QJEEM23Ma91f2: JEE Main 2023 Mathematics - Class 11 JEEM23Ma91f2 Solution: Let and be positive real numbers such that . If the maximum…

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 $\frac{a}{5}, \frac{a}{5}, \frac{a}{5}, \frac{a}{5}, \frac{a}{5}, \frac{b}{3}, \frac{b}{3}, \frac{b}{3}, \frac{c}{2}, \frac{c}{2}, d$ .

Applying the AM-GM inequality:

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

Since $a+b+c+d=11$ , we have:

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

Now, raising both sides to the power of 11:

$1^{11} \geq \frac{a^5 b^3 c^2 d}{5^5 3^3 2^2 1}$

From the given information, we know that $a^5 b^3 c^2 d \geq 3750\beta$ :

$5^5 3^3 2^2 \geq 3750\beta$

Now, we can solve for $\beta$ :

$\beta \leq \frac{1}{3750} \cdot 5^5 3^3 2^2$

Since we are looking for the maximum value of $\beta$ , we take the equality case:

$\beta = \frac{1}{3750} \cdot 5^5 3^3 2^2$

Calculating the value, we find that:

$\beta = 90$

So, the value of $\beta$ is 90.