Let a, b, c and d be positive real numbers such that a+b+c+d=11.

Question

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

Accepted Answer

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.

Solutions · 1