Let $$a, b, c$$ and $$d$$ be positive real numbers such that $$a + b + c + d = 11$$. If the maximum value of $$a^5b^3c^2d$$ is $$3750\beta$$, then the value of $$\beta$$ is

Given $$a + b + c + d = 11$$ with $$a, b, c, d > 0$$, we maximize $$a^5 b^3 c^2 d$$.

Using the AM-GM inequality, we split the sum into $$5 + 3 + 2 + 1 = 11$$ equal parts:

$$\frac{\frac{a}{5} + \frac{a}{5} + \cdots(5\text{ times}) + \frac{b}{3} + \frac{b}{3} + \frac{b}{3} + \frac{c}{2} + \frac{c}{2} + d}{11} \geq \left(\left(\frac{a}{5}\right)^5 \left(\frac{b}{3}\right)^3 \left(\frac{c}{2}\right)^2 d\right)^{1/11}$$

$$1 \geq \left(\frac{a^5 b^3 c^2 d}{5^5 \cdot 3^3 \cdot 2^2}\right)^{1/11}$$

Equality holds when $$\frac{a}{5} = \frac{b}{3} = \frac{c}{2} = d$$, i.e., $$a = 5, b = 3, c = 2, d = 1$$.

Maximum value:

$$a^5 b^3 c^2 d = 5^5 \cdot 3^3 \cdot 2^2 \cdot 1 = 3125 \times 27 \times 4 = 337500$$

Given $$337500 = 3750\beta$$:

$$\beta = \frac{337500}{3750} = 90$$

The answer is Option A: $$90$$.