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

Given that $a+b+c+d=11$ and the maximum value of $a^5 b^3 c^2 d$ is $3750\beta$, you assumed the numbers to be $$\frac{a}{5}, \frac{a}{5}, \frac{a}{5}, \frac{a}{5}, \frac{a}{5}, \frac{b}{3}, \frac{b}{3}, \frac{b}{3}, \frac{c}{2}, \frac{c}{2}, d$$. <br/><br/>Applying the AM-GM inequality: <br/><br/>$$\frac{\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}+\frac{c}{2}+\frac{c}{2}+d}{11} \geq\left(\frac{\left(a^5 b^3 c^2 d\right)}{5^5 3^3 2^2 1}\right)^{\frac{1}{11}}$$ <br/><br/>Since $a+b+c+d=11$, we have: <br/><br/>$$1 \geq\left(\frac{\left(a^5 b^3 c^2 d\right)}{5^5 3^3 2^2 1}\right)^{\frac{1}{11}}$$ <br/><br/>Now, raising both sides to the power of 11: <br/><br/>$1^{11} \geq \frac{a^5 b^3 c^2 d}{5^5 3^3 2^2 1}$ <br/><br/>From the given information, we know that $a^5 b^3 c^2 d \geq 3750\beta$: <br/><br/>$5^5 3^3 2^2 \geq 3750\beta$ <br/><br/>Now, we can solve for $\beta$: <br/><br/>$\beta \leq \frac{1}{3750} \cdot 5^5 3^3 2^2$ <br/><br/>Since we are looking for the maximum value of $\beta$, we take the equality case: <br/><br/>$\beta = \frac{1}{3750} \cdot 5^5 3^3 2^2$ <br/><br/>Calculating the value, we find that: <br/><br/>$\beta = 90$ <br/><br/>So, the value of $\beta$ is 90.