If circumradius of triangle is 2, then the maximum value of `(abc)/(a+b+c)` is

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To find the maximum value of \(\frac{abc}{a+b+c}\) given that the circumradius \(R\) of the triangle is 2, we can follow these steps:

### Step 1: Use the relationship between circumradius and sides of the triangle
The circumradius \(R\) of a triangle is related to its sides \(a\), \(b\), and \(c\) by the formula:
\[ R = \frac{abc}{4\Delta} \]
where \(\Delta\) is the area of the triangle.

Given \(R = 2\), we can write:
\[ 2 = \frac{abc}{4\Delta} \]
From this, we can express \(abc\) in terms of \(\Delta\):
\[ abc = 8\Delta \]

### Step 2: Express \(a + b + c\) in terms of the semi-perimeter
The semi-perimeter \(s\) of the triangle is given by:
\[ s = \frac{a+b+c}{2} \]
Thus, we can express \(a + b + c\) as:
\[ a + b + c = 2s \]

### Step 3: Substitute into the expression
Now we can substitute \(abc\) and \(a+b+c\) into our expression:
\[ \frac{abc}{a+b+c} = \frac{8\Delta}{2s} = \frac{4\Delta}{s} \]

### Step 4: Use the area formula
Using the formula for the area of a triangle in terms of its circumradius:
\[ \Delta = \frac{abc}{4R} \]
Substituting \(R = 2\):
\[ \Delta = \frac{abc}{8} \]

### Step 5: Relate \(\Delta\) and \(s\)
We can apply the inequality of the area and the semi-perimeter:
\[ \Delta \leq \sqrt{s(s-a)(s-b)(s-c)} \]
For maximum area, we can consider the case of an equilateral triangle where \(a = b = c\).

### Step 6: Find the maximum value
In the case of an equilateral triangle with circumradius \(R = 2\), the side length \(a\) can be found using the formula:
\[ R = \frac{a}{\sqrt{3}} \implies a = 2\sqrt{3} \]
Thus, \(a + b + c = 6\sqrt{3}\) and:
\[ abc = (2\sqrt{3})^3 = 24\sqrt{3} \]

Now substituting these values:
\[ \frac{abc}{a+b+c} = \frac{24\sqrt{3}}{6\sqrt{3}} = 4 \]

### Conclusion
Thus, the maximum value of \(\frac{abc}{a+b+c}\) when the circumradius \(R = 2\) is:
\[ \boxed{4} \]