of the cone be \( h \).
- The slant height of the cone be \( l \).
- The angle of the sector removed be \( 2\theta \).

### Step 2: Relate the Variables
Since the circular sheet has a radius of 1 unit, the slant height \( l \) of the cone is also 1 unit. The relationship between the height, radius, and slant height of the cone is given by the Pythagorean theorem:
\[ l^2 = h^2 + r^2 \]
Substituting \( l = 1 \):
\[ 1 = h^2 + r^2 \implies h = \sqrt{1 - r^2} \]

### Step 3: Find the Volume of the Cone
The volume \( V \) of the cone is given by:
\[ V = \frac{1}{3} \pi r^2 h \]
Substituting \( h \):
\[ V = \frac{1}{3} \pi r^2 \sqrt{1 - r^2} \]

### Step 4: Differentiate the Volume
To find the maximum volume, we differentiate \( V \) with respect to \( r \):
\[ \frac{dV}{dr} = \frac{1}{3} \pi \left( 2r \sqrt{1 - r^2} + r^2 \cdot \frac{-r}{\sqrt{1 - r^2}} \right) \]
This simplifies to:
\[ \frac{dV}{dr} = \frac{\pi}{3} r \left( 2\sqrt{1 - r^2} - \frac{r^2}{\sqrt{1 - r^2}} \right) \]

### Step 5: Set the Derivative to Zero
Setting \( \frac{dV}{dr} = 0 \):
\[ r \left( 2\sqrt{1 - r^2} - \frac{r^2}{\sqrt{1 - r^2}} \right) = 0 \]
This gives two cases:
1. \( r = 0 \)
2. \( 2\sqrt{1 - r^2} - \frac{r^2}{\sqrt{1 - r^2}} = 0 \)

### Step 6: Solve for \( r \)
From the second case:
\[ 2(1 - r^2) = r^2 \implies 2 - 2r^2 = r^2 \implies 3r^2 = 2 \implies r^2 = \frac{2}{3} \implies r = \sqrt{\frac{2}{3}} \]

### Step 7: Find the Corresponding \( \theta \)
Using the relationship \( r = 1 - \frac{\theta}{\pi} \):
\[ \theta = \pi(1 - r) = \pi\left(1 - \sqrt{\frac{2}{3}}\right) \]

### Step 8: Calculate the Sector Area to be Removed
The area of the sector removed is given by:
\[ \text{Area} = \frac{1}{2} \cdot r^2 \cdot \text{angle} = \frac{1}{2} \cdot 1^2 \cdot 2\theta = \theta \]
Substituting the value of \( \theta \):
\[ \text{Area} = \pi\left(1 - \sqrt{\frac{2}{3}}\right) \]

### Final Answer
The sectorial area to be removed from the circular sheet is:
\[ \text{Area} = \pi\left(1 - \sqrt{\frac{2}{3}}\right) \text{ square units.} \]