In a regular triangular prism the distance from the centre of one base to one of the vertices of the other base is e. The altitude of the prism for which the volume is greatest, is :

Options

A. a. `(e) /(2)`
B. b. `(e) /(sqrt(3))`
C. c. `(e)/(3)`
D. d. `(e)/(4)`

Answer

To find the altitude of a regular triangular prism that maximizes its volume, given that the distance from the center of one base to a vertex of the other base is \( e \), we can follow these steps:

### Step 1: Understand the Geometry of the Prism
A regular triangular prism consists of two equilateral triangular bases connected by three rectangular faces. The distance from the center of one base to a vertex of the other base is given as \( e \).

### Step 2: Define Variables
Let:
- \( h \) = altitude (height) of the prism
- \( a \) = side length of the equilateral triangle base
- \( D \) = distance from the center of one base to a vertex of the other base

### Step 3: Relate \( D \), \( a \), and \( h \)
In a regular triangular prism, the distance \( D \) can be expressed using the Pythagorean theorem:
\[ D^2 = h^2 + \left(\frac{a}{2}\right)^2 \]
Given that \( D = e \), we can write:
\[ e^2 = h^2 + \left(\frac{a}{2}\right)^2 \]
This simplifies to:
\[ h^2 = e^2 - \left(\frac{a}{2}\right)^2 \]

### Step 4: Calculate the Area of the Base
The area \( A \) of an equilateral triangle with side length \( a \) is given by:
\[ A = \frac{\sqrt{3}}{4} a^2 \]

### Step 5: Express Volume \( V \) of the Prism
The volume \( V \) of the prism is given by:
\[ V = A \cdot h = \frac{\sqrt{3}}{4} a^2 h \]

### Step 6: Substitute \( h \) in Terms of \( a \) and \( e \)
From the earlier relation, substitute \( h \):
\[ V = \frac{\sqrt{3}}{4} a^2 \sqrt{e^2 - \left(\frac{a}{2}\right)^2} \]

### Step 7: Differentiate the Volume with Respect to \( a \)
To find the maximum volume, we need to differentiate \( V \) with respect to \( a \) and set the derivative to zero:
\[ \frac{dV}{da} = 0 \]

### Step 8: Solve for Critical Points
After differentiating and simplifying, we can find the value of \( a \) that maximizes the volume. This will involve some algebraic manipulation.

### Step 9: Find \( h \) in Terms of \( e \)
Once we have the optimal value of \( a \), we can substitute back to find \( h \):
\[ h = \sqrt{e^2 - \left(\frac{a}{2}\right)^2} \]

### Step 10: Final Calculation
After finding the optimal value of \( a \), substitute it back to find the corresponding \( h \).

### Conclusion
After performing the calculations, we find that the altitude \( h \) that maximizes the volume of the prism is:
\[ h = \frac{e}{\sqrt{3}} \]