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 \(\frac{abc}{a+b+c}\) when the circumradius \(R\) of a triangle is 2, we can follow these steps:

### Step 1: Understanding the Circumradius
The circumradius \(R\) of a triangle is related to its sides \(a\), \(b\), and \(c\) by the formula:
\[ R = \frac{abc}{4K} \]
where \(K\) is the area of the triangle. Given that \(R = 2\), we have:
\[ 2 = \frac{abc}{4K} \implies abc = 8K \]

### Step 2: Using the Homogeneity of the Expression
The expression \(\frac{abc}{a+b+c}\) is homogeneous of degree 0. This means that if we scale \(a\), \(b\), and \(c\) by a factor \(k\), the value of the expression remains unchanged. Therefore, we can assume that the triangle is equilateral to maximize the expression.

### Step 3: Finding the Side Length of the Equilateral Triangle
For an equilateral triangle with circumradius \(R = 2\), the side length \(s\) can be calculated using the relation:
\[ R = \frac{s}{\sqrt{3}} \]
Thus, we have:
\[ 2 = \frac{s}{\sqrt{3}} \implies s = 2\sqrt{3} \]

### Step 4: Assigning Values to \(a\), \(b\), and \(c\)
Since the triangle is equilateral, we have:
\[ a = b = c = 2\sqrt{3} \]

### Step 5: Calculating the Expression
Now we can substitute \(a\), \(b\), and \(c\) into the expression \(\frac{abc}{a+b+c}\):
\[ abc = (2\sqrt{3})(2\sqrt{3})(2\sqrt{3}) = 8 \cdot 3 = 24 \]
\[ a + b + c = 2\sqrt{3} + 2\sqrt{3} + 2\sqrt{3} = 6\sqrt{3} \]
Thus, we have:
\[ \frac{abc}{a+b+c} = \frac{24}{6\sqrt{3}} = \frac{4}{\sqrt{3}} \cdot \sqrt{3} = 4 \]

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